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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