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What is foolmath?

If you are a math enthusiast or aficionado, have you ever encountered mathematical foolish proofs like $1=2$ or $2+5=8$ or even evaluating $\infty$? Those seem so strange, queer, quirky and absurd. Many of them are foolish, while some of them are backed by impeccable logic. I intend this repository to be a treasure trove or collection of mind-bending demonstrations that challenge your concepts about numbers, algebra, trigonometry, logarithm, calculus and more. They will leave you scratching your head dissecting each equation to find errors, incorrectness or oversights in each proof. I also include some valid proofs and strange mathematical properties, which are considered interesting and worth reading. That's what foolmath is all about!

Notes:

1. Almost all mathematical proofs here are foolish, unless explicitly noted as valid.
2. Plain-text LaTeX .tex code is in src/.
3. Best to view foolmath on GitHub.com using Firefox web browser on any devices, GitHub app on mobile devices or GitHub.io doesn't render LaTeX or MathJax. GitHub.com itself is also updated every now and then. Sometimes it appears to have rendered incorrectly on other web browsers.
4. There are many pages in the repository. Other than README.md in the root directory of the repository, GitHub will show the left pane displaying the source tree when viewing the other files. For the best viewing experience, you can simply close such left pane, in order to see the contents in full screen.

Proofs speak louder than words. Let's start!

Table of Contents

• Page 1 |view it|
  • Ramanujan summation |view it|
  • Ramanujan alternative 1 |view it|
  • Ramanujan alternative 2 |view it|
  • Summing the power of two |view it|
  • Summing all natural odd numbers |view it|
  • Summing all natural even numbers |view it|
  • Why is $0=1$? |view it|
  • Why is $1=2$? |view it|
  • Let's see a little higher numbers, $4=5$. |view it|
  • Are all intergers equal? |view it|
  • Why is $2+5=8$? |view it|
  • Just another freak, $9=17$. |view it|
  • Very silly solutions to find $\frac{0}{0}$. |view it|
  • Who said $i$ is imaginary, why is $i=1$ then? |view it|
  • Let's talk more about $i$. |view it|
  • $i=\pm1$, isn't it? |view it|
  • $i=0$, probably |view it|
  • How much is $\infty$? |view it|
  • How much is $0.\infty$? |view it|
  • Is $0$ an even number? |view it|
  • Solve $x$ for $x+2=x$. |view it|
  • $e^{i\pi}=$?, also known as Euler's Identity |view it|
  • $e^x=1$, no matter what $x$ maybe. |view it|
  • How much is $\pi$? |view it|
  • How much is $i^i$? |view it|
  • How much is $\sqrt[^i]{i}$? |view it|
  • Who said $n^0=1$? |view it|
  • Why is $\frac{d}{dx}e^x=e^x$? |view it|
  • Why is multiplying factor converting the power $1\phi$ to $3\phi$ is $\sqrt{3}$? |view it|
  • Magic numbers |view it|
  • Vortex math |view it|
  • A property of prime numbers |view it|
  • Relation between $e$ and $\pi$ a.k.a. Gaussian integral |view it|
  • Golden ratio from Fibonacci sequence |view it|
  • Isn't it $\infty=\Phi$? |view it|
  • How much is $1^i$? |view it|
  • How much is $\sqrt[^i]{1}$? |view it|
  • Whether $e=1$ or $\pi=0$ |view it|
  • How much is $e^i$? |view it|
  • Simple continued fraction expansion of $\pi$ |view it|
  • $\sqrt{-\ln{(-1)}\ln{(-1)}}=\pi$, how? |view it|
  • How is $6$ afraid of $7$? It ain't and will never be. |view it|
• Page 2 |view it|
  • The first equation we learnt in kindergartens |view it|
  • Production of all multiple of $2$ |view it|
  • Production of all natural odd numbers |view it|
  • Production of all natural numbers |view it|
  • Production of all natural even numbers |view it|
  • Infinite numbers of primes |view it|
  • Magic squares |view it|
  • Why is $\sqrt{2}$ irrational? |view it|
  • How do programmers increase a variable? |view it|
  • $\frac{dx}{dx}=0$, really? |view it|
  • $\pi=-\pi$, what? |view it|
  • $2=3$, isn't it absurd? |view it|
  • What if $3=0$? |view it|
  • Do you know that $1=-1$? |view it|
  • Oh what, $\pm2\pi=0$? |view it|
  • Is $1$ defined in mathematics? |view it|
  • How much is $e$? |view it|
  • How much is $\frac{a}{0}$? |view it|
  • $\pi$ can be any real numbers. |view it|
  • $\infty=0$, the universe is empty. |view it|
  • $\pi=e$, why do engineers say so? |view it|
  • Freak again, $i=\infty$ |view it|
  • Whether $i$ or $\pi$ is $0$. |view it|
  • Infinite nested radical |view it|
• Page 3 |view it|
  • Sum of Cubes Identity |view it|
  • What if $9^\infty=1$? |view it|
  • A gimmick of multiple roots |view it|
  • A play with inequality |view it|
• Support foolmath |here|

Ramanujan summation

We firstly start from the well-known Ramanujan Summation, which is known to most mathematicians.

$\qquad1+2+3+4+5+6+...\quad=\quad?$

$$\begin{alignat*}{5} &\,&S\quad&=\quad&&1+2+3+4+5+6+...\\\ &\small\text{Let}&S_1\quad&=&&1-1+1-1+...\\\ &\,&1-S_1\quad&=&&1-(1+1-1+1-...)\\\ &\,&\,&=&&1-1+1-1+...\\\ &\,&\,&=&& S_1\\\ &\,&2S_1\quad&=&&1\\\ &\,&S_1\quad&=&&\frac{1}{2}\\\ &\small\text{Let}&S_2\quad&=&&1-2+3-4+5-6+...\\\ &\,&2S_2\quad&=&&1-2+3-4+5-6+...\\\ &\,&\,&\,&&\quad+1-1+3-4+5-6\\\ &\,&\,&=&&1-1+1-1+...\\\ &\,&2S_2\quad&=&& S_1\\\ &\,&S_2\quad&=&&\frac{S_1}{2}\\\ &\,&S_2\quad&=&&\frac{1}{4}\\\ &\,&S-S_2\quad&=&&1+2+3+4+5+6+7+8+...\\\ &\,&\,&\,&&\,\,\,-(1-2+3-4+5-6+7-...)\\\ &\,&\,&=&&4+8+12+16+...\\\ &\,&\,&=&&4(1+2+3+4+...)\\\ &\,&S-S_2\quad&=&&4S\\\ &\,&3S\quad&=&&-S_2\\\ &\,&S\quad&=&&-\frac{S_2}{3}\\\ &\,&\,&=&&-\frac{1}{3*4}\\\ &\small\text{Thus}&S\quad&=&&-\frac{1}{12} \end{alignat*}$$

source code: rama_sum.tex | Go to top | TOC

There are still a few Ramanujan alternatives.

Ramanujan alternative 1

$$\begin{alignat*}{5} &\,&S\quad&=\quad&&1+2+3+4+5+6+7+8+9+10+...\\\ &\,&\,&=&&1+(2+3+4)+(5+6+7)+(8+9+10)+...\\\ &\,&\,&=&&1+(9+18+27+...)\\\ &\,&\,&=&&1+9(1+2+3+...)\\\ &\,&S\quad&=&&1+9S\\\ &\,&8S\quad&=&&-1\\\ &\small\text{Thus}&\qquad S\quad&=&&-\frac{1}{8} \end{alignat*}$$

source code: rama_alt_1.tex | Go to top | TOC

Hold on! There is still another alternative.

Ramanujan alternative 2

$$\begin{alignat*}{5} &\,&S\quad&=\quad&&1+2+3+4+5+6+7+8+...\\\ &\,&\,&=&&(2+4+6+8+10+12+14+...)+\\\ &\,&\,&\,&&(1+3+5+7+9+11+13+...)\\\ &\,&\,&=&&2(1+2+3+4+5+6+7+...)+\\\ &\,&\,&\,&&~\,(1+(3+5)+(7+9)+(11+13)+...)\\\ &\,&\,&=&&2S+(1+8+16+24+32+...)\\\ &\,&\,&=&&2S+(1+8(1+2+3+4+...))\\\ &\,&\,&=&&2S+(1+8S)\\\ &\,&S\quad&=&&10S+1\\\ &\small\text{Thus}\normalsize\qquad&S\quad&=&&-\frac{1}{9} \end{alignat*}$$

source code: rama_alt_2.tex | Go to top | TOC

Oops, was Ramanujan wrong?

What if summing the power of two?

$$\begin{alignat*}{5} &\small\text{Let}&S\quad&=\quad&&\sum_{n=0}^\infty 2^n\\\ &\,&\,&=&&1+2+4+8+16+32+...\\\ &\,&\,&=&&1+2(1+2+4+8+16+...)\\\ &\,&S\quad&=&&1+2S\\\ &\small\text{Thus}\normalsize\qquad&S\quad&=&&-1 \end{alignat*}$$

source code: sum_power_of_2.tex | Go to top | TOC

Negative! once again.

What if summing all natural odd numbers?

$$\begin{alignat*}{5} &\,&S\quad&=\quad&&1+3+5+7+9+...\\\ &\,&-\frac{1}{12}\quad&=&&1+2+3+4+5+6+7+8+9+10+...\\\ &\,&\,&=&&(1+3+5+7+9+...)+(2+4+6+8+10+...)\\\ &\,&\,&=&&(1+3+5+7+9+...)+2(1+2+3+4+5+...)\\\ &\,&\,&=&&(1+3+5+7+9+...)+2\left(-\frac{1}{12}\right)\\\ &\,&-\frac{1}{12}\quad&=&&(1+3+5+7+9+...)-\frac{2}{12}\\\ &\,&\frac{2}{12}-\frac{1}{12}\quad&=&&1+3+5+7+9+...\\\ &\small\text{Thus}&\frac{1}{12}\quad&=&&1+3+5+7+9+... \end{alignat*}$$

source code: sum_of_odd.tex | Go to top | TOC

Wow! this time the summation is positive.

