# Spivak's Calculus: The Basic Properties of Numbers

1. Aug 29, 2006

### Tasaio

Hi there,

I'm using Spivak's Calculus. Some of these questions are giving me trouble...

6.(a)
Prove that if 0 <= x < y, then x^n < y^n, n = 1, 2, 3, ...

Suppose 0 <= x < y.

...

Since this is Chapter 1, I shouldn't be using induction. ;-)

I'll try anyway.

Let S be the set of positive integers n for which:
If 0 <= x < y then x^n < y^n

Case: n = 1

Since x < y, then x^(1) = x < y = y^(1)

So 1 in S.

Assume k in S.
Then if 0 <= x < y, then x^k < y^k.

Case: n = k + 1.
Suppose 0 <= x < y.
...

Here, I get stuck.

Does anyone have an idea about what to do next? Also, is there possibly a way to do with question without using induction?

Thanks in advance for any assistance,

Tasaio

2. Aug 29, 2006

### dmoravec

well thats a good approach. It would go like this.

Let P(k) be the statement: 0<=x^k<y^k

P(1): 0<=x<y is given

Assume P(n)... ie: 0<=x^n < y^n
Show P(n+1) is true based on P(n)
ie: show 0<=x^(n+1) < y^(n+1) assuming P(n) is true.

Thats a quick way to do it with induction.

3. Aug 29, 2006

### Tasaio

As you wrote, I need to show that P(k) --> P(k + 1).

Assume P(k).

Then x < y --> x^k < y^k.

Now show: P(k + 1).

Suppose x < y.
.
. <-- Proof that x^(k + 1) < y^(k + 1)
.

Any tips on how I can prove this?

4. Aug 29, 2006

### dmoravec

look at it this way.

since x^k<y^k and x<y

x^k*x < x^k*y<y^k*y

which is equivalent to
x^(k+1)<y^(k+1)

5. Aug 29, 2006

### Tasaio

Wow -- I can't believe I missed that. :rofl:

Here's one I'm trying to figure out, that's much more difficult.
Prove:

If 0 < a < b,
Then

a < sqrt(ab) < (a + b) / 2 < b

---
My solution
It's easy to show:

a = sqrt(a*a), since a > 0
sqrt(a*a) < sqrt(a*b) since a < b

So a < sqrt(ab)

b = 2b/2 = (b + b) / 2 > (a + b) / 2, since a < b

So (a + b) /2 < b

Now we come to the crux of the problem:

Prove
sqrt(ab) < (a + b) / 2

Work backwards -- multiply both sides by 2
2*sqrt(ab) < a + b

Square both sides
4(ab) < (a + b)^2
4ab < a^2 + 2ab + b^2
2ab < a^2 + b^2
ab < (a^2 + b^2) / 2

Clearly the expression on the right is the "average" of the squares of a and b.

I've tried and tried, but can't figure out the next step. Any hints you can give me?

Thanks!

6. Aug 29, 2006

### matt grime

since a,b etc are all positive, you let x^2=a, y^2=b, for positive x and y. Then what are you asked to prove about x and y?

7. Aug 29, 2006

### Tasaio

Let x^2 = a
Let y^2 = b

sqrt(ab) < (a + b) / 2

Subbing in x and y:
sqrt(x^2 * y^2) < (x^2 + y^2) / 2
sqrt((xy)^2) < (x^2 + y^2) / 2
xy < (x^2 + y^2) / 2
2xy < (x^2 + y^2)

This is just like the result I obtained using a and b. How can I use this?

8. Aug 29, 2006

### matt grime

you mean you don't recognize x^2-2xy+y^2? (prepare for another of those 'how didn't I see that' moments.)

9. Aug 29, 2006

### Tasaio

ARGH! You're right, how didn't I see that?! :rofl: :rofl: :rofl:

2ab < a^2 + b^2

0 < a^2 - 2ab + b^2
0 < (a - b) ^2

0 < (a - b)(a - b)

This is true iff a != b

Since a < b, it is true.

Thus we just need to work backwards from here.

Proof

Since a < b
Then 0 < b - a
So
0 < (b - a)^2
0 < (a - b)^2
0 < a^2 - 2ab + b^2
2ab < a^2 + b^2
4ab < a^2 + 2ab + b^2
4ab < (a + b)^2
2sqrt(ab) < a + b
sqrt(ab) < (a + b) / 2

QED

Thanks so much for your help with that question -- it had be going for at least an hour!

This next one seems trickier:

Prove that if x < y and n is odd, then x^n < y^n.

My solution

Suppose x < y.
Suppose n is odd.

