Irrational Number Approximation Error Explained

[XJellieBX]

Question:
Using the fact that \sqrt{2} is irrational, we can actually come up with some interesting facts about other numbers. Consider the number t=1/\sqrt{2}, which is also irrational. Let a and b be positive integers, and a<b. We will prove that any rational approximation a/b of t will differ by at least 1/(4b^2), that is, for all integers 0<a<b.
|(1/\sqrt{2})-(a/b)|\geq1/(4b^2)
Notice that a<b because (1/\sqrt{2})<1.
(a) We prove the statement by contradiction. Assume the conclusion is false. Show it must follow that
|b-a\sqrt{2}|<\sqrt{2}/4b.
(b) Show that, as a result,
|b^2-2a^2|<(\sqrt{2}/4)+(a/2b).

The Attempt:
(a) |(1/\sqrt{2})-(a/b)<1/(4b^2)
Combining the fractions on the left side, we get
|(b-a\sqrt{2})/b\sqrt{2}|<1/(4b^2)
Multiplying the denominator of the left side to both sides gives
|b-a\sqrt{2}| < \sqrt{2}/4b

(b) I started part b by squaring both sides of the inequality I found in part a.
|b-a\sqrt{2}|^2 < (\sqrt{2}/4b)^2
|b^2-2(a\sqrt{2})+2a^2| < 2/(16b^2)

And then I'm not sure about where to go from here to get the result indicated in part b. Any tips?

 

[Dick]

Multiply both sides of the result of a) by b+a*sqrt(2).

 

[XJellieBX]

thanks =) it worked out

However, I have an alternate part to this question.

For any integers 0<a<b, prove that |b^2-2a^2|>1

My Attempt:
I used the axioms of ordering to find that a^2<b^2.
Rearranging the inequality gave me b^2-a^2>0.
Subtracting a^2 from both sides, b^2-2a^2>-a^2

I'm not sure where to go from here or if this is the right way to go. Any hints?

 

[HallsofIvy]

b^2- 2a^2\ge -a^2 means that b^2- 2a^2|\ge a^2 doesn't it? And how small can a² be if a is an integer?

 

[Dick]

b^2- 2a^2\ge -a^2 means that b^2- 2a^2|\ge a^2 doesn't it? And how small can a² be if a is an integer?

No, I don't think that follows. It's even simpler than that. |b^2-2a^2| is also a nonnegative integer. If it's not greater than or equal to one, then it's zero.