User:Jake95/Maths

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

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.

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