User:Jake95/Maths

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.

Squaring Formulae

Mine

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

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

Oliver's

$\sum _{?}?$

Oliver's TV Joke

• 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, $P=tm$.

A saying says, 'time is money', so $t=m$.

Therefore $p=m^{2}$.

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

Therefore $m^{2}=e$.

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

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

The television programme is evil.

QED.

Infinity = Negative Infinity

This attempts to prove that $\infty =-\infty$.
Take a vertical line graph (g1). The gradient of this line is $\infty$.
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 $-\infty$.
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 $\infty =-\infty$.
QED

i = -1

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

Divisibility

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.

By 3 or 9

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

By 5 and 10

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. $({\frac {n}{10}}-[{\frac {n}{10}}])\times 10$

By 7

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

Something-and-a-Half-Squared

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

Quadratics Formula

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

Footnote

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

The Zero Proof!!

Take two numbers that are the same.

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

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