Irrational Number Approximation Error Explained

[XJellieBX]

Question:
Using the fact that [tex]\sqrt{2}[/tex] is irrational, we can actually come up with some interesting facts about other numbers. Consider the number t=1/[tex]\sqrt{2}[/tex], 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/[tex]\sqrt{2}[/tex])-(a/b)|[tex]\geq[/tex]1/(4b^2)
Notice that a<b because (1/[tex]\sqrt{2}[/tex])<1.
(a) We prove the statement by contradiction. Assume the conclusion is false. Show it must follow that
|b-a[tex]\sqrt{2}[/tex]|<[tex]\sqrt{2}[/tex]/4b.
(b) Show that, as a result,
|b^2-2a^2|<([tex]\sqrt{2}[/tex]/4)+(a/2b).

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

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

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

 

[Dick]

[itex]b^2- 2a^2\ge -a^2[/itex] means that [itex]b^2- 2a^2|\ge a^2[/itex] 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.