Trying summing all natural even numbers

Solution 1

$$\begin{alignat*}{5} &\,&S\quad&=\quad&&2+4+6+8+10+...\\\ &\,&-\frac{1}{12}\quad&=&&1+2+3+4+5+6+7+8+9+10+...\\\ &\,&\,&=&&(1+3+5+7+9+...)+(2+4+6+8+10+...)\\\ &\,&\,&=&&\frac{1}{12}+(2+4+6+8+10+...)\\\ &\,&-\frac{1}{12}\quad&=&&\frac{1}{12}+S\\\ &\small\text{Thus}\normalsize\qquad&S\quad&=&&-\frac{1}{6} \end{alignat*}$$

source code: sum_of_even_0.tex | Go to top | TOC

Why is it negative again, who know? Probably something wrong.
Try the next proof, which is simpler.

Solution 2

$$\begin{alignat*}{5} &\,&S\quad&=\quad&&2+4+6+8+10+...\\\ &\,&\,&=&&2(1+2+3+4+5+...)\\\ &\,&\,&=&&2\left(-\frac{1}{12}\right)\\\ &\small\text{Thus}\normalsize\qquad&S\quad&=&&-\frac{1}{6} \end{alignat*}$$

source code: sum_of_even_1.tex | Go to top | TOC

Same result, do you start to believe?

Why is $0=1$?

Here come the most foolish proof!

Proof 1

$$\begin{alignat*}{5} &\,&0\quad&=\quad&&0+0+0+0+...\\\ &\,&\,&=&&(1-1)+(1-1)+(1-1)+(1-1)+...\\\ &\,&\,&=&&1+(-1+1)+(-1+1)+(-1+1)+...\\\ &\,&\,&=&&1+0+0+0+0+...\\\ &\small\text{Thus}\normalsize\qquad&0\quad&=&&1 \end{alignat*}$$

source code: 0eq1_0.tex | Go to top | TOC

What about the second most foolish proof?

Proof 2

$$\begin{alignat*}{5} &\small\text{Let}\normalsize\qquad&S\quad&=\quad&&1+1+1+1++1...\\\ &\,&\,&=&&1+(1+1+1+1+...)\\\ &\,&\cancel{S}\quad&=&&1+\cancel{S}\\\ &\small\text{Thus}\normalsize\qquad&0\quad&=&&1 \end{alignat*}$$

source code: 0eq1_1.tex | Go to top | TOC

Is there any proof looking more advanced than these?

Proof 3

$$\begin{alignat*}{5} &\,&\int u\,dv\quad&=\quad&&uv-\int v\,du\\\ &\,&\int\frac{1}{x}\,dx\quad&=&&\cancel{x}\frac{1}{\cancel{x}}-\int x\,d\left(\frac{1}{x}\right)\\\ &\,&\,&=&&1-\int x\,d(x^{-1})\\\ &\,&\,&=&&1-\int x\,(-1x^{-2})\,dx\\\ &\,&\,&=&&1+\int x.x^{-2}\,dx\\\ &\,&\cancel{\int\frac{1}{x}\,dx}\quad&=&&1+\cancel{\int\frac{1}{x}\,dx}\\\ &\small\text{Thus}\normalsize\qquad&0\quad&=&&1 \end{alignat*}$$

source code: 0eq1_2.tex | Go to top | TOC

Can you find an error? Hmm, binary no longer exists.

Why is $1=2$?

Here, the proof I learnt in junior high school.

Proof 1

$$\begin{alignat*}{5} &\,&\small\text{Let}\normalsize\qquad a\quad&=\quad&&b\\\ &\times b\small\text{ both sides}\quad&a*b\quad&=&&b*b\\\ &\,&a*b\quad&=&&b^2\\\ &-a^2\small\text{ both sides}\quad&a*b-a^2\quad&=&&b*b-a^2\\\ &\,&a(b-a)\quad&=&&b^2-a^2\\\ &\,&a\cancel{(b-a)}\quad&=&&\cancel{(b-a)}(b+a)\\\ &\small\text{since }b=a&a\quad&=&&a+a\\\ &\,&\cancel{a}\quad&=&&2\cancel{a}\\\ &\,&\small\text{Thus}\normalsize\qquad1\quad&=&&2 \end{alignat*}$$

source code: 1eq2_0.tex | Go to top | TOC

Nah, there is another proof in high school using trigonometry.

Proof 2

$$\begin{alignat*}{5} &\small\text{where }x=\frac{\pi}{4}~or~\frac{5\pi}{4}&\cos{x}\quad&=\quad&&\sin{x}\\\ &\,&\cos{2x}\quad&=&&\sin{2x}\\\ &\,&1-2\sin^2{x}\quad&=&&2\sin{x}\cos{x}\\\ &\small\text{as }\sin{x}=\cos{x}&\,&=&&2\cos{x}\cos{x}\\\ &\,&1-2\sin^2{x}\quad&=&&2\cos^2{x}\\\ &\,&1\quad&=&&2\sin^2{x}+2\cos^2{x}\\\ &\,&\,&=&&2\cancelto{1}{(\sin^2{x+\cos^2{x}})}\\\ &\,&\small\text{Thus}\normalsize\qquad1\quad&=&&2 \end{alignat*}$$

source code: 1eq2_1.tex | Go to top | TOC

Yet, there is another proof using calculus.

Proof 3

$$\begin{alignat*}{5} &\,&\underbrace{x+x+x+...+x}_{x\text{ terms}}\quad&=\quad&&x*x\\\ &\,&\,&=&&x^2\\ \\\ &\small\text{diff both sides }&\underbrace{1+1+1+...+1}_{x\text{ terms}}\quad&=&&2x\\\ &\,&\cancel{x}\quad&=&&2\cancel{x}\\ \\\ &\,&\small\text{Thus}\normalsize\qquad1\quad&=&&2 \end{alignat*}$$

source code: 1eq2_2.tex | Go to top | TOC

Do you find any clues?

Let's see a little higher numbers, $4=5$.

$$\begin{alignat*}{5} &\,&-20\quad&=\quad&&-20\\\ &\,&16-36\quad&=&&25-45\\\ &\,&4^2-4*9\quad&=&&5^2-5*9\\\ &\small+\frac{81}{4}\text{both sides }&4^2-4*9+\frac{81}{4}\quad&=&&5^2-5*9+\frac{81}{4}\\\ &\,&4^2-2*4*\frac{9}{2}+\left(\frac{9}{2}\right)^2\quad&=&&5^2-2*5*\frac{9}{2}+ \left(\frac{9}{2}\right)^2\\\ &\,&\left(4-\frac{9}{2}\right)^2\quad&=&&\left(5-\frac{9}{2}\right)^2\\\ &\small\sqrt{}~\text{both sides }&4-\cancel{\frac{9}{2}}\quad&=&&5-\cancel{\frac{9}{2}}\\\ &\,&\small\text{Thus}\normalsize\qquad4\quad&=&&5 \end{alignat*}$$

source code: 4eq5_0.tex | Go to top | TOC

Hey, what? How come, $4=5$?

Wait, there are something more.

Are all intergers equal?

$$\begin{alignat*}{5} &\,&\frac{-1}{1}\quad&=\quad&&\frac{1}{-1}&(1)\\\ &\,&\sqrt{\frac{-1}{1}}\quad&=&&\sqrt{\frac{1}{-1}}&(2)\\\ &\,&\frac{\sqrt{-1}}{\sqrt{1}}\quad&=&&\frac{\sqrt{1}}{\sqrt{-1}}&(3)\\\ &\,&\frac{i}{1}\quad&=&&\frac{1}{i}&(4)\\\ &\small\times\frac{1}{2}&\frac{i}{2}\quad&=&&\frac{1}{2i}&(5)\\\ &\small+\frac{3}{2i}&\frac{i}{2}+\frac{3}{2i}\quad&=&&\frac{1}{2i}+\frac{3}{2i}&(6)\\\ &\times i&i\left(\frac{i}{2}+\frac{3}{2i}\right)\quad&=&&i\left(\frac{1}{2i}+\frac{3}{2i}\right)\qquad&(7)\\\ &\,&\frac{i^2}{2}+\frac{3\cancel{i}}{2\cancel{i}}\quad&=&&\frac{\cancel{i}}{2\cancel{i}}+\frac{3\cancel{i}}{2\cancel{i}}&(8)\\\ &\,&-\frac{1}{2}+\frac{3}{2}\quad&=&&\frac{1}{2}+\frac{3}{2}&(9)\\\ &\,&\frac{-1+3}{2}\quad&=&&\frac{1+3}{2}&(10)\\\ &\,&\frac{\cancel{2}}{\cancel{2}}\quad&=&&\cancelto{2}{\frac{4}{2}}&(11)\\\ &\,&1\quad&=&&2&(12)\\\ &\small\text{from (9)}&-\frac{1}{2}+\cancel{\frac{3}{2}}\quad&=&&\frac{1}{2}+\cancel{\frac{3}{2}}&(13)\\\ % line (14): Can't add `\cancel` inside `\frac`, MathJax bug? &\,&-\frac{1}{2}\quad&=&&\frac{1}{2}&(14)\\\ &\,&-1\quad&=&&1&(15)\\\ &\small+1&-1+1\quad&=&&1+1&(16)\\\ &\,&0\quad&=&&2&(17)\\\ &\small\text{from (12), (15), (17)}&-1~=~0\quad&=&&1~=~2&(18) \end{alignat*}$$

source code: all_int_eq.tex | Go to top | TOC

It is very articulate, indeed.

There is one more simple equation. Have a look.

Was I wrongly taught? Why is $2+5=8$?

$$\begin{alignat*}{5} &\,&2+5\quad&=\quad&&4+3\\\ &\,&\,&=&&4-\frac{9}{2}+\frac{9}{2}+3\\\ &\,&\,&=&&\sqrt{\left(4-\frac{9}{2}\right)^2}+\frac{9}{2}+3\\\ &\,&\,&=&&\sqrt{16-2.4.\frac{9}{2}+\left(\frac{9}{2}\right)^2}+\frac{9}{2}+3\\\ &\,&\,&=&&\sqrt{16-36+\left(\frac{9}{2}\right)^2}+\frac{9}{2}+3\\\ &\,&\,&=&&\sqrt{-20+\left(\frac{9}{2}\right)^2}+\frac{9}{2}+3\\\ &\,&\,&=&&\sqrt{25-45+\left(\frac{9}{2}\right)^2}+\frac{9}{2}+3\\\ &\,&\,&=&&\sqrt{25-2.5.\frac{9}{2}+\left(\frac{9}{2}\right)^2}+\frac{9}{2}+3\\\ &\,&\,&=&&\sqrt{5^2-2.5.\frac{9}{2}+\left(\frac{9}{2}\right)^2}+\frac{9}{2}+3\\\ &\,&\,&=&&\sqrt{\left(5-\frac{9}{2}\right)^2}+\frac{9}{2}+3\\\ &\,&\,&=&&5-\cancel{\frac{9}{2}}+\cancel{\frac{9}{2}}+3\\\ &\small\text{Thus}\normalsize\qquad&2+5\quad&=&&8 \end{alignat*}$$

source code: 2plus5eq8_0.tex | Go to top | TOC

There is one more, it is tricky.

Just another freak, $9=17$.

$$\begin{alignat*}{5} &\small\text{Let}&x\quad&=\quad&&9\\\ &\,&x^2-26x+169\quad&=&&9^2-26(9)+169\\\ &\,&\,&=&&16\\\ &\,&x^2-2(13)x+13^2\quad&=&&16\\\ &\,&(x-13)^2\quad&=&&16\\\ &\,&x-13\quad&=&&4\\\ &\,&x\quad&=&&17\\\ &\small\text{Thus}&9\quad&=&&17 \end{alignat*}$$

source code: 9eq17_0.tex | Go to top | TOC

Square of negative numbers really does the trick.

Who said $0$ couldn't be a denominator? See the following foolish proofs.

Very silly solutions to find $\frac{0}{0}$.

Let's see the first fool.

$$\begin{alignat*}{5} &\,&\small\text{Let}\normalsize\qquad\frac{2}{0}\quad&=\quad&&\frac{x}{1}\\\ &\times\small\text{0 both sides}&\frac{2*0}{0}\quad&=&&\frac{x*0}{1}\\\ &\,\small &divide\text{ 2 both sides}&\frac{\cancel{2}*0}{0*\cancel{2}}\quad&=&&\cancelto{0}{\frac{x*0}{1*2}}\\\ &\,&\small\text{Thus}\normalsize\qquad\frac{0}{0}\quad&=&&0 \end{alignat*}$$

source code: 0by0_0.tex | Go to top | TOC

Here, the second fool, which is very silly.

$$\begin{alignat*}{5} &\,&\frac{0}{0}\quad&=\quad&&\frac{100-100}{100-100}\\\ &\,&\,&=&&\frac{10*10-10*10}{10*10-10*10}\\\ &\,&\,&=&&\frac{10^2-10^2}{10(10-10)}\\\ &\,&\,&=&&\frac{(10+10)\cancel{(10-10)}}{10\cancel{(10-10)}}\\\ &\,&\,&=&&\frac{20}{10}\\\ &\small\text{Thus}\normalsize\qquad&\frac{0}{0}\quad&=&&2 \end{alignat*}$$

source code: 0by0_1.tex | Go to top | TOC

Bruh, how can you divide $(10-10)$ with $(10-10)$?

Who said $i$ is imaginary, why is $i=1$, then?

$$\begin{alignat*}{5} &\,&i\quad&=\quad&&\sqrt{-1}\\\ &\,&i^2\quad&=&&\sqrt{-1}.\sqrt{-1}\\\ &\,&\,&=&&\sqrt{(-1)(-1)}\\\ &\,&\,&=&&\sqrt{1}\\\ &\,&i^2\quad&=&&1\\\ &\small\text{Thus}\normalsize\qquad&i\quad&=&&1 \end{alignat*}$$

source code: ieq1_0.tex | Go to top | TOC

Let's talk more about $i$.

This is the valid proof of $\frac{1}{i}$.

$$\begin{alignat*}{5} &\,&-1\quad&=\quad&&i^2\\\ &\small &divide(-i)\text{ both sides}&\frac{-1}{-i}\quad&=&&\frac{i^2}{-i}\\\ &\,&\small\text{Thus}\normalsize\qquad\frac{1}{i}\quad&=&&-i\qquad\small\text{(valid proof)} \end{alignat*}$$

source code: inv_i_valid.tex | Go to top | TOC

Nah, there is another $\frac{1}{i}$, but it is foolish.

This is the foolish proof of $\frac{1}{i}$.

$$\begin{alignat*}{5} &\,&\frac{1}{i}\quad&=\quad&&i^{-1}\\\ &\,&\,&=&&\sqrt{-1}^{-1}\\\ &\,&\,&=&&\left((-1)^\frac{1}{2}\right)^{-1}\\\ &\,&\,&=&&\left((-1)^{-1}\right)^\frac{1}{2}\\\ &\,&\,&=&&\left(\frac{1}{-1}\right)^\frac{1}{2}\\\ &\,&\,&=&&-1^\frac{1}{2}\\\ &\,&\,&=&&\sqrt{-1}\\\ &\small\text{Thus}\normalsize\qquad&\frac{1}{i}\quad&=&& i\qquad\small\text{(foolish proof)} \end{alignat*}$$

source code: inv_i_fool.tex | Go to top | TOC

Which one will you believe?

This is the foolish proof of $-i$.

$$\begin{alignat*}{5} &\,&i\quad&=\quad&&\sqrt{-1}\\\ &\,&-i\quad&=&&-\sqrt{-1}\\\ &\,&\,&=&&(-1)(-1)^\frac{1}{2}\\\ &\,&\,&=&&(-1)^{1+\frac{1}{2}}\\\ &\,&\,&=&&(-1)^\frac{3}{2}\\\ &\,&\,&=&&\sqrt{-1^3}\\\ &\,&\,&=&&\sqrt{-1}\\\ &\small\text{Thus}\normalsize\qquad&-i\quad&=&&i&\small\text{(foolish proof)} \end{alignat*}$$

source code: i_eq-i.tex | Go to top | TOC

It is exactly imaginary.

$i=\pm1$, isn't it?

$$\begin{alignat*}{5} &\,&i^2\quad&=\quad&&-1\\\ &\,&i*i\quad&=&&-1\\\ &\,&i\quad&=&&\frac{-1}{i}\\\ &\rlap{i=\frac{-1}{i}=\frac{-1}{\frac{-1}{i}}=\frac{-1}{\frac{-1}{\frac{-1}{i}}}=...}\\\ &\,&i\quad&=&&(-1)^\infty\\\ &\,&\small\text{Thus}\normalsize\qquad i\quad&=&&\pm1 \end{alignat*}$$

source code: i_eq_pm1.tex | Go to top | TOC

Hold on! $i$ is probably something else.

$i=0$, probably

$$\begin{alignat*}{5} &\,&0\quad&=\quad&&i-i\\\ &\,&\,&=&&i-\sqrt{-1}\\\ &\,&\,&=&&i-\sqrt{(-1)(1)}\\\ &\,&\,&=&&i-\sqrt{(-1)(-1)^2}\\\ &\,&\,&=&&i-\sqrt{-1}\sqrt{(-1)^2}\\\ &\,&\,&=&&i-(i)(-1)\\\ &\,&\,&=&&i+i\\\ &\,&0\quad&=&&2i\\\ &\small\text{Thus}\normalsize\qquad&i\quad&=&&0\\\ \end{alignat*}$$

source code: i0.tex | Go to top | TOC

Whatever it is, it is not imaginary anyway.

How much is $\infty$?

Solution 1

$$\begin{alignat*}{5} &\,&-\frac{1}{12}\quad&=\quad&&1+2+3+...+\infty\\\ &\,&\,&=&&\frac{\infty(\infty+1)}{2}\\\ &\,&-1\quad&=&&6\infty(\infty+1)\\\ &\,&0\quad&=&&6\infty^2+6\infty+1\\\ &\,&\infty\quad&=&&\frac{-6\pm\sqrt{6^2-4(6)(1)}}{2(6)}\\\ &\,&\,&=&&\frac{-6\pm\sqrt{36-24}}{12}\\\ &\,&\,&=&&\frac{-6\pm\sqrt{12}}{12}\\\ &\,&\,&=&&\frac{-6\pm\sqrt{2^2*3}}{12}\\\ &\,&\,&=&&\frac{-6\pm2\sqrt{3}}{12}\\\ &\small\text{Thus}&\infty\quad&=&&\frac{-3\pm\sqrt{3}}{6} \end{alignat*}$$

source code: infty_0.tex | Go to top | TOC

Hold on, $\infty$ is probably something else.

Solution 2

$$\begin{alignat*}{5} &\,&x\quad&=\quad&&\bar{9}&&\small\text{(1)}\\\ &\,&x\quad&=&&\infty&&\small\text{(2)}\\\ &\small10*\text{(1)}&10x\quad&=&&\bar{9}0\\\ &\small+9\text{ both sides}\qquad&10x+9\quad&=&&\bar{9}\\\ &\small\text{from (1)}&10x+9\quad&=&&x\\\ &\,&9x\quad&=&&-9\\\ &\,&x\quad&=&&-1\qquad&&\small\text{(3)}\\\ &\small\text{(2)=(3)}&\infty\quad&=&&-1&&\small\text{(4)}\\\ &\small1&divide\text{(4)}&\frac{1}{\infty}\quad&=&&\frac{1}{-1}\\\ &\,&0\quad&=&&-1\\\ &\small\text{from (3)}&x\quad&=&&0&&\small\text{(5)}\\\ &\small\text{Thus (2)=(3)=(5)}&\infty\quad&=&&-1\quad=\quad0 \end{alignat*}$$

source code: infty_1.tex | Go to top | TOC

Here, $\infty$ is so small, you see?

Solution 3

$$\begin{alignat*}{5} &\,&\infty\quad&=\quad&&1+1+1+1+...\\\ &\small\text{regroup in multiple of 2}&\,&=&&2+4+6+8+...\\\ &\,&\,&=&&2(1+2+3+4+...)\\\ &\,&\,&=&&2\left(-\frac{1}{12}\right)\\\ &\qquad\qquad\small\text{Thus}&\infty\quad&=&&-\frac{1}{6}\\\ &\small\text{or regroup in multiple of 3}&\,&=&&3+6+9+12+...\\\ &\,&\,&=&&3(1+2+3+4+...)\\\ &\,&\,&=&&3\left(-\frac{1}{12}\right)\\\ &\qquad\qquad\small\text{Thus}&\infty\quad&=&&-\frac{1}{4}\\\ &\small\text{or regroup in multiple of 96}\normalsize\qquad&\,&=&&96+192+288+384+...\\\ &\,&\,&=&&96(1+2+3+4+...)\\\ &\,&\,&=&&96\left(-\frac{1}{12}\right)\\\ &\qquad\qquad\small\text{Thus}&\infty\quad&=&&-8\\\ \end{alignat*}$$

source code: infty_2.tex | Go to top | TOC

Well, $\infty$ can be anything negative.

How much is $0.\infty$?

$$\begin{alignat*}{5} &\,&S\quad&=\quad&&1+1+1+1+1+1+...\\\ &\,&\,&=&&(1+1)+(1+1)+(1+1)+...\\\ &\,&\,&=&&2+2+2+...\\\ &\,&\,&=&&2(1+1+1+...)\\\ &\,&S\quad&=&&2S\\\ &\,&S\quad&=&&0\qquad&\small\text{(1)}\\\ &\,&2S\quad&=&&1+1+1+1+1+1+...\\\ &\,&\,&\,&&\quad~~~1+1+1+1+1+...\\\ &\,&2S\quad&=&&1+2+2+2+2+2+...\\\ &\,&3S\quad&=&&1+1+1+1+1+1+...\\\ &\,&\,&\,&&\quad~~~1+1+1+1+1+...\\\ &\,&\,&\,&&\qquad\quad~~1+1+1+1+...\\\ &\,&3S\quad&=&&1+2+3+3+3+3+...\\\ &\,&\,&\,&&...\\\ &\,&\infty S\quad&=\quad&&1+2+3+4+5+6+...\\\ &\small\text{from (1) }S=0\normalsize\quad&\infty0\quad&=&&1+2+3+4+5+6+...\\\ &\qquad\small\text{Thus}&0\infty\quad&=&&-\frac{1}{12} \end{alignat*}$$

source code: 0infty.tex | Go to top | TOC

Hmm, I will never believe.

Is $0$ an even number?

This proof is believed to be valid.

$$\begin{alignat*}{5} &\,&0\quad&=\quad&&0\\\ &\,&\,&=&&1-1\\\ &\,&\,&=&&1^2-1^2\\\ &\,&\,&=&&(1+1)(1-1)\\\ &\,&\,&=&&(1+1)(1^2-1^2)\\\ &\,&\,&=&&(1+1)(1+1)(1-1)\\\ &\small\text{repeat the last term}&\,&=&&(1+1)(1+1)(1+1)...(1-1)\\\ &\,&0\quad&=&&(1+1)^\infty*(1-1)\\\ &\qquad\qquad\small\text{Thus}\normalsize\qquad&0\quad&=&&2^\infty*0\qquad\small\text{(valid proof)}\\\ &\rlap{\qquad\qquad\small\text{Any number being a multiple of 2 is always even.}} \end{alignat*}$$

source code: 0_even.tex | Go to top | TOC

Solve $x$ for $x+2=x$.

$$\begin{alignat*}{5} &\,&x+2\quad&=\quad&&x\\\ &\,&(x+2)^2\quad&=&& x^2\\\ &\,&\cancel{x^2}+4x+4\quad&=&&\cancel{x^2}\\\ &\,&4x\quad&=&&-4\\\ &\,&\small\text{Thus}\normalsize\qquad x\quad&=&&-1 \end{alignat*}$$

source code: x+2eqx.tex | Go to top | TOC

A silly equation always has the solution.

Now, have a look at the valid proof of Euler's identity.

$e^{i\pi}=?$, also known as Euler's identity

$$\begin{alignat*}{7} &\,&\sin{x}&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}&&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\qquad&\small\text{for all x}\\\ &\,&\cos{x}&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}&&=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...\qquad&\small\text{for all x}\\\ &\,&e^x&=\sum_{n=0}^{\infty}\frac{x^n}{n!}&&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\qquad&\small\text{for all x}\\\ &\,&e^{ix}&=\rlap{1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}+\frac{(ix)^6}{6!}+\frac{(ix)^7}{7!}+\frac{(ix)^8}{8!}+...}\\\ &\,&\,&=\rlap{1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-\frac{x^6}{6!}-\frac{ix^7}{7!}+\frac{x^8}{8!}+...}\\\ &\,&\,&=\rlap{\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-...\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\right)}\\\ &\,&e^{ix}&=\cos{x}+i\sin{x}&&\small\text{***}\\\ &\,&e^{i\pi}&=\cos{\pi}+\cancelto{0}{i\sin{\pi}}\\\ &\small\text{Thus}\normalsize\qquad&e^{i\pi}&=-1\qquad\small\text{(valid proof)} \end{alignat*}$$

source code: euler.tex | Go to top | TOC

Euler's identity shall be used in the other proofs (whether valid or foolish).
Let's see the first foolish proofs using Euler's identity.

$e^x=1$, no matter what $x$ maybe.

Proof 1

$$\begin{alignat*}{5} &\,&e^x\quad&=\quad&&e^{i\pi.2\frac{x}{i\pi.2}}\\\ &\,&\,&=&&(e^{i\pi})^{2\frac{x}{2i\pi}}\\\ &\,&\,&=&&(-1)^{2\frac{x}{2i\pi}}\\\ &\,&e^x\quad&=&&1^\frac{x}{2i\pi}\qquad&\small(1)\\\ &\small\text{as}&e^{i\pi}\quad&=&&-1\\\ &\,&\ln{e^{i\pi}}\quad&=&&\ln{(-1)}\\\ &\,&i\pi\cancelto{1}{\ln{e}}&=&&\ln{(-1)}\\\ &\,&i\pi\quad&=&&\ln{(-1)}\qquad&\small(2)\\\ &\small\text{from (1)}&e^x\quad&=&&1^\frac{x}{2\ln(-1)}\\\ &\,&\,&=&&1^\frac{x}{\ln(-1^2)}\\\ &\,&\,&=&&1^\frac{x}{\cancelto{0}{\ln{1}}}\\\ &\,&\,&=&&1^\frac{x}{0}\\\ &\,&\,&=&&1^\infty\\\ &\small\text{Thus}&e^x\quad&=&&1\qquad&\small\text{(so foolish)} \end{alignat*}$$

source code: ex_eq_1_0.tex | Go to top | TOC

Q: Bruh, how can you divide by $0$ and evaluate $1^\infty$ to $1$?
A: Hold on, there is another neater proof.

Proof 2

$$\begin{alignat*}{5} &\,&e^{i\pi}\quad&=\quad&&-1\\\ &\,&\left(e^{i\pi}\right)^2\quad&=&&-1^2\\\ &\,&e^{2i\pi}\quad&=&&1\\\ &\small\text{power }n\text{ both sides}&e^{2ni\pi}\quad&=&&1^n\qquad&\small\text{for all n}\\\ &\,&e^{2ni\pi}\quad&=&&1\\\ &\small\times e\text{ both sides}&e^{2ni\pi}.e\quad&=&&e\\\ &\,&e^{2ni\pi+1}\quad&=&&e&\small(1)\\\ &\small\text{power }(2ni\pi+1)\text{ both sides}&\left(e^{2ni\pi+1}\right)^{2ni\pi+1}\quad&=&&e^{2ni\pi+1}\\\ &\small\text{from (1)}\normalsize\quad\,&e^{(2ni\pi+1)(2ni\pi+1)}\quad&=&&e\\\ &\,&e^{(2ni\pi+1)^2}\quad&=&&e\\\ &\,&e^{(2ni\pi)^2+2(2ni\pi)(1)+1^2}\quad&=&&e\\\ &\,&e^{-4n^2\pi^2+4ni\pi+1}\quad&=&&e\\\ &\,&e^{-4n^2\pi^2}.e^{4ni\pi}.\cancel{e}\quad&=&&\cancel{e}\\\ &\,&e^{-4n^2\pi^2}.e^{4ni\pi}\quad&=&&1\\\ &\,&e^{-4n^2\pi^2}.\left(e^{i\pi}\right)^{4n}\quad&=&&1\\\ &\,&e^{-4n^2\pi^2}.(-1)^{2.2n}\quad&=&&1\\\ &\,&e^{-4n^2\pi^2}.\cancelto{1}{1^{2n}}\quad&=&&1\\\ &\,&e^{-4n^2\pi^2}\quad&=&&1\\\ &\rlap{\small\text{as }n\text{ can be any real values, }x=-4n^2\pi^2\text{, can also be any real values.}}\\\ &\,&\small\text{Thus}\normalsize\qquad e^x\quad&\,=\quad1 \end{alignat*}$$

source code: ex_eq_1_1.tex | Go to top | TOC

Now, let's see the value of $\frac{\text{circumference}}{\text{diameter}}$, you know.

How much is $\pi$?

Solution 1

$$\begin{alignat*}{5} &\,&e^{i\pi}\quad&=\quad&&-1\\\ &\,&\ln{e^{i\pi}}\quad&=&&\ln{(-1)}\\\ &\,&i\pi\cancelto{1}{\ln e}\quad&=&&\ln{(-1)}\\\ &\,&\pi\quad&=&&\frac{\ln{(-1)}}{i}\\\ &\,&\,&=&&\frac{2\ln{(-1)}}{2i}\\\ &\,&\,&=&&\frac{\ln{(-1)^2}}{2i}\\\ &\,&\,&=&&\frac{\cancelto{0}{\ln{1}}}{2i}\\\ &\small\text{Thus}&\pi\quad&=&&0 \end{alignat*}$$

source code: pi_eq_0_0.tex | Go to top | TOC

If you don't believe, yet there are another solutions.

Solution 2

$$\begin{alignat*}{5} &\,&e^{i\pi}\quad&=\quad&&-1\\\ &\,&(e^{i\pi})^2\quad&=&&(-1)^2\qquad&\,\\\ &\,&e^{2i\pi}\quad&=&&1&\small(1)\\\ &\,&e^{2i\pi}.e\quad&=&&e\\\ &\,&e^{2i\pi+1}\quad&=&&e\\\ &\,&\left(e^{2i\pi+1}\right)^{2i\pi+1}\quad&=&&e^{2i\pi+1}\\\ &\,&e^{(2i\pi+1)(2i\pi+1)}\quad&=&&e^{2i\pi}.e\\\ &\small\text{from (1)}\normalsize\qquad&e^{(2i\pi+1)(2i\pi+1)}\quad&=&&1e\\\ &\,&e^{(2i\pi)^2+2.2i\pi+1^2}\quad&=&&e\\\ &\,&e^{(-4\pi^2+4i\pi+1)}\quad&=&&e\\\ &\,&e^{-4\pi^2}.e^{4i\pi}.\cancel{e}\quad&=&&\cancel{e}\\\ &\,&e^{-4\pi^2}.e^{4i\pi}\quad&=&&1\\\ &\,&e^{-4\pi^2}.e^{i\pi.2.2}\quad&=&&1\\\ &\,&e^{-4\pi^2}.(-1)^{2.2}\quad&=&&1\\\ &\,&e^{-4\pi^2}.\cancel{1}^2\quad&=&&\cancel{1}\\\ &\,&\ln{(e^{-4\pi^2})}\quad&=&&\ln1\\\ &\,&-4\pi^2\cancelto{1}{\ln{(e)}}\quad&=&&0\\\ &\,&\pi^2\quad&=&&0\\\ &\,&\small\text{Thus}\normalsize\qquad\pi\quad&=&&0 \end{alignat*}$$

source code: pi_eq_0_1.tex | Go to top | TOC

$\pi$ is really zero, let's confirm with the next solution.

Solution 3

$$\begin{alignat*}{5} &\,&e^{ix}\quad&=\quad&&\cos{x}+i\sin{x}\\\ &\,&e^{2i\pi}\quad&=&&\cos{2\pi}+\cancelto{0}{i\sin{2\pi}}\\\ &\,&\,&=&&1\\\ &\small\text{ln() both sides}\normalsize\qquad&\ln{e^{2i\pi}}\quad&=&&\cancelto{0}{\ln{1}}\\\ &\,&2i\pi\quad&=&&0\\\ &\small&divide\,i\text{ both sides}&\frac{2\cancel{i}\pi}{\cancel{i}}\quad&=&&\cancelto{0}{\frac{0}{i}}\\\ &\,&2\pi\quad&=&&0\\\ &\qquad\small\text{Thus}&\pi\quad&=&&0 \end{alignat*}$$

source code: pi_eq_0_2.tex | Go to top | TOC

Solution 4

$$\begin{alignat*}{5} &\,&e^{i\pi}\quad&=\quad&&-1\\\ &\,&\left(e^{i\pi}\right)^2\quad&=\quad&&(-1)^2\\\ &\,&e^{2i\pi}\quad&=&&1\qquad\qquad&\small\text{(1)}\\\ &\,&\sqrt[^i]{2}\quad&=&&\sqrt[^i]{2}\\\ &\small\text{from (1)}&\sqrt[^i]{2\cdot e^{2i\pi}}\quad&=&&\sqrt[^i]{2}\\\ &\,&\cancel{\sqrt[^i]{2}}\cdot\sqrt[^i]{e^{2i\pi}}\quad&=&&\cancel{\sqrt[^i]{2}}\\\ &\,&e^{\frac{2\cancel{i}\pi}{\cancel{i}}}\quad&=&&1\\\ &\small\ln{}\text{ both sides}\normalsize\qquad&\ln{e^{2\pi}}\quad&=&&\ln{1}\\\ &\,&2\pi\cancelto{1}{\ln{e}}\quad&=&&0\\\ &\qquad\small\text{Thus}&\pi\quad&=&&0 \end{alignat*}$$

source code: pi_eq_0_3.tex | Go to top | TOC

Hold on, $\pi$ is probably something else.

Solution 5

$$\begin{alignat*}{5} &\,&e^{i\pi}\quad&=\quad&&-1\\\ &\,&\ln{e^{i\pi}}\quad&=&&\ln{(-1)}\\\ &\,&i\pi\cancelto{1}{\ln{e}}\quad&=&&\ln{(-1)}\\\ &\,&1^{i\pi}\quad&=&&1^{\ln{(-1)}}\\\ &\,&(1^\pi)^i\quad&=&&1^{\ln{(-1)}}\\\ &\,&1^i\quad&=&&1^{\ln{(-1)}}\\\ &\,&\log_{1}1^i\quad&=&&\log_{1}1^{\ln{(-1)}}\\\ &\,&i\log_{1}1\quad&=&&\ln{(-1)}\log_{1}1\\\ &\,&i\quad&=&&\ln{(-1)}\\\ &\,&i\quad&=&&\ln{(e^{i\pi})}\\\ &\,&\cancel{i}\quad&=&&\cancel{i}\pi\cancelto{1}{\ln{e}}\\\ &\small\text{Thus}&\pi\quad&=&&1 \end{alignat*}$$

source code: pi_eq_1_0.tex | Go to top | TOC

Yet, there are another silly solutions.

Solution 6

$$\begin{alignat*}{5} &\rlap{\text{Rotate 6 by 180}^\circ\text{, it will be 9.}}\\\ &\rlap{\text{Rotate 6 by }\pi~\text{radians, it will be 9.}}\\\ &\,&6+\pi\quad&=\quad&&9\\\ &\,&\pi\quad&=&&9-6\\\ &\small\text{Thus}&\pi\quad&=&&3 \end{alignat*}$$

source code: pi_eq_3_0.tex | Go to top | TOC

Oops! engineers also say that!

Solution 7

$$\begin{alignat*}{7} &\small\text{Let}&x\quad&=\quad&&\frac{\pi+3}{2}\\\ &\,&2x\quad&=&&\pi+3\\\ &\small\times(\pi-3)\qquad&2x(\pi-3)\quad&=&&(\pi+3)(\pi-3)\\\ &\,&2\pi x-6x\quad&=&&\pi^2-9\\\ &\,&9-6x\quad&=&&\pi^2-2\pi x\\\ &\small+x^2&9-6x+x^2\quad&=&&\pi^2-2\pi x+x^2\\\ &\,&(3-x)^2\quad&=&&(\pi-x)^2\\\ &\,&3-\cancel{x}\quad&=&&\pi-\cancel{x}\\\ &\small\text{Thus}&\pi\quad&=&&3 \end{alignat*}$$

source code: pi_eq_3_1.tex | Go to top | TOC

We have seen several foolish proofs so far, let's see valid proofs.

How much is $i^i$?

$$\begin{alignat*}{5} &\small\text{from Euler's identity}&e^{ix}\quad&=\quad&&\cos{x}+i\sin{x}\\\ &\,&i\quad&=&&0+i\\\ &\,&\,&=&&\cos{\left(\frac{\pi}{2}\right)}+i\sin{\left(\frac{\pi}{2}\right)}\\\ &\,&\,&=&&e^{i\frac{\pi}{2}}\\\ &\small\text{power }i\text{ both sides}&i^i\quad&=&& e^{i^2\frac{\pi}{2}}\\\ &\,&\small\text{Thus}\normalsize\qquad i^i\quad&=&&e^{-\frac{\pi}{2}}\qquad\small\text{(valid proof)}\\\ &\rlap{\qquad\qquad\qquad\qquad\small\text{And it is a real number.}} \end{alignat*}$$

source code: i_power_i.tex | Go to top | TOC

How much is $\sqrt[^i]{i}$?

$$\begin{alignat*}{5} &\,&e^{i\pi}\quad&=\quad&&-1\\\ &\,&\,&=&&i^2\\\ &\small\text{take ln both sides}&\ln{e^{i\pi}}\quad&=&&\ln{i^2}\\\ &\,&i\pi\cancelto{1}{\ln{e}}\quad&=&&2\ln{i}\\\ &\,&\quad\frac{\pi}{2}\quad&=&&\frac{{1}}{i}\ln{i}\\\ &\,&\quad\ln{i^\frac{1}{i}}\quad&=&&\frac{\pi}{2}\\\ &\,&\quad i^\frac{1}{i}\quad&=&&e^\frac{\pi}{2}\\\ &\,&\small\text{Thus}\normalsize\qquad\sqrt[^i]{i}\quad&=&&e^\frac{\pi}{2}\qquad\small\text{(valid proof)}\\\ &\rlap{\qquad\qquad\qquad\qquad\small\text{And it is a real number.}} \end{alignat*}$$

source code: i_root_i.tex | Go to top | TOC

Who said $n^0=1$?

$$\begin{alignat*}{5} &\,&n^{-1}\quad&=\quad&&\frac{1}{n}\\\ &\,&n^{\left(i^2\right)}\quad&=&&\frac{1}{n}\\\ &\small\text{power }\frac{1}{i}\text{ both sides}&n^{\left(\frac{i^2}{i}\right)}\quad&=&&\left(\frac{1}{n}\right)^\frac{1}{i}\\\ &\,&n^i\quad&=&&\frac{1^\frac{1}{i}}{n^\frac{1}{i}}\\\ &\,&n^i.n^\frac{1}{i}\quad&=&&1^\frac{1}{i}\\\ &\,&n^{i+\frac{1}{i}}\quad&=&&1^\frac{1}{i}\\\ &\small\text{as }i=\frac{1}{-i}&n^{-\frac{1}{i}+\frac{1}{i}}\quad&=&&1^\frac{1}{i}\\\ &\,&n^0\quad&=&&(-1)^{2.\frac{1}{i}}\\\ &\small\text{as }e^{i\pi}=-1&\,&=&&(e^{\cancel{i}\pi})^\frac{2}{\cancel{i}}\\\ &\,&\small\text{Thus}\normalsize\qquad n^0\quad&=&&e^{2\pi}\\\ \end{alignat*}$$

source code: nto0neq1.tex | Go to top | TOC

It is still a constant, but not 1, not even an integer.
It is even an irrational and transcendental number.

