This is my Maths page of suggested proofs, etc. This is NOT the Maths Sandbox^{Here it is!}. Please do not add to it unless you are me or my friend Oliver.

${\displaystyle a=}$ a number whose square is to be found out. ${\displaystyle a-1=b,b-1=c}$... etc. ${\displaystyle x=}$ a number whose square is known.

${\displaystyle a^{2}=(b2+1)+(c2+1)+...+(x2+1)+x^{2}}$

${\displaystyle \sum _{?}?}$

• All television programmes are evil.

Explanation

Take a television programme (P).

It is the product of time (t) and money (m), and because product also means multiply, ${\displaystyle P=tm}$.

A saying says, 'time is money', so ${\displaystyle t=m}$.

Therefore ${\displaystyle p=m^{2}}$.

Another saying says, 'money is the root of all evil [e].' So ${\displaystyle {\sqrt {e}}=m}$

Therefore ${\displaystyle m^{2}=e}$.

And as ${\displaystyle P=m^{2}}$, ${\displaystyle P=e}$.

And as P = the television programme and e = evil ...

The television programme is evil.

QED.

This attempts to prove that ${\displaystyle \infty =-\infty }$.
Take a vertical line graph (g1). The gradient of this line is ${\displaystyle \infty }$.
Take a diagonal line graph (g2). The gradient of this line is 2.
Mirror (g2). The gradient of the new line is -2, so mirroring a line multiplies its gradient by -1.
So mirroring (g1) means the new gradient is ${\displaystyle -\infty }$.
Now mirroring a straight vertical line will give you a straight vertical line, so (-g1) is vertical, and so equal to (g1).
Because the graphs (g1) and (-g1) are equal, their gradients must be equal.
So ${\displaystyle \infty =-\infty }$.
QED

This attempts to prove that ${\displaystyle i=-1}$, destroying the very fabric of the existance of ${\displaystyle i}$.
If ${\displaystyle i={\sqrt {-1}}}$
Then ${\displaystyle -i={\sqrt {1}}}$
So ${\displaystyle -i=1}$
Therefore ${\displaystyle i=-1}$.
QED

n is an at least 2-digit positive integer. It is presumed that one would be in the automatic knowledge that a smaller positive number is thus divisible. It is not known (someone test it?) if these formulas will work for negative numbers. Notice positive not non-negative.

To find if n is divisible by 3 (or 9), find if x is. Square brackets represent giving an integral answer to bracketed term.
${\displaystyle x=[({\frac {n}{10}}-[{\frac {n}{10}}])\times 10]+[({\frac {n}{100}}-[{\frac {n}{100}}])\times 10]+...}$
Seeing that this is taking the digital root, one might satirically write instead:
${\displaystyle x={\sqrt[{digit}]{n}}}$

n is divisible by 5 (but not 10) if the answer to the following expression is 5, but if the answer is 0, it is also divisible by 10. ${\displaystyle ({\frac {n}{10}}-[{\frac {n}{10}}])\times 10}$

To find if n is divisible by 7, find if x is. Square brackets represent giving an integral answer to bracketed term.
${\displaystyle x=([{\frac {n}{10}}]\times 3)+(({\frac {n}{10}}-[{\frac {n}{10}}])\times 10)}$

${\displaystyle (x+{\frac {1}{2}})^{2}=x(x+1)+{\frac {1}{4}}}$

${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}{\mbox{ when }}ax^{2}+bx+c=0{\mbox{ and }}a\neq 0}$

a must not = 0 because then the formula would be ${\displaystyle x={\frac {\cdots }{2\times 0}}}$ and you can't divide by 0.

Take two numbers that are the same.

Call them a and b.	${\displaystyle a=b}$
Both sides multiplied by a:	${\displaystyle a^{2}=ab}$
Subtract b2 from both sides:	${\displaystyle a^{2}-b^{2}=ab-b^{2}}$
Or	${\displaystyle (a+b)(a-b)=b(a-b)}$
Divide each side by (a-b):	${\displaystyle (a+b)=b}$
Or	${\displaystyle a+b=b}$
Subtract b from both sides:	${\displaystyle a+b-b=b-b}$
Or	${\displaystyle a=0}$.

We didn't say what a was to start with, so we've proved that anything = 0.