phrontister wrote:

0<x<0

Looks like I messed up there.

I meant that x is non-zero, either positive or negative.

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As you've used less than signs rather than less than or equal, your set of values is very small indeed.

Mathegocart

If I show you my proof you'll understand this, I think.

Bob

]]>But can you prove it always works?

I found a proof for 0<x<0, but it doesn't quite satisfy 'always'.

]]>1 + 1= 2 1 - 1=0 1 * 1 is 1 and 1 divided by 1 is 1 take all four answers and add them together and you get four. Four is the first square, two times two equals four...... Now two plus two is four 2 - 2 is zero though 2 * 2 is 4 2 / 2 x 1... When adding a squared number 3 * 3you could do this for any number and it'll give you the square up the following number

Can you state this in clearer terms? It is... a little hard to comprehend.

]]>How interesting! But can you prove it always works?

Bob

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