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On page $10$, Rudin makes the following assertions

  1. If $0\leq t<1$ then $t^n\leq t$.

  2. If $0<y_1<y_2$, then $y_1^n<y_2^n$.

Now, I understand this is true, but I can't get the grasp of how we can prove this with the previous content in the book, there are some propositions about multiplication and inequalities, but they proved not useful.

Does anyone know how to prove this fact with the content available in the book up to page 10?

I think both of these could be proved using induction, but I don't think that's available yet, as Rudin hasn't used it up to this point.

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YoTengoUnLCD
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  • Sorry, forgot that $t$ had to be $\geq 0$. – YoTengoUnLCD Dec 22 '15 at 13:30
  • Alright, can you show that $0\le x\le y$ implies $0\le tx\le ty$? – justanotherhagman Dec 22 '15 at 13:32
  • Take a look at Proposition 1.18. – J.G Dec 22 '15 at 13:37
  • @justanotherhagman Yes, I thought I could use that to say that $t\leq 1\rightarrow t^2\leq t\rightarrow \cdots\rightarrow t^n\leq t^{n-1}\leq \cdots \leq t$, but I thought I'd need induction to fill in the "..." Details, and I'm not sure if induction is available yet, as Rudin hasn't used it so far. – YoTengoUnLCD Dec 22 '15 at 15:26
  • The very irritating thing about Rudin, is he doesn't specify what you can and can't use. The student is confronting for the very first time that she can't use her basic assumptions. But here's what I figured: Rudin is formulizing the Real numbers as a complete ordered field. That must be treated as new. Everything else, logic and arithmetic of indexing notation can be assumed. It can be assumed very glibly. So induction is fine (he's never going to define it). .. more to come .... – fleablood Dec 23 '15 at 23:54
  • On page 6 he very quickly states "one usually writes" $x^2$ for $x\cdot x$. And that's it. That is the entirity of exponents to integer powers. Then in excercise 6. He wants you to expand the definition to rational powers and then to real powers and always show the definitions are well defined and $b^{x+y} = b^xb^y$. But he never showed that for integer powers In fact $b^n; n \le 1$ was never defined. After several readings I determined that $b^n$ is just notation and the n is simple counting. From that Rudin assumes $b^nb^m=b^{n+m}$ is simply a matter of counting and writing. – fleablood Dec 24 '15 at 00:01
  • Thanks for your comments! I agree with your vision: For proving theorem 1.20 it seems he uses the well-ordering principle (see: http://math.stackexchange.com/questions/476689/baby-rudin-theorem-1-20-b-proof) or induction, but that was not really clear to me. According to your comment on $x^2 = x\cdot x$, I think that's not really refering to exponentiation per se, I think it's like when we have a group in which we denote the law of composition by $+$, we write $2a$ for $a+a$, not meaning the product of $2$ with $a$, because there may not even be a $2$ in our group. – YoTengoUnLCD Dec 24 '15 at 01:49
  • @fleablood This book is making me crazy: right after proving that every positive number has an $n$-th root, he starts 'taking square roots' on both sides of inequalities (proving the triangle inequality for complex numbers, for example) with no justification as to why that's ok! – YoTengoUnLCD Dec 24 '15 at 06:18
  • I hated Rudin when I first read it but I really, really like it now. You have to read and read it again Taking the square root of both sides is okay. For b >0 and c > 0 then $b^{1/2} < c^{1/2} <=> b = {b^{1/2}}^2 < b^{1/2}c^{1/2} < c^{1/2}c^{1/2} = c$. – fleablood Dec 24 '15 at 06:42
  • That was my point about $x^2 = x\cdot x$ The 2 is simply an index and therefore we don't have to prove $b^{n+m}=b^nb^m$ as that's a consequence of writing. But $b^{m/n}$ is a definition and relies on properties of the real. It's bit subtle. And glib. – fleablood Dec 24 '15 at 06:48

1 Answers1

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Prop 1.18 b

If $x > 0$ and $y < z$ then $xy < xz$

$$0 \le t < 1\text{ so }0 = 0\cdot t \le t\cdot t = t^2 < 1\cdot t = t$$

For any $\,t^{n-1} < t$, $t^n = t^{n-1}t < t^2 < t$. So the result follows inductively.

2) is similar. $y_1 < y_2$ so $y_1^2 < y_1y_2 < y_2^2$ and the rest follows inductively.

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