1. ## Rational Numbers

Hey fellas. I'm stuck on the following question. I'm quite ashamed to say I've spent about an hour and half on it, and my paper is virtually blank.

Question:

If $z$ is a real number, then $z^n$ is defined recursively for non-negative integers n: $z^0 = 1$, and for $n \ge 0, z^{n+1}=z . z^n$. Also $z^{-n} = \frac{1}{z^n}$, for any natural number n.

If $z$ is a positive real number, and $b$ a non-zero integer, then we define the positive $b^{th}$ root $y=z^{\frac{1}{b}}$ to be the positive real number $y$ such that $y^b = z$.

1) Is it possible for a positive real number to have two different $b^{th}$ roots?

For integers a,b, where b doesn't equal 0, and a real number $z>0$, we define $z^{\frac{a}{b}} = (z^a)^{\frac{1}{b}}$

2) If a,b,c,d are integers and b,d do not equal 0, $ad=bc$, and $z>0$, use these definitions to prove that $z^{\frac{a}{b}} = z^{\frac{c}{d}}$

2. Originally Posted by WWTL@WHL
Question:

If $z$ is a real number, then $z^n$ is defined recursively for non-negative integers n: $z^0 = 1$, and for $n \ge 0, z^{n+1}=z . z^n$. Also $z^{-n} = \frac{1}{z^n}$, for any natural number n.

If $z$ is a positive real number, and $b$ a non-zero integer, then we define the positive $b^{th}$ root $y=z^{\frac{1}{b}}$ to be the positive real number $y$ such that $y^b = z$.

1) Is it possible for a positive real number to have two different $b^{th}$ roots? Briefly justify your answer.
But of course! Let z = 4 and let's find the "2th root." Then $4^{1/2} = 2, -2$. (The proof of this requires the next definition in the series, so technically I suppose you can't say this yet. That will be up to your instructor.)

Originally Posted by WWTL@WHL
For integers a,b, where b doesn't equal 0, and a real number $z>0$, we define $z^{\frac{a}{b}} = (z^a)^{\frac{1}{b}}$

2) If a,b,c,d are integers and b,d do not equal 0, $ab=cd$, and $z>0$, use these definitions to prove that $z^{\frac{a}{b}} = z^{\frac{c}{d}}$
Hmmmm... Not true this time. Consider z = 12, a = 2, b = 3, c = 1, d = 6. Note that ab = cd = 6. But
$12^{2/3} \neq 12^{1/6}$.

Did you possibly mean that given "ad = bc" that $z^{a/b} = z^{c/d}$? This one is almost true. We have to add one more definition:
Given a positive number z and a rational number q, then the "principle value" of $z^q$ is also positive. (If we speak of principle values, then the answer to part a) is that we can only have one principle value to $z^{1/b}$.)

So if we are speaking of principle values then we have $\frac{a}{b} = \frac{c}{d}$ and have to prove that $p = q$ implies that $z^p = z^q$, which is a simple proof that I leave to you.

-Dan

3. I disagree with Dan on this one. I do agree that you must follow your instructor/text.
But I think that $4^{1/2} = 2$ period; while $- (4^{1/2}) = -2$.
Just like $\sqrt 9 = 3\,\& \,\sqrt 9 \ne - 3$.
From reading the statement in the question, it seems clear to me that is what is meant when it says “the positive…”.

If $b > 0$ then it is easy to show that $\begin{array}{l} f:\Re \mapsto (0,\infty ) \\ f(x) = b^x \\ \end{array}$ is an injection.

4. ARGH! Sorry, that's meant to be ad=bc on the 2nd question. Really sorry about that. Still not much luck though. I'll just look through your posts thoroughly now, and I'll add my attempts. Thanks.

5. Originally Posted by WWTL@WHL
that's meant to be ad=bc on the 2nd question.
Consider, $ad = bc \Rightarrow \quad z^{ad} = z^{cb} \Rightarrow \quad \left( {z^{ad} } \right)^{\frac{1}{{bd}}} = \left( {z^{cb} } \right)^{\frac{1}{{bd}}} \Rightarrow \quad z^{\frac{a}{b}} = z^{\frac{c}{d}}$.

6. Right. Thanks for the help - here is my attempt. I realise my answers are not very formal or perhaps even mathematically sinful, so if you don't mind, please point out anything that's 'wrong'. Hopefully the jist of the argument is right.

__________________________________________________ ______________

My attempts:

1) No. Let $f: R \to (0, \infty)$ represent a positive real number z. Then $z = f(x) = b^x$ is the function that transforms a real number z to its $x^{th}$ root.
So we need to show that this function is an injection for b>0, since that would mean that each real number z has at most one $x^{th}$ root. In order to prove $f$ is an injection, we have to show there cannot be distinct numbers $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$

$f(x_1) = b^{x_1}$ and $f(x_2) = b^{x_2}$
$b^{x_1}= b^{x_2}$ (take log of both sides)
$x_1 ln(b) = x_2 ln(b)$
$x_1 = x_2$

Hence f is an injection. Therefore there is at most one positive $x^th$ root for any positive real number z.

__________________________________________________ ____________

2) Dan said that p=q implies $z^p = z^q$ is a simple proof, but I can't even do it through 'truth tables' and I'm not very comfortable with the logical framework. Is there a general formula, of sorts, for writing proofs for 'implies' statements?

What you've written, Plato, makes sense to me. I just have no idea how to prove it.

7. Originally Posted by WWTL@WHL
What you've written, Plato, makes sense to me.
I just have no idea how to prove it.
I don’t understand that at all. I proved it! What more do you require?

As for proving that if $b > 0$ then the function $f(x) = b^x$ being an injection here is the proof that I like.
$\begin{array}{l}

8. Originally Posted by Plato
I don’t understand that at all. I proved it! What more do you require?
Yeah, thanks, I guess you have. But don't you need to prove, for example, a=b implies $z^a = z^b$ ?

Is there a way I could go about showing that? I know it's obvious, but still...

And thanks for your clever method on the injection, I would've never seen that 'trick'.

9. Do you agree that $z^5 = z^5$ and that $z^x = z^x$?

Well: $a = b\quad \Rightarrow \quad z^a = z^a = z^b$.

10. Yes, ok, it really isn't needed. It's just that in the past I've been penalised for not being thorough enough.

Thanks so much for the help topsquark and Plato - very kind of you.