I have edited and reframed the final question. (I have left the first two questions, but I consider those answered.)

The problem is to prove ${x}^{n} < {y}^{n}$ given that ${x} < {y}$ and $n$ is odd. Now, I understand that Spivak relies heavily on what has been previously proved, and also what has been defined in the text up to now. Concentrating specifically on the case where $ x < y \leq 0$, Spivak simply states

then $ 0\leq -y < -x$

Question 1
Is this simply then that $$ y - x > 0 = -x > -y \geq 0 $$
Now, if I may, I would like to clarify the following assertions.

So, $ (-y)^n < (-x)^n $ by part(a): which means $ -y^n < -x^n $ (since n is odd) and hence $ x^n < y^n $

Question 2.
The assertion that 'n is odd'. To the seasoned mathematicians as are on this board, this must indeed seem obvious. But, maybe I should ask it this - somewhat clumsy way. What allows Spivak to state so categorically that

... n is odd

Final Question (edited and reframed),

If $x < y \leq 0$ then $0 \leq -y < -x$ so $(-y)^{n} < (-x)^{n} <$ by part(a).

Up to now, perfect. It is simply restating what was covered earlier.

Now two points of clarification, hoping this is not nitpicking.

this means that $-y^{n} < -x^{n} <$ (since $n$ is odd)

Why does Spivak restate what he simply said earlier (minus the parentheses)? There is clearly a subtle point I am missing here.

hence $x^{n} < y^{n} $

which of course is the asked for proof. But, this last assertion of the proof. Is this merely restating in essence what has been said earlier ie if $ a > b $ then $-b > -a$

I know these are a lot of questions, but I think it will go a long way to help many at this early stage of this journey.

  • $\begingroup$ If $a<b$, then $-a>-b$. Proof: $a<b\implies a+(-a)+(-b)<b+(-a)+(-b)\implies-b<-a$. This proof uses the fact that $a<b\implies a+c<b+c$, which I believe is proven in the first chapter. $\endgroup$
    – Joe
    Sep 11 '21 at 21:27
  • 1
    $\begingroup$ Your displayed line needs editing. I totally do not get your question 2. What is strange? To enunciate a crucial hypothesis? $\endgroup$ Sep 11 '21 at 21:33
  • $\begingroup$ If n is even it does not hold. Start with $x\gt 0$ and $y=-x-1$, Then for even n, $y^n \gt x^n$. $\endgroup$ Sep 11 '21 at 21:45
  • $\begingroup$ If $f(x) = x^n$ then $f'(x) = nx^{n-1}.$ This means that when $n$ is odd and $x \neq 0$, then $f'(x) > 0.$ $\endgroup$ Sep 11 '21 at 22:23
  • $\begingroup$ @Joe Thank you. That is indeed in the first chapter and puts it into perspective. $\endgroup$ Sep 12 '21 at 2:55

Question 1

Spivak simply multiplies the inequality by $-1$. When you do that, the sign on the inequality flips (you can look at Joe's excellent comment to see why). If you'd like, you can consider the compound inequality in parts:

$$x<y \implies -x>-y$$

$$y \leq 0 \implies -y \geq 0$$

Question 2

I'm not sure what you mean. The claim states that $n$ is odd; that's why Spivak mentions it. If you're wondering why $n$ has to be odd, it's because if you have $x<y$ but $|x|>|y|$ (for example, $-4<3$), an even $n$ will go against the claim ($16>9$, $256>81$, etc.).

Final Question

He isn't restating it: he first has $(-y)^n<(-x)^n$. Then, because $n$ is odd, we can write this as $-y^n<-x^n$ (i.e. odd powers of $-1$ are $-1$).

And yes; see my response to question 1. If $-y^n<-x^n$ then $x^n<y^n$.

  • $\begingroup$ Thank you for that. Just one clarification, and I hope this is not seen as being too pedantic. When you say he is not restating it ( The Final question answer) I want to understand why $(-y)^{n} < (-x)^{n}$ is different without the parentheses. Is this a fair analogy? $(-3)^{3} \neq -3^{3}$. $\endgroup$ Sep 24 '21 at 0:55
  • $\begingroup$ @user1115542 Don't forget that $n$ is odd. $2$ isn't odd. A better analogy would be $(-3)^3 = -3^3$. $\endgroup$
    – Kman3
    Sep 24 '21 at 0:59
  • $\begingroup$ Sorry about that ... I edited my comment before you had a chance to see that. $\endgroup$ Sep 24 '21 at 1:18

Spivak makes use of the following, but does not provide justification. Probably an oversight.

Lemma: For any real number $a$,

\begin{align} (-a)^n & = a^n \text{ if n is even}, \\ & = -a^n \text{ if n is odd}. \\ \end{align}

Note here that $-a^n$ is the number $-(a^n)$. It is not the same as $(-a)^n$.

Lemma Proof

Case 1: $n$ is even.

$n = 2k$, where $k$ is a natural number.

\begin{align} (-a)^n &= (-a)^{2k} \\ &= [(-a)^2]^k \\ &= (a^2)^k \\ &= a^{2k} \\ &= a^n. \end{align}

Case 2: $n$ is odd.

$$n = 2k + 1$$.

\begin{align} (-a)^n &= (-a)^{2k+1} \\ &= (-a)^{2k} \cdot (-a) \\ &= a^{2k} \cdot (-a) \text{ (from Case 1.)}\\ &= -(a^{2k} \cdot a) \\ &= -a^{2k + 1} \\ &= -a^n. \end{align}

This may be easier to follow after chapter 2, which introduces natural numbers and inductive arguments. You might benefit from returning to this question after reading chapter 2.

  • 1
    $\begingroup$ Aha… thank you for that. And, especially for pointing out that he provided no justification for his assertion, as up to this point, he would usually do so. I will come back to this. $\endgroup$ Sep 26 '21 at 16:31

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