- #1
gazzo
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Hey everyone!
This question came up in my analysis assignment, we're studying continuity and differentiability at the moment so I'm unsure of my answer! It seems too short :(
For a), it seems to follow immediately,
If [itex]mq = np[/itex] then [itex]m = \frac{np}{q}[/itex] and so we have
[itex](x^{\frac{1}{n}})^m = (x^{\frac{1}{n}})^{\frac{np}{q}} = ((x^{\frac{1}{n}})^n)^{\frac{p}{q}} = (x^{\frac{p}{q}}) = x^{\frac{1}{n}} = (x^{\frac{1}{q}})^p[/itex] as required.
And b) is similar (there are too many exponents to latexify before my stuff gets stolen upstairs in the study floors).
I went about it proving if [itex]a \in \mathbb{R}[/itex] and [itex]m,n \in \mathbb{N}[/itex] then [itex]a^{m+n} = {a^m}{a^n}[/itex] and [itex](a^m)^n = a^{mn}[/itex] by induction. (Can you use induction over m and n by doing it twice fixing m the first time and inducting over n and the second time fixing n?).
Or it's probably better to first prove, if [itex]m\in \mathbb{Z}, n \in \mathbb{N}, x \in (0,\infty)[/itex] then [itex]x^{\frac{m}{n}} = (x^m)^{\frac{1}{n}}[/itex].
If [itex]x \in (0,\infty)[/itex] and [itex]m,n \in \mathbb{Z}[/itex] then [itex](x^m)^n = x^{mn} = (x^n)^m[/itex]. Now let [itex] y:=x^{\frac{m}{n}} = (x^{\frac{1}{n}})^m > 0[/itex]. So [itex]y^n = ((x^{\frac{1}{n}})^m)^n = ((x^{\frac{1}{n}})^n)^m = x^m[/itex]. Therefore it follows that [itex]y = (x^m)^{\frac{1}{n}}[/itex].
If we proved it for the naturals, how can we prove it for the rationals? It seems to follow naturally.
These analysis proofs seem so empty, although subtle. I have huge trouble sponging it all together into a valid proof.
Thanks a lot!
-Gareth
1. Let [itex]x \in (0, \infty)[/itex]
a) Prove that [itex](x^{\frac{1}{n}})^m = (x^{\frac{1}{q}})^p [/itex] for all [itex]m,p \in \mathbb{Z} [/itex] such that [itex]mq = np[/itex]
Because of the previous we can define [itex]x^r \in \mathbb{R}[/itex] for all [itex]r \in \mathbb{Q}[/itex] by [itex]x^r = (x^{\frac{1}{n}})^m[/itex] if [itex]r = \frac{m}{n}[/itex] with [itex]m \in \mathbb{Z}[/itex] and [iitex]n \in \mathbb{N}[/itex]
b) Prove that [itex]x^{r+s} = {x^r}{x^s}[/itex] and [itex](x^r)^s = x^{rs}[/itex] for all [itex]r,s \in \mathbb{Q}[/itex]
This question came up in my analysis assignment, we're studying continuity and differentiability at the moment so I'm unsure of my answer! It seems too short :(
For a), it seems to follow immediately,
If [itex]mq = np[/itex] then [itex]m = \frac{np}{q}[/itex] and so we have
[itex](x^{\frac{1}{n}})^m = (x^{\frac{1}{n}})^{\frac{np}{q}} = ((x^{\frac{1}{n}})^n)^{\frac{p}{q}} = (x^{\frac{p}{q}}) = x^{\frac{1}{n}} = (x^{\frac{1}{q}})^p[/itex] as required.
And b) is similar (there are too many exponents to latexify before my stuff gets stolen upstairs in the study floors).
I went about it proving if [itex]a \in \mathbb{R}[/itex] and [itex]m,n \in \mathbb{N}[/itex] then [itex]a^{m+n} = {a^m}{a^n}[/itex] and [itex](a^m)^n = a^{mn}[/itex] by induction. (Can you use induction over m and n by doing it twice fixing m the first time and inducting over n and the second time fixing n?).
Or it's probably better to first prove, if [itex]m\in \mathbb{Z}, n \in \mathbb{N}, x \in (0,\infty)[/itex] then [itex]x^{\frac{m}{n}} = (x^m)^{\frac{1}{n}}[/itex].
If [itex]x \in (0,\infty)[/itex] and [itex]m,n \in \mathbb{Z}[/itex] then [itex](x^m)^n = x^{mn} = (x^n)^m[/itex]. Now let [itex] y:=x^{\frac{m}{n}} = (x^{\frac{1}{n}})^m > 0[/itex]. So [itex]y^n = ((x^{\frac{1}{n}})^m)^n = ((x^{\frac{1}{n}})^n)^m = x^m[/itex]. Therefore it follows that [itex]y = (x^m)^{\frac{1}{n}}[/itex].
If we proved it for the naturals, how can we prove it for the rationals? It seems to follow naturally.
These analysis proofs seem so empty, although subtle. I have huge trouble sponging it all together into a valid proof.
Thanks a lot!
-Gareth
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