The set of squares of rational numbers is inductive

[mrchris]

Homework Statement

The set of squares of rational numbers is inductive

Homework Equations

definition of an inductive set

The Attempt at a Solution

sorry i know this is probably very easy to most but I am just learning analysis. Okay, so we can see that 1 is in the set because it is rational and 1² exists. Am I correct in thinking that it doesn't matter whether or not 1² is rational? I mean obviously it is, but aren't we really just checking to make sure that S(1) is defined? Then for the second part, we can take S(x) to be x² whenever x=m/n, where m,n are integers and either m or n is odd. Again, aren't we basically proving that if S(x) exists, or (m²/n²) exists, then S(x+1), or [(m+n)²/n²] also exists. But then don't I also need to show that if x is rational, then so is x+1? Maybe i am reading way too much into this, but again I am new to these and I'm trying to understand exactly what I need to show.

 

[SP90]

No, you're pretty much correct. Let S be the set of the squares of rationals. To show it's inductive you need to show that 0 is in S (it is because 0 is 0^{2} and 0 is rational). Then you need to show that if x is in S, then x+1 is in S.

Then for two integers m,n x=\frac{m^{2}}{n^{2}} so x+1=\frac{m^{2} + n^{2}}{n^{2}}. The question then boils down to showing that this can be written in the form \frac{p^{2}}{q^{2}}, which isn't the same as \frac{(m+n)^{2}}{n^2}. They're rational by your definition of x, but the latter can't be assumed to be rational squared, unless you show it.

 

[SP90]

I'm fairly sure it's not inductive by the way.

Take x=\frac{1}{4}. Then x+1=\frac{5}{2^{2}}. This cannot be written in the form \frac{p^2}{q^2}.

Every time you multiply the fraction by some multiple \frac{5}{5} you get that the index of 5 increases to an even number on the top but then an odd number on the bottom. So you can never have an even index of 5 on the top and bottom. So the fraction is not part of the rationals squared.