We want: x^n < y^n
0 < y^n - x^n
0 < (y - x)(y^n-1 + y^n-2*x + ... + yx^n-2 + x^n-1)

Since n is odd, then
n = 2k + 1, where k is an integer.

But n >= 1, so k >= 0

Let's try induction, again:

Proof by Induction
Let S be the set of positive integers k for which:
x < y and k >= 0 --> x^(2k + 1) < y^(2k + 1)

Suppose x < y.

Case: k = 0

Then 2k + 1 = 2(0) + 1 = 1
So x^(2k + 1) = x^1 = x < y = y^1 = y^(2k + 1)

So 1 in S.

Asssume t in S.

Then x^(2t + 1) < y^(2t + 1)

Case: k = t + 1
Then
x^(2t + 1) = x^(2t) * x < y^(2t) * x < y^(2t) * y = y^(2t + 1)

Thus x^(2t + 1) < y^(2t + 1)

So t in S --> (t + 1) in S.

Thus S = N.

QED

Does that look right?

10. Aug 29, 2006

### matt grime

Why not just use the first result in the thread? If n is odd x^n is the same as |x|^n (|x| means the abs value of x, i.e. forget the sign.)

11. Aug 29, 2006

### Tasaio

The first result I can think of was:
0 <= x < y --> x^n < y^n, n= 1, 2, 3, ...

I can't use this, since the hypothesis has the additional requirement that 0<=x.

I'm trying to prove x < y --> x^n < y^n for n odd. (There is no requirement that 0<= x)

---

Are you sure that x^n = |x|^n for n odd?

What about x = -1, n = 3?

Then
(-1)^3 = -1 < 1 = |-1|^3

They are not equal.

12. Aug 29, 2006

### Tasaio

6(c)

Prove that if x^n = y^n and n is odd, then x = y.

Suppose x^n = y^n.
Suppose n is odd.

Then there exists k in Z, such that n = 2k+ 1.

But n >= 1, so k >= 0.

We have x^(2k + 1) = y^(2k + 1)

We need to prove x = y for all k >= 0. Let's use Proof By Induction.

Proof by Induction

Case: k = 0
Suppose x^(2k + 1) = y^(2k + 1).
Then x^(2(0) + 1) = y^(2(0) + 1)
x^(1) = y^(1)
x = y

Thus 1 in S.

Assume t in S.

Then x^(2t + 1) = y^(2t + 1) --> x = y

Case: k = t + 1

Suppose x^(2(t + 1) + 1) = y^(2(t + 1) + 1)

Then
x^(2t + 2 + 1) = y^(2t + 2 + 1)
x^(2t + 1 + 2) = y^(2t + 1 + 2)
x^2(x^(2t + 1)) = y^2(y^(2t + 1))

Where should I go from here? Any advice?

13. Aug 30, 2006

### nocturnal

Try breaking into the following 3 cases
1) $$0 \leq x < y$$, this one should be obvious from the first result of the thread
2) $$x < 0 \leq y$$
3) $$x < y \leq 0$$, hint: consider (-x) and (-y) and the first result of the thread.

Try a proof by contradiction. First assume x<y and use the above result. Then do the same assuming y<x.

14. Aug 30, 2006

### matt grime

Of course that is wrong: there is a minus sign missing.

THe point is you dont' need to do anything else other than look at some cases and invoke the first result you proved. If 0<=x<y then the first argument applies, if x<0<=y then there is nothing to prove (x^n is all ways negative for odd n), and if x<y<0 then you can again use the first argument since 0<-y<-x.

Just because something isn't exactly in the form you're used to doesn't mean you can't put it in the form you're used to.

15. Aug 30, 2006

### Tasaio

I solved the other 2 cases, but number 2 is giving me trouble.

Case: $$x < 0 \leq y$$

Then
$$x - y < -y \leq 0$$

Since x < 0 and $$-y \leq 0$$
Then $$x + (-y) \leq x$$
So
$$x-y \leq x$$

Where should I go from here?

16. Aug 30, 2006

### matt grime

The proof of case 2 was given in my post: a negative n umber to an odd power is a negative number.

17. Aug 30, 2006

### Tasaio

Yes, I solved it by breaking it into 2 cases: y > 0 and y = 0.

18. Aug 31, 2006

### Tasaio

6. (d)
Prove that if x^n = y^n and n is even, then x = y or x = -y

My attempt
Suppose x^n = y^n and n is even.
Then x^n = y^n.
And n is even.

Proof by Contradiction: Suppose !(x = y or x = -y)
Then (x !=y and x != -y) by DeMorgan's Theorