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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
Notes:
Almost all mathematical proofs here are foolish , unless explicitly noted as valid .
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Proofs speak louder than words. Let's start!
We firstly start from the well-known Ramanujan Summation
$\qquad1+2+3+4+5+6+...\quad=\quad?$
$$\begin{alignat*}{5}
&\,&S\quad&=\quad&&1+2+3+4+5+6+...\\\
&\small\text{Let}&S_1\quad&=&&1-1+1-1+...\\\
&\,&1-S_1\quad&=&&1-(1+1-1+1-...)\\\
&\,&\,&=&&1-1+1-1+...\\\
&\,&\,&=&& S_1\\\
&\,&2S_1\quad&=&&1\\\
&\,&S_1\quad&=&&\frac{1}{2}\\\
&\small\text{Let}&S_2\quad&=&&1-2+3-4+5-6+...\\\
&\,&2S_2\quad&=&&1-2+3-4+5-6+...\\\
&\,&\,&\,&&\quad+1-1+3-4+5-6\\\
&\,&\,&=&&1-1+1-1+...\\\
&\,&2S_2\quad&=&& S_1\\\
&\,&S_2\quad&=&&\frac{S_1}{2}\\\
&\,&S_2\quad&=&&\frac{1}{4}\\\
&\,&S-S_2\quad&=&&1+2+3+4+5+6+7+8+...\\\
&\,&\,&\,&&\,\,\,-(1-2+3-4+5-6+7-...)\\\
&\,&\,&=&&4+8+12+16+...\\\
&\,&\,&=&&4(1+2+3+4+...)\\\
&\,&S-S_2\quad&=&&4S\\\
&\,&3S\quad&=&&-S_2\\\
&\,&S\quad&=&&-\frac{S_2}{3}\\\
&\,&\,&=&&-\frac{1}{3*4}\\\
&\small\text{Thus}&S\quad&=&&-\frac{1}{12}
\end{alignat*}$$
source code: rama_sum.tex | Go to top | TOC
There are still a few Ramanujan alternatives.
$$\begin{alignat*}{5}
&\,&S\quad&=\quad&&1+2+3+4+5+6+7+8+9+10+...\\\
&\,&\,&=&&1+(2+3+4)+(5+6+7)+(8+9+10)+...\\\
&\,&\,&=&&1+(9+18+27+...)\\\
&\,&\,&=&&1+9(1+2+3+...)\\\
&\,&S\quad&=&&1+9S\\\
&\,&8S\quad&=&&-1\\\
&\small\text{Thus}&\qquad S\quad&=&&-\frac{1}{8}
\end{alignat*}$$
source code: rama_alt_1.tex | Go to top | TOC
Hold on! There is still another alternative.
$$\begin{alignat*}{5}
&\,&S\quad&=\quad&&1+2+3+4+5+6+7+8+...\\\
&\,&\,&=&&(2+4+6+8+10+12+14+...)+\\\
&\,&\,&\,&&(1+3+5+7+9+11+13+...)\\\
&\,&\,&=&&2(1+2+3+4+5+6+7+...)+\\\
&\,&\,&\,&&~\,(1+(3+5)+(7+9)+(11+13)+...)\\\
&\,&\,&=&&2S+(1+8+16+24+32+...)\\\
&\,&\,&=&&2S+(1+8(1+2+3+4+...))\\\
&\,&\,&=&&2S+(1+8S)\\\
&\,&S\quad&=&&10S+1\\\
&\small\text{Thus}\normalsize\qquad&S\quad&=&&-\frac{1}{9}
\end{alignat*}$$
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 $$\begin{alignat*}{5}
&\,&S\quad&=\quad&&2+4+6+8+10+...\\\
&\,&-\frac{1}{12}\quad&=&&1+2+3+4+5+6+7+8+9+10+...\\\
&\,&\,&=&&(1+3+5+7+9+...)+(2+4+6+8+10+...)\\\
&\,&\,&=&&\frac{1}{12}+(2+4+6+8+10+...)\\\
&\,&-\frac{1}{12}\quad&=&&\frac{1}{12}+S\\\
&\small\text{Thus}\normalsize\qquad&S\quad&=&&-\frac{1}{6}
\end{alignat*}$$
source code: sum_of_even_0.tex | Go to top | TOC
Why is it negative again, who know? Probably something wrong.
$$\begin{alignat*}{5}
&\,&S\quad&=\quad&&2+4+6+8+10+...\\\
&\,&\,&=&&2(1+2+3+4+5+...)\\\
&\,&\,&=&&2\left(-\frac{1}{12}\right)\\\
&\small\text{Thus}\normalsize\qquad&S\quad&=&&-\frac{1}{6}
\end{alignat*}$$
source code: sum_of_even_1.tex | Go to top | TOC
Same result, do you start to believe?
Here come the most foolish proof!
$$\begin{alignat*}{5}
&\,&0\quad&=\quad&&0+0+0+0+...\\\
&\,&\,&=&&(1-1)+(1-1)+(1-1)+(1-1)+...\\\
&\,&\,&=&&1+(-1+1)+(-1+1)+(-1+1)+...\\\
&\,&\,&=&&1+0+0+0+0+...\\\
&\small\text{Thus}\normalsize\qquad&0\quad&=&&1
\end{alignat*}$$
source code: 0eq1_0.tex | Go to top | TOC
What about the second most foolish proof?
$$\begin{alignat*}{5}
&\small\text{Let}\normalsize\qquad&S\quad&=\quad&&1+1+1+1++1...\\\
&\,&\,&=&&1+(1+1+1+1+...)\\\
&\,&\cancel{S}\quad&=&&1+\cancel{S}\\\
&\small\text{Thus}\normalsize\qquad&0\quad&=&&1
\end{alignat*}$$
source code: 0eq1_1.tex | Go to top | TOC
Is there any proof looking more advanced than these?
$$\begin{alignat*}{5}
&\,&\int u\,dv\quad&=\quad&&uv-\int v\,du\\\
&\,&\int\frac{1}{x}\,dx\quad&=&&\cancel{x}\frac{1}{\cancel{x}}-\int x\,d\left(\frac{1}{x}\right)\\\
&\,&\,&=&&1-\int x\,d(x^{-1})\\\
&\,&\,&=&&1-\int x\,(-1x^{-2})\,dx\\\
&\,&\,&=&&1+\int x.x^{-2}\,dx\\\
&\,&\cancel{\int\frac{1}{x}\,dx}\quad&=&&1+\cancel{\int\frac{1}{x}\,dx}\\\
&\small\text{Thus}\normalsize\qquad&0\quad&=&&1
\end{alignat*}$$
source code: 0eq1_2.tex | Go to top | TOC
Can you find an error? Hmm, binary no longer exists.
Here, the proof I learnt in junior high school.
$$\begin{alignat*}{5}
&\,&\small\text{Let}\normalsize\qquad a\quad&=\quad&&b\\\
&\times b\small\text{ both sides}\quad&a*b\quad&=&&b*b\\\
&\,&a*b\quad&=&&b^2\\\
&-a^2\small\text{ both sides}\quad&a*b-a^2\quad&=&&b*b-a^2\\\
&\,&a(b-a)\quad&=&&b^2-a^2\\\
&\,&a\cancel{(b-a)}\quad&=&&\cancel{(b-a)}(b+a)\\\
&\small\text{since }b=a&a\quad&=&&a+a\\\
&\,&\cancel{a}\quad&=&&2\cancel{a}\\\
&\,&\small\text{Thus}\normalsize\qquad1\quad&=&&2
\end{alignat*}$$
source code: 1eq2_0.tex | Go to top | TOC
Nah, there is another proof in high school using trigonometry.
$$\begin{alignat*}{5}
&\small\text{where }x=\frac{\pi}{4}~or~\frac{5\pi}{4}&\cos{x}\quad&=\quad&&\sin{x}\\\
&\,&\cos{2x}\quad&=&&\sin{2x}\\\
&\,&1-2\sin^2{x}\quad&=&&2\sin{x}\cos{x}\\\
&\small\text{as }\sin{x}=\cos{x}&\,&=&&2\cos{x}\cos{x}\\\
&\,&1-2\sin^2{x}\quad&=&&2\cos^2{x}\\\
&\,&1\quad&=&&2\sin^2{x}+2\cos^2{x}\\\
&\,&\,&=&&2\cancelto{1}{(\sin^2{x+\cos^2{x}})}\\\
&\,&\small\text{Thus}\normalsize\qquad1\quad&=&&2
\end{alignat*}$$
source code: 1eq2_1.tex | Go to top | TOC
Yet, there is another proof using calculus.
$$\begin{alignat*}{5}
&\,&\underbrace{x+x+x+...+x}_{x\text{ terms}}\quad&=\quad&&x*x\\\
&\,&\,&=&&x^2\\ \\\
&\small\text{diff both sides }&\underbrace{1+1+1+...+1}_{x\text{ terms}}\quad&=&&2x\\\
&\,&\cancel{x}\quad&=&&2\cancel{x}\\ \\\
&\,&\small\text{Thus}\normalsize\qquad1\quad&=&&2
\end{alignat*}$$
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, $4=5$ ?
$$\begin{alignat*}{5}
&\,&\frac{-1}{1}\quad&=\quad&&\frac{1}{-1}&(1)\\\
&\,&\sqrt{\frac{-1}{1}}\quad&=&&\sqrt{\frac{1}{-1}}&(2)\\\
&\,&\frac{\sqrt{-1}}{\sqrt{1}}\quad&=&&\frac{\sqrt{1}}{\sqrt{-1}}&(3)\\\
&\,&\frac{i}{1}\quad&=&&\frac{1}{i}&(4)\\\
&\small\times\frac{1}{2}&\frac{i}{2}\quad&=&&\frac{1}{2i}&(5)\\\
&\small+\frac{3}{2i}&\frac{i}{2}+\frac{3}{2i}\quad&=&&\frac{1}{2i}+\frac{3}{2i}&(6)\\\
&\times i&i\left(\frac{i}{2}+\frac{3}{2i}\right)\quad&=&&i\left(\frac{1}{2i}+\frac{3}{2i}\right)\qquad&(7)\\\
&\,&\frac{i^2}{2}+\frac{3\cancel{i}}{2\cancel{i}}\quad&=&&\frac{\cancel{i}}{2\cancel{i}}+\frac{3\cancel{i}}{2\cancel{i}}&(8)\\\
&\,&-\frac{1}{2}+\frac{3}{2}\quad&=&&\frac{1}{2}+\frac{3}{2}&(9)\\\
&\,&\frac{-1+3}{2}\quad&=&&\frac{1+3}{2}&(10)\\\
&\,&\frac{\cancel{2}}{\cancel{2}}\quad&=&&\cancelto{2}{\frac{4}{2}}&(11)\\\
&\,&1\quad&=&&2&(12)\\\
&\small\text{from (9)}&-\frac{1}{2}+\cancel{\frac{3}{2}}\quad&=&&\frac{1}{2}+\cancel{\frac{3}{2}}&(13)\\\
% line (14): Can't add `\cancel` inside `\frac`, MathJax bug?
&\,&-\frac{1}{2}\quad&=&&\frac{1}{2}&(14)\\\
&\,&-1\quad&=&&1&(15)\\\
&\small+1&-1+1\quad&=&&1+1&(16)\\\
&\,&0\quad&=&&2&(17)\\\
&\small\text{from (12), (15), (17)}&-1~=~0\quad&=&&1~=~2&(18)
\end{alignat*}$$
source code: all_int_eq.tex | Go to top | TOC
It is very articulate, indeed.
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.$0$ couldn't be a denominator? See the following foolish proofs.
Very silly solutions to find $\frac{0}{0}$ . Let's see the first fool. $$\begin{alignat*}{5}
&\,&\small\text{Let}\normalsize\qquad\frac{2}{0}\quad&=\quad&&\frac{x}{1}\\\
&\times\small\text{0 both sides}&\frac{2*0}{0}\quad&=&&\frac{x*0}{1}\\\
&\,\small ÷\text{ 2 both sides}&\frac{\cancel{2}*0}{0*\cancel{2}}\quad&=&&\cancelto{0}{\frac{x*0}{1*2}}\\\
&\,&\small\text{Thus}\normalsize\qquad\frac{0}{0}\quad&=&&0
\end{alignat*}$$
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 $\frac{1}{i}$ . $$\begin{alignat*}{5}
&\,&-1\quad&=\quad&&i^2\\\
&\small ÷(-i)\text{ both sides}&\frac{-1}{-i}\quad&=&&\frac{i^2}{-i}\\\
&\,&\small\text{Thus}\normalsize\qquad\frac{1}{i}\quad&=&&-i\qquad\small\text{(valid proof)}
\end{alignat*}$$
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.
$$\begin{alignat*}{5}
&\,&i^2\quad&=\quad&&-1\\\
&\,&i*i\quad&=&&-1\\\
&\,&i\quad&=&&\frac{-1}{i}\\\
&\rlap{i=\frac{-1}{i}=\frac{-1}{\frac{-1}{i}}=\frac{-1}{\frac{-1}{\frac{-1}{i}}}=...}\\\
&\,&i\quad&=&&(-1)^\infty\\\
&\,&\small\text{Thus}\normalsize\qquad i\quad&=&&\pm1
\end{alignat*}$$
source code: i_eq_pm1.tex | Go to top | TOC
Hold on! $i$ is probably something else.
$$\begin{alignat*}{5}
&\,&0\quad&=\quad&&i-i\\\
&\,&\,&=&&i-\sqrt{-1}\\\
&\,&\,&=&&i-\sqrt{(-1)(1)}\\\
&\,&\,&=&&i-\sqrt{(-1)(-1)^2}\\\
&\,&\,&=&&i-\sqrt{-1}\sqrt{(-1)^2}\\\
&\,&\,&=&&i-(i)(-1)\\\
&\,&\,&=&&i+i\\\
&\,&0\quad&=&&2i\\\
&\small\text{Thus}\normalsize\qquad&i\quad&=&&0\\\
\end{alignat*}$$
source code: i0.tex | Go to top | TOC
Whatever it is, it is not imaginary anyway.
$$\begin{alignat*}{5}
&\,&-\frac{1}{12}\quad&=\quad&&1+2+3+...+\infty\\\
&\,&\,&=&&\frac{\infty(\infty+1)}{2}\\\
&\,&-1\quad&=&&6\infty(\infty+1)\\\
&\,&0\quad&=&&6\infty^2+6\infty+1\\\
&\,&\infty\quad&=&&\frac{-6\pm\sqrt{6^2-4(6)(1)}}{2(6)}\\\
&\,&\,&=&&\frac{-6\pm\sqrt{36-24}}{12}\\\
&\,&\,&=&&\frac{-6\pm\sqrt{12}}{12}\\\
&\,&\,&=&&\frac{-6\pm\sqrt{2^2*3}}{12}\\\
&\,&\,&=&&\frac{-6\pm2\sqrt{3}}{12}\\\
&\small\text{Thus}&\infty\quad&=&&\frac{-3\pm\sqrt{3}}{6}
\end{alignat*}$$
source code: infty_0.tex | Go to top | TOC
Hold on, $\infty$ is probably something else.
$$\begin{alignat*}{5}
&\,&x\quad&=\quad&&\bar{9}&&\small\text{(1)}\\\
&\,&x\quad&=&&\infty&&\small\text{(2)}\\\
&\small10*\text{(1)}&10x\quad&=&&\bar{9}0\\\
&\small+9\text{ both sides}\qquad&10x+9\quad&=&&\bar{9}\\\
&\small\text{from (1)}&10x+9\quad&=&&x\\\
&\,&9x\quad&=&&-9\\\
&\,&x\quad&=&&-1\qquad&&\small\text{(3)}\\\
&\small\text{(2)=(3)}&\infty\quad&=&&-1&&\small\text{(4)}\\\
&\small1÷\text{(4)}&\frac{1}{\infty}\quad&=&&\frac{1}{-1}\\\
&\,&0\quad&=&&-1\\\
&\small\text{from (3)}&x\quad&=&&0&&\small\text{(5)}\\\
&\small\text{Thus (2)=(3)=(5)}&\infty\quad&=&&-1\quad=\quad0
\end{alignat*}$$
source code: infty_1.tex | Go to top | TOC
Here, $\infty$ is so small, you see?
$$\begin{alignat*}{5}
&\,&\infty\quad&=\quad&&1+1+1+1+...\\\
&\small\text{regroup in multiple of 2}&\,&=&&2+4+6+8+...\\\
&\,&\,&=&&2(1+2+3+4+...)\\\
&\,&\,&=&&2\left(-\frac{1}{12}\right)\\\
&\qquad\qquad\small\text{Thus}&\infty\quad&=&&-\frac{1}{6}\\\
&\small\text{or regroup in multiple of 3}&\,&=&&3+6+9+12+...\\\
&\,&\,&=&&3(1+2+3+4+...)\\\
&\,&\,&=&&3\left(-\frac{1}{12}\right)\\\
&\qquad\qquad\small\text{Thus}&\infty\quad&=&&-\frac{1}{4}\\\
&\small\text{or regroup in multiple of 96}\normalsize\qquad&\,&=&&96+192+288+384+...\\\
&\,&\,&=&&96(1+2+3+4+...)\\\
&\,&\,&=&&96\left(-\frac{1}{12}\right)\\\
&\qquad\qquad\small\text{Thus}&\infty\quad&=&&-8\\\
\end{alignat*}$$
source code: infty_2.tex | Go to top | TOC
Well, $\infty$ can be anything negative.
$$\begin{alignat*}{5}
&\,&S\quad&=\quad&&1+1+1+1+1+1+...\\\
&\,&\,&=&&(1+1)+(1+1)+(1+1)+...\\\
&\,&\,&=&&2+2+2+...\\\
&\,&\,&=&&2(1+1+1+...)\\\
&\,&S\quad&=&&2S\\\
&\,&S\quad&=&&0\qquad&\small\text{(1)}\\\
&\,&2S\quad&=&&1+1+1+1+1+1+...\\\
&\,&\,&\,&&\quad~~~1+1+1+1+1+...\\\
&\,&2S\quad&=&&1+2+2+2+2+2+...\\\
&\,&3S\quad&=&&1+1+1+1+1+1+...\\\
&\,&\,&\,&&\quad~~~1+1+1+1+1+...\\\
&\,&\,&\,&&\qquad\quad~~1+1+1+1+...\\\
&\,&3S\quad&=&&1+2+3+3+3+3+...\\\
&\,&\,&\,&&...\\\
&\,&\infty S\quad&=\quad&&1+2+3+4+5+6+...\\\
&\small\text{from (1) }S=0\normalsize\quad&\infty0\quad&=&&1+2+3+4+5+6+...\\\
&\qquad\small\text{Thus}&0\infty\quad&=&&-\frac{1}{12}
\end{alignat*}$$
source code: 0infty.tex | Go to top | TOC
Hmm, I will never believe.
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
I keep adding these kinds of freaks or topics, which are considered interesting, no matter what they are foolish or valid math. foolmath
If you find foolmath buy me a coffee . I will highly appreciate and will be very grateful.
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