Let x and y be real numbers. Then x=y if and only if abs(x-y)< epsilon for every epsilon>0
LOL, well I suppose one could prove it somehow. It seems to belabor the obvious though.
I honestly have no clue how to even start this proof. Other than assume x=y and then assume that abs(x-y)<epsilon
well, draw two points on a number line maybe a distance epsilon apart then let epsilon go to zero.
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
I'm still not seeing it. I have not used epsilon before today
I know that if x=y, for any epsilon>0, abs(x-y)=0<epsilon
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.
Okay, so since epsilon has to be the smallest number passible x has to equal y.
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.