Next, let's see valid proofs.

You might be wondering, why is $\frac{d}{dx}e^x=e^x$?

There are many solutions out there, let's have a look.

Proof 1: Limit Definition

$$\begin{alignat*}{5} &\,&f'(x)\quad&=\quad&&\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\\ &\,&\frac{d}{dx}e^x\quad&=&&\lim_{h\to0}\frac{e^{x+h}-e^x}{h}\\\ &\,&\frac{d}{dx}e^x\quad&=&&e^x.\frac{e^h-1}{h}\qquad&\small(1)\\\ &\,&\small\text{Let}\normalsize\qquad t\quad&=&&e^h-1\qquad&\small(2)\\\ &\,&\ln{t}\quad&=&&\ln{(e^h-1)}\\\ &\,&\,&=&&\frac{\ln{e^h}}{\ln{1}}\\\ &\,&\ln{t}\quad&=&&h.\frac{\cancelto{1}{\ln{e}}}{\ln{1}}\\\ &\,&h\quad&=&&\ln{t}.\ln{1}\\\ &\,&h\quad&=&&\ln{(t+1)}\qquad&\small(3)\\\ &\rlap{\qquad\qquad\small\text{since }h\text{ approaches }0\text{ , in order to balance the equation,}}\\\ &\rlap{\qquad\qquad\qquad\qquad\quad\small\text{so }t\text{ must also approach }0.}\\\ &\small\text{from (1), (2) and (3)}&\frac{d}{dx}e^x\quad&=&& e^x\lim_{t\to0}\frac{t}{\ln{(t+1)}}\\\ &\,&\,&=&&e^x\lim_{t\to0}\frac{1}{\frac{1}{t}\ln{(t+1)}}\\\ &\,&\frac{d}{dx}e^x\quad&=&&e^x\lim_{t\to0}\frac{1}{\ln{(t+1)}^\frac{1}{t}}\qquad&\small(4)\\\ &\rlap{\small\text{recall}\normalsize\qquad\qquad\qquad e=\lim_{n\to0}(1+n)^\frac{1}{n}=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n}\\\ &\small\text{from (4)}&\frac{d}{dx}e^x\quad&=&&e^x\frac{1}{\cancelto{1}{\ln{e}}}\\\ &\,&\small\text{Thus}\normalsize\qquad\frac{d}{dx}e^x\quad&=&&e^x\qquad\small\text{(valid proof)} \end{alignat*}$$

source code: d_e_power_x_0.tex | Go to top | TOC

Proof 2: Taylor Series

$$\begin{alignat*}{5} &\,&e^x\quad&=\quad&&\sum_{i=0}^\infty\frac{x^n}{n!}\\\ &\,&\,&=&&\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\\\ &\,&e^x\quad&=&&1+\frac{x^1}{1}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\qquad&\small\text{(1)}\\\ &\small\text{diff}&\frac{d}{dx}e^x\quad&=&&0+1+\frac{2x^1}{2!}+\frac{3x^2}{3!}+\frac{4x^3}{4!}+...\\\ &\,&\,&=&&1+\frac{\cancel{2}x^1}{\cancel{2}.1!}+\frac{\cancel{3}x^2}{\cancel{3}.2!}+\frac{\cancel{4}x^3}{\cancel{4}.3!}+...\\\ &\,&\,&=&&(1)\\\ &\small\text{Thus}\normalsize\qquad&\frac{d}{dx}e^x\quad&=&&e^x\qquad\small\text{(valid proof)} \end{alignat*}$$

source code: d_e_power_x_1.tex | Go to top | TOC

Proof 3: Implicit Differentiation

$$\begin{alignat*}{5} &\,&y\quad&=\quad&&e^x\\\ &\,&\ln{y}\quad&=&&\ln{e^x}\\\ &\,&\,&=&&x\cancelto{1}{\ln{e}}\\\ &\,&\ln{y}\quad&=&&x\\\ &\small\text{diff both sides}\normalsize\quad&\frac{d}{dx}\ln{y}\quad&=&&\frac{dx}{dx}\\\ &\,&\frac{1}{y}\frac{dy}{dx}\quad&=&&1\\\ &\,&\frac{dy}{dx}\quad&=&&y\\\ &\qquad\small\text{Thus}\normalsize\qquad&\frac{d}{dx}e^x\quad&=&&e^x\qquad\small\text{(valid proof)} \end{alignat*}$$

source code: d_e_power_x_2.tex | Go to top | TOC

Proof 4: Differential Equation

$$\begin{alignat*}{5} &\,&\frac{dy}{dx}\quad&=\quad&&y\\\ &\,&\frac{dy}{y}\quad&=&&dx\\\ &\small\text{integral}\quad&\int\frac{dy}{y}\quad&=&&\int{dx}\\\ &\,&\ln{y}\quad&=&&x\qquad&\small\text{omit}~C\\\ &\,&y\quad&=&&e^x\\\ &\small\text{Thus}&\frac{d}{dx}e^x\quad&=&&e^x\qquad&\small\text{(valid proof)} \end{alignat*}$$

source code: d_e_power_x_3.tex | Go to top | TOC

Four proofs should suffice.
$e^x$ is the only expression, you can't kill with differentiation.

Multiplying factor to change power voltage from $1\phi$ to $3\phi$

You may notice that the electrical power $1\phi\cdot110V$ is branched from $3\phi\cdot190V$. Or $1\phi\cdot220V$ is branched from $3\phi\cdot380V$ in some countries. This is because $110\sqrt{3}=190$ or $220\sqrt{3}=380$. You may wonder, why the factor is $\sqrt{3}$. Here's the proof.

This proof is kinda engineering, so $j$ denotes an imaginary number.

In $1\phi$ electricity, the voltage is measured across the live line and neutral, which is always 0 and the amplitude of live line is always a constant. But in $3\phi$ electricity, the voltage is measured across two live lines. And each live line is a sine curve (plus offset). How can we measure the voltage across two live lines? Let's see.

A full circle covers the angle of $2\pi$ radians or $360^{\circ}$. And it is split into $3\phi$. So each phase will set an angle by $\frac{2\pi}{3}$ radians or $120^\circ$ apart from one another. The voltage across phases can be computed by the difference between phases. In this case we will subtract the voltage at $120^\circ$ from the voltage at $0^\circ$.

$$\begin{alignat*}{5} &\,&V_{ab}\quad&=\quad&&V_{an}\measuredangle0^\circ-V_{bn}\measuredangle120^\circ\\\ &\,&\,&=&& V_{an}(\cos{0^\circ}+j\sin{{0^\circ}})-V_{bn}(\cos{120^\circ}+j\sin{{120^\circ}})\\\ &\,&\,&=&&V_{an}(1+0)-V_{bn}(-\frac{1}{2}+j\cdot\frac{\sqrt{3}}{2})\\\ &\,&V_{ab}\quad&=&&V_{an}-V_{bn}(-\frac{1}{2}+j\cdot\frac{\sqrt{3}}{2})\qquad&\small\text(1)\\\ &\small\text{take magnitude}&|V_{an}|\quad&=&&|V_{bn}|\quad=\quad|V|\quad\small\text{(measure line to neutral)}\\\ &\small\text{from (1)}&|V_{ab}|\quad&=&&|V|-|V|(-\frac{1}{2}+j\cdot\frac{\sqrt{3}}{2})\\\ &\,&|V_{ab}|\quad&=&&|V|+\frac{1}{2}|V|-j|V|\cdot\frac{\sqrt{3}}{2}\\\ &\,&\,&=&&|V|\frac{3}{2}-j|V|\cdot\frac{\sqrt{3}}{2}\\\ &\,&\,&=&&|V|\left(\frac{3}{2}-j\cdot\frac{\sqrt{3}}{2}\right)\\\ &\small\text{polar to rectangular}\normalsize\quad&|V_{ab}|\quad&=&&|V|\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\\\ &\,&\,&=&&|V|\sqrt{\frac{9}{4}+\frac{3}{4}}\\\ &\,&\,&=&&|V|\sqrt{\frac{12}{4}}\\\ &\qquad\qquad\small\text{Thus}\normalsize\qquad&|V_{ab}|\quad&=&&|V|\sqrt{3}\qquad\small\text{(valid proof)}\\\ \end{alignat*}$$

source code: 3p_power.tex | Go to top | TOC

So now you know that the power $1\phi\cdot110V$ is branched from $3\phi\cdot190V$. Or $1\phi\cdot220V$ is branched from $3\phi\cdot380V$ in some countries. The multiplying factor is $\sqrt{3}$.

Let's see some weirdness of mathematics.

The following math properties are not the proofs, but they are the demonstrations showing the valid weirdness of mathematics.

Magic numbers

Have you ever know a magic number 9? Whatever multiplying with 9 will result number of digits. We keep adding each digit until the final result is one digit. The final result is always 9 for examples ...

5 * 9 = 45,
then 4 + 5 = 9

or

135686 * 9 = 1221174
then 1 + 2 + 2 + 1 + 1 + 7 + 4 = 18,
and then 1 + 8 = 9.

