# Math Proof

Let x and y be real numbers. Then x=y if and only if abs(x-y)< epsilon for every epsilon>0

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1. LOL, well I suppose one could prove it somehow. It seems to belabor the obvious though.

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posted by Damon
2. I honestly have no clue how to even start this proof. Other than assume x=y and then assume that abs(x-y)<epsilon

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posted by Ashley
3. well, draw two points on a number line maybe a distance epsilon apart then let epsilon go to zero.

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posted by Damon
4. on second thought draw x as a point on a number line, then y points epsilon to the right and to the left. Then let epsilon ---> 0

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posted by Damon
5. I'm still not seeing it. I have not used epsilon before today

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posted by Ashley
6. I know that if x=y, for any epsilon>0, abs(x-y)=0<epsilon

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posted by Ashley
7. well epsilon is anything you want it to be, but usually you make it a small number and let it approach zero.
If x = 5
and y = 7
then |x-y| = epsilon
now we would call epsilon = 2
BUT
x is only equal to y if |x-y|< ANY epsilon bigger than zero, including e = .0000000001
so when could |x-y| be that small?
Only if x is awfully close to y
but epsilon could be much smaller than .0000000000001, in fact our equality is only true if epsilon is so small we can not tell it is there at all.

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posted by Damon
8. Okay, so since epsilon has to be the smallest number passible x has to equal y.

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posted by Ashley
9. yes
it has to be for any old epsilon, including one that is vanishingly small so that the difference between x and y is so tiny it does not exist.

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posted by Damon
10. How would I write that though, since epsilon has to be greater than 0 and abs(x-y) has to be less than epsilon. would i suppose that x doesn't equal y, then epsilon <0.. then epsilon is less than abs(x-y)/2, then abs(x-y)<epsilon<abs(x-y)/2. which contradicts abs(x-y)< epsilon for all epsilon<0.

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posted by Ashley
11. You say that as epsilon ---> 0, |x-y|--->0

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posted by Damon
12. Thank you

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posted by Ashley
13. The key word here is ** EVERY ** epsilon, including one vanishingly small .

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posted by Damon
14. You are welcome.

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posted by Damon

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