let xy be real numbers such that x is less than or equal to

let x,y be real numbers such that x is less than or equal to y+ epsilon for every epsilon greater than 0. then x is less than or equal to y

Solution

Assume x>y

So choose epsilon,e=(x-y)/2, this is greater than 0 because x>y by our assumption

y+e=y+(x-y)/2=(x+y)/2

But since ,x>y ,(x+y)/2>(x+x)/2=x

ie ,y+e>x

which contradicts what we are given in the problem. HEnce contradiction.

Hence, x is less than equal to y

let x,y be real numbers such that x is less than or equal to y+ epsilon for every epsilon greater than 0. then x is less than or equal to ySolutionAssume x>y

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