All explaination above is on base 10. However this concept also applies to all bases. The magic number for each base is base - 1. This is a valid weirdness.

base 10: magic number = 10 - 1 = 9
1 * 9 = 9
2 * 9 = 18 --> 1 + 8 = 9
3 * 9 = 27 --> 2 + 7 = 9
4 * 9 = 36 --> 3 + 6 = 9
5 * 9 = 45 --> 4 + 5 = 9
6 * 9 = 54 --> 5 + 4 = 9
7 * 9 = 63 --> 6 + 3 = 9
8 * 9 = 72 --> 7 + 2 = 9
9 * 9 = 81 --> 8 + 1 = 9
The final result will always be 9.

base 8: magic number = 010 - 01 = 07
01 * 07 = 07
02 * 07 = 016 --> 01 + 06 = 07
03 * 07 = 025 --> 02 + 05 = 07
04 * 07 = 034 --> 03 + 04 = 07
05 * 07 = 043 --> 04 + 03 = 07
06 * 07 = 052 --> 05 + 02 = 07
07 * 07 = 061 --> 06 + 01 = 07
The final result will always be 07.

base 16: magic number = 0x10 - 0x1 = 0xf
0x1 * 0xf = 0xf
0x2 * 0xf = 0x1e --> 0x1 + 0xe = 0xf
0x3 * 0xf = 0x2d --> 0x2 + 0xd = 0xf
0x4 * 0xf = 0x3c --> 0x3 + 0xc = 0xf
0x5 * 0xf = 0x4b --> 0x4 + 0xb = 0xf
0x6 * 0xf = 0x5a --> 0x5 + 0xa = 0xf
0x7 * 0xf = 0x69 --> 0x6 + 0x9 = 0xf
0x8 * 0xf = 0x78 --> 0x7 + 0x8 = 0xf
0x9 * 0xf = 0x87 --> 0x8 + 0x7 = 0xf
0xa * 0xf = 0x96 --> 0x9 + 0x6 = 0xf
0xb * 0xf = 0xa5 --> 0xa + 0x5 = 0xf
0xc * 0xf = 0xb4 --> 0xb + 0x4 = 0xf
0xd * 0xf = 0xc3 --> 0xc + 0x3 = 0xf
0xe * 0xf = 0xd2 --> 0xd + 0x2 = 0xf
0xf * 0xf = 0xe1 --> 0xe + 0x1 = 0xf
The final result will always be 0xf.

You can try different bases. The results will be the same.

Vortex math

Consider the following series, each number is the power of 2. This is a valid math property.

$$\begin{alignat*}{15} &1,&\quad2,&\quad4,&\quad8,&~~~16,&~~32,&~~64,&~128,&~256,&~512,&~1024,&~2048,&~4096, ...\\\ \rlap{\text{Keep adding each digit, until the final result is one digit, ...}}\\\ &1,&\quad2,&\quad4,&\quad8,&\quad~7,&\quad5,&\quad1,&\quad~2,&\quad~4,&\quad~8,&\quad~~~7,&\quad~~~5,&\quad~~~1, ...\\\ \rlap{\text{The series will repeat with the following sub-series.}}\\\ &1,&\quad2,&\quad4,&\quad8,&\quad~7,&\quad5~ \end{alignat*}$$

source code: vortex.tex | Go to top | TOC

Do you feel that math is so strange?

A property of prime numbers

The square of a prime number other than 2 or 3 is always a multiple of 24 pluses 1. Let's see.

$$\begin{alignat*}{3} &~~5^2&=~~25&=(24*~~1)+1\\\ &~~7^2&=~~49&=(24*~~2)+1\\\ &11^2&=121&=(24*~~5)+1\\\ &13^2&=169&=(24*~~7)+1\\\ &17^2&=289&=(24*12)+1\\\ &19^2&=361&=(24*15)+1\\\ &23^2&=529&=(24*22)+1\\\ &29^2&=841&=(24*35)+1\\\ &31^2&=961&=(24*40)+1 \end{alignat*}$$

source code: prime_0.tex | Go to top | TOC

Primes (other than 2 and 3) can only have remainders of 1 or 5 when they are divided by 6. If the remainder was 0, 2 or 4 it would be even, and if it was 3, it would be divisible by 3.

$$\begin{alignat*}{5} &\,&p\quad&=\quad&&6n+1\\\ &\,&p^2\quad&=&&(6n+1)^2\\\ &\,&p^2\quad&=&&36n^2+12n+1&&\qquad\small\text{(1)}\\\ &\qquad\qquad\qquad\rlap{\text{If }n\text{ is even, }n(3n+1)\text{ is also even.}}\\\ &\qquad\rlap{\text{If }n\text{ is odd, }3n+1\text{ is even, so }n(3n+1)\text{ is still even.}}\\\ &\small\text{so for all }n\normalsize\quad&n(3n+1)\quad&=&&2k\qquad\small\text{for some integer }k\\\ &\small\text{from (1)}&p^2\quad&=&&12n(3n+1)+1\\\ &\,&\,&=&&12(2k)+1\\\ &\qquad\qquad\small\text{Thus}&p^2\quad&=&&24k+1\qquad\small\text{(valid proof)} \end{alignat*}$$

source code: prime_1.tex | Go to top | TOC

The proof for the case that p has a remainder of 5 is almost the same, just start with p = 6n - 1 instead of 6n + 1.

Proof of $\int_{-\infty}^\infty e^{-x^2}dx=\sqrt\pi$

This valid proof is called Gaussian intergral. It demonstrates the relation between $e$ and $\pi$.

$$\begin{alignat*}{5} &\,&I^2\quad&=\quad&&\int_{-\infty}^\infty\,e^{-x^2}\,dx.\int_{-\infty}^\infty\,e^{-y^2}\,dy\\\ &\,&\,&=&&\iint_{-\infty}^\infty\,e^{-(x^2+y^2)}\,dx\,dy\\\ &\rlap{\small\text{convert the double integral into polar coordinates}}\\\ &\,&I^2\quad&=&&\int_0^{2\pi}\int_0^\infty\,e^{-r^2}rdr\,d\theta\\\ &\rlap{\small\text{the inner integral}}\\\ &\,&\int_0^\infty\,e^{-r^2}rdr\quad&=&&\left.-\frac{1}{2}e^{-r^2}\right\vert_0^\infty\\\ &\,&\,&=&&-\frac{1}{2}e^{-\infty^2}-\left(\frac{1}{2}e^{-0^2}\right)\\\ &\,&\,&=&&-0-\left(-\frac{1}{2}\right)\\\ &\,&\int_0^\infty\,e^{-r^2}rdr\quad&=&&\frac{1}{2}\\\ &\rlap{\small\text{substituting this result back into the double integral}}\\\ &\,&I^2\quad&=&&\int_0^{2\pi}\,\frac{1}{2}\,d\theta\\\ &\,&\,&=&&\left.\frac{\theta}{2}\right\vert_0^{2\pi}\\\ &\,&\,&=&&\frac{\cancel{2}\pi}{\cancel{2}}-\cancelto{0}{\frac{0}{2}}\\\ &\,&I^2\quad&=&&\pi\\\ &\,&\small\text{Thus}\normalsize\qquad I~~\quad&=&&\sqrt{\pi}\qquad\small\text{(valid proof)} \end{alignat*}$$

source code: e_pi_relation.tex | Go to top | TOC

Gaussian integral is one of many relations between $e$ and $\pi$.

Golden ratio from Fibonacci sequence

$$\begin{alignat*}{5} &\,&F_0\quad&=\quad&&0\\\ &\,&F_1\quad&=&&1\\\ &\small\text{where }n\geqslant2&F_n\quad&=&&F_{n-1}+F_{n-2}\\\ &\small\text{Let }&\Phi\quad&=&&\lim_{n\to\infty}\frac{F_{n+1}}{F_n}\\\ &\,&\,&=&&\lim_{n\to\infty}\frac{F_n+F_{n-1}}{F_n}\\\ &\,&\,&=&&1+\lim_{n\to\infty}\frac{F_{n-1}}{F_n}\\\ &\,&\,&=&&1+\frac{1}{\Phi}\\\ &\,&\Phi\quad&=&&\frac{\Phi+1}{\Phi}\\\ &\,&\Phi^2\quad&=&&\Phi+1\\\ &\,&\Phi^2-\Phi-1\quad&=&&0\\\ &\,&\Phi\quad&=&&\frac{-(-1)\pm\sqrt{(-1^2)-4(1)(-1)}}{2(1)}\\\ &\small\Phi\text{ is positive}&\,&=&&\frac{1+\sqrt{1+4}}{2}\\\ &\,&\small\text{Thus}\normalsize\qquad\Phi\quad&=&&\frac{1+\sqrt{5}}{2}\qquad\small\text{(valid proof)} \end{alignat*}$$

source code: golden_fib.tex | Go to top | TOC

Well, we have been talking about valid proofs for long.
Let's start freaks again.

Isn't it $\infty=\Phi$?

$$\begin{alignat*}{5} &\,&\infty^2\quad&=\quad&&\infty\qquad&\small\text{(1)}\\\ &\,&\infty+1\quad&=&&\infty&\small\text{(2)}\\\ &\small\text{(1)=(2)}\normalsize\quad&\infty^2\quad&=&&\infty+1\\\ &\,&\infty^2-\infty-1\quad&=&&0\\\ &\,&\infty\quad&=&&\frac{-(-1)\pm\sqrt{(-1^2)-4(1)(-1)}}{2(1)}\\\ &\,&\,&=&&\frac{1\pm\sqrt{1+4}}{2}\\\ &\small\text{take positive}&\infty\quad&=&&\frac{1+\sqrt{5}}{2}\\\ &\,&\small\text{Thus}\normalsize\qquad\infty\quad&=&&\Phi\quad\small\text{a.k.a. Fibonacci golden ratio} \end{alignat*}$$

source code: infty_fibo.tex | Go to top | TOC

Now you can see, $\infty$ is as small as $1.618034$.

How much is $1^i$?

$$\begin{alignat*}{5} &\,&-1\quad&=\quad&&e^{i\pi}\\\ &\small\text{power }2i&-1^{2i}\normalsize\quad&=&&\left(e^{i\pi}\right)^{2i}\\\ &\,&\left(-1^2\right)^i\quad&=&&e^{2\pi.i^2}\\\ &\,&1^i\quad&=&&e^{2\pi.(-1)}\\\ &\small\text{Thus}&1^i\quad&=&&e^{-2\pi} \end{alignat*}$$

source code: 1poweri.tex | Go to top | TOC

Oops, a unit real number raises a unit imaginary number is real. The number is so small 0.001867.... And it is not 1, like what being said by WolframAlpha. Why, who know?

Let's see more about a unit real number and a unit imaginary number.

How much is $\sqrt[^i]{1}$?

$$\begin{alignat*}{5} &\,&-1\quad&=\quad&&e^{i\pi}\\\ &\small\text{power }i^3&\left(-1\right)^{i^3}\quad&=&&(e^{i\pi})^{i^3}\\\ &\,&-1^{i^2i}\quad&=&&e^{i^4\pi}\\\ &\,&-1^{-1i}\quad&=&&e^{(i^2)^2\pi}\\\ &\,&-1^{-i}\quad&=&&e^{-1^2\pi}\\\ &\,&-1^{-i}\quad&=&&e^{1\pi}\\\ &\small\text{as }-i=\frac{1}{i}&-1^\frac{1}{i}\quad&=&&e^\pi\\\ &\small\text{power }2&-1^\frac{2}{i}\quad&=&&e^{2\pi}\\\ &\,&\left(-1^2\right)^\frac{1}{i}\quad&=&&e^{2\pi}\\\ &\,&\quad1^{\frac{1}{i}}\quad&=&&e^{2\pi}\\\ &\qquad\small\text{Thus}&\sqrt[^i]{1}\quad&=&&e^{2\pi} \end{alignat*}$$

source code: sqrti1.tex | Go to top | TOC

You can see a unit imaginary root of a unit real number is real and indeed a big real. It is as big as 535.5. Once again, it is not 1, why does WolframAlpha say so?

Whether $e=1$ or $\pi=0$

$$\begin{alignat*}{5} &\,&e^{i\pi}\quad&=\quad&&-1\\\ &\small\text{power }i^3&\left(e^{i\pi}\right)^{i^3}\quad&=&&-1^{i^3}\\\ &\,&\left(e^\pi\right)^{i^4}\quad&=&&\left(-1^{i^2}\right)^i\\\ &\,&\left(e^\pi\right)^{i^2.i^2}\quad&=&&\left(-1^{-1}\right)^i\\\ &\,&\left(e^\pi\right)^{(-1)(-1)}\quad&=&&\left(\frac{1}{-1}\right)^i\\\ &\,&\left(e^\pi\right)^1\quad&=&&(-1)^i\\\ &\small\text{power }2&e^{2\pi}\quad&=&&(-1)^{2i}\\\ &\,&\,&=&&1^i\\\ &\small\text{from the previous proof}&\,&=&&e^{-2\pi}\\\ &\,&\small\text{Thus}\normalsize\qquad e^{2\pi}\quad&=&&e^{-2\pi} \end{alignat*}$$

source code: e1pi0.tex | Go to top | TOC

Oops, whether $e=1$ or $\pi=0$, indeed!

How much is $e^i$?

$$\begin{alignat*}{5} &\,&e^i\quad&=\quad&&\left(e^i\right)^\frac{2\pi}{2\pi}\\\ &\,&\,&=&&\left(e^{i\pi}\right)^{2.\frac{1}{2\pi}}\\\ &\,&\,&=&&(-1)^{2.\frac{1}{2\pi}}\\\ &\,&\,&=&&1^\frac{1}{2\pi}\\\ &\small\text{Thus}\normalsize\qquad&e^i\quad&=&&1 \end{alignat*}$$

source code: etoi.tex | Go to top | TOC

Simple continued fraction expansion of $\pi$

In our early primary school, we were taught that $\pi$ is approximately a rational number 22/7. But actually $\pi$ is an irrational number. We can have higher degree of accuracy and precision using higher numerator and denomerator. This method is called a simple continued fraction expansion of $\pi$, of which the series has been evaluated by matheticians over the centuries. That is ...

3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1, 1, 15, 3, 13, 1, 4, 2, 6, 6, 99, 1, 2, 2, 6, 3, 5, 1, 1, 6, 8, 1, 7, 1, 2, 3, 7, 1, 2, 1, 1, 12, 1, 1, 1, 3, 1, 1, 8, 1, 1, 2, 1, 6, 1, 1, 5, 2, 2, 3, 1, 2, 4, 4, 16, 1, 161, 45, 1, 22, 1, 2, 2, 1, 4, 1, 2, 24, 1, 2, 1, 3, 1, 2, 1, ...

(valid $\pi$ fraction expansion)

Further reference

We might start with the first two terms.

$$\begin{alignat*}{5} &\,&\pi\quad&\approx\quad&&3+\frac{1}{7}\\\ &\,&\,&\approx&&\frac{21+1}{7}\\\ &\small\text{Thus}\qquad&\pi\quad&\approx&&\frac{22}{7}\rlap{\quad=\quad3.14}\\\ &\rlap{\qquad\quad\small\text{(valid approximation)}} \end{alignat*}$$

source code: pi_exp_2_terms.tex | Go to top | TOC

Or we might want higher accuracy and precision. Let's take three terms.

$$\begin{alignat*}{5} &\,&\pi\quad&\approx\quad&&3+\frac{1}{7+\frac{1}{15}}\\\ &\,&\,&\approx&&3+\frac{1}{\frac{106}{15}}\\\ &\,&\,&\approx&&3+\frac{15}{106}\\\ &\small\text{Thus}\qquad&\pi\quad&\approx&&\frac{333}{106}\rlap{\quad=\quad3.1415}\\\ &\rlap{\qquad\qquad\small\text{(valid approximation)}} \end{alignat*}$$

source code: pi_exp_3_terms.tex | Go to top | TOC

And we go a little further, Let's take five terms.

$$\begin{alignat*}{5} &\,&\pi\quad&\approx\quad&&3+\frac{1}{7+\frac{1}{15+\frac{1}{1+\frac{1}{292}}}}\\\ &\,&\,&\approx\quad&&3+\frac{1}{7+\frac{1}{15+\frac{1}{\frac{293}{292}}}}\\\ &\,&\,&\approx\quad&&3+\frac{1}{7+\frac{1}{15+\frac{292}{293}}}\\\ &\,&\,&\approx\quad&&3+\frac{1}{7+\frac{1}{\frac{4687}{293}}}\\\ &\,&\,&\approx\quad&&3+\frac{1}{7+\frac{293}{4687}}\\\ &\,&\,&\approx\quad&&3+\frac{1}{\frac{33102}{4687}}\\\ &\,&\,&\approx\quad&&3+\frac{4687}{33102}\\\ &\small\text{Thus}\quad&\pi\quad&\approx\quad&&\frac{103993}{33102}\rlap{\quad=\quad3.141592653}\\\ &\rlap{\qquad\qquad\quad\small\text{(valid approximation)}} \end{alignat*}$$

source code: pi_exp_5_terms.tex | Go to top | TOC

$\sqrt{-\ln{(-1)}\ln{(-1)}}=\pi$, how?

In terms of complex numbers, this proof is believed to be valid.

$$\begin{alignat*}{5} &\,&-1\quad&=\quad&&e^{i\pi}\\\ &\small\text{take ln both sides}&\ln{(-1)}\quad&=&&\ln{e^{i\pi}}\\\ &\,&\,&=&&i\pi\cancelto{1}{\ln{e}}\\\ &\,&\ln{(-1)}\quad&=&&i\pi&\small\text{(1)}\\\ &\small\text{from (1)}&\sqrt{-\ln{(-1)}\ln{(-1)}}\quad&=&&\sqrt{-(i\pi)(i\pi)}\\\ &\,&\,&=&&\sqrt{-(i\pi)^2}\\\ &\,&\,&=&&\sqrt{-i^2\pi^2}\\\ &\,&\,&=&&\sqrt{-(-1)(\pi^2)}\\\ &\,&\,&=&&\sqrt{\pi^2}\\\ &\small\text{Thus}&\sqrt{-\ln{(-1)}\ln{(-1)}}\quad&=&&\pi\qquad\small\text{(valid proof)} \end{alignat*}$$

source code: ln-1.tex | Go to top | TOC

Well, WolframAlpha confirms that!

How is $6$ afraid of $7$? It ain't and will never be.

There was a joke from our childhood, why $6$ is afraid of $7$. Because $7$ ate $9$ or someone say $\cos{(789)}$. Moreover, now 7 turns around and it opens its mouth while facing 6.

Nah, 6 ain't afraid of 7, here's the proof. I can't believe, I simply thought about this joke, then discovered this foolish proof within a few hours. ;-p

$$\begin{alignat*}{5} &\,&6-7\quad&=\quad&&-1\\\ &\,&\,&=&&2-3\\\ &\,&\,&=&&2-\frac{5}{2}+\frac{5}{2}-3\\\ &\,&\,&=&&\sqrt{\left(2-\frac{5}{2}\right)^2}+\frac{5}{2}-3\\\ &\,&\,&=&&\sqrt{2^2-2(\cancel{2})\left(\frac{5}{\cancel{2}}\right)+\left(\frac{5}{2}\right)^2}+\frac{5}{2}-3\\\ &\,&\,&=&&\sqrt{4-10+\left(\frac{5}{2}\right)^2}+\frac{5}{2}-3\\\ &\,&\,&=&&\sqrt{-6+\left(\frac{5}{2}\right)^2}+\frac{5}{2}-3\\\ &\,&\,&=&&\sqrt{9-15+\left(\frac{5}{2}\right)^2}+\frac{5}{2}-3\\\ &\,&\,&=&&\sqrt{3^2-2(3)\left(\frac{5}{2}\right)+\left(\frac{5}{2}\right)^2}+\frac{5}{2}-3\\\ &\,&\,&=&&\sqrt{\left(3-\frac{5}{2}\right)^2}+\frac{5}{2}-3\\\ &\,&\,&=&&3-\frac{5}{2}+\frac{5}{2}-3\\\ &\,&6-7\quad&=&&0\\\ &\small\text{Thus}&6\quad&=&&7 \end{alignat*}$$

source code: 6_not_afraid_7.tex | Go to top | TOC

So, $6$ is not afraid of $7$ and will never be. They are on par.

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