How do you prove that?

Question

Prove that if y is not equal to 0 and |y-x| < min [(|x|/2),(ε|x|^2,2)], then y is not eqaul to 0 and |(1/y)-(1/x)| < ε

Answer 1

I have assumed the term &quot;min [(|x|/2),(ε|x|^2,2)]&quot; is meant to say &quot;min [(|x|/2),(ε|x|^2/2)]&quot;.

Right bear with me for this first part where I show you a formula for the minimum of two numbers, I learnt this result in analysis and it is obviously true once you know it, but its an odd way of writing it.

min (a, b) = (|a+b| - |b-a|)/2.

When a is smaller this expression becomes

((a+b) - (b-a))/2 = 2a/2 = a

When b is smaller this expression becomes

((a+b) - (a-b))/2 = 2b/2 = b

Now using this formula for min (a=|x|/2,b=ε|x|^2/2):

|y-x| &lt;min [(|x|/2),(ε|x|^2/2)]

        &lt; (| |x|/2+ ε|x|^2/2 | - | ε|x|^2/2 -|x|/2 |)/2

Dividing both sides by |x|:

|y-x|/|x| &lt; (|1+ ε|x|| - | ε|x| - 1|)/4

|y/x- 1| &lt; (|1+ ε|x|| - | ε|x| - 1|)/4

Dividing both sides by |y|:

|1/x- 1/y| = |1/y - 1/x | &lt; (|1+ ε|x|| - | ε|x| - 1|)/4|y|

By the triangle inequality &quot;|a| - |b| &lt; |a+b|&quot; this becomes:

|1/x- 1/y| &lt; |1+ ε|x| + ε|x| - 1|/4|y|

              &lt; |     ε|x| +  ÃŽÂµ|x|     |/4|y|

              &lt; 2| ε|x| | / 4|y|

              &lt; ε|x|/2|y| (*Need this to be less than ε)

Now we know that |y-x| &lt; |x|/2  by the inital assumption. The triangle inequality can be rearranged as follows:

|x|-|y| = |-x| - |y| &lt; |- x + y|

                         &lt;|x|/2

So |x|/2 &lt; |y|

So |x|/2|y| &lt; 1

So ε|x|/2|y| &lt; ε

as required*.