Question

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

Answer 1

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

Answer 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

Answer 3

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

Answer 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

Answer 5

I'm still not seeing it. I have not used epsilon before today

Answer 6

I know that if x=y, for any epsilon>0, abs(x-y)=0<epsilon

Answer 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.

Answer 8

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

Answer 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.

Answer 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.

Answer 11

You say that as epsilon ---> 0, |x-y|--->0

Answer 12

Thank you

Answer 13

The key word here is ** EVERY ** epsilon, including one vanishingly small .

Answer 14

You are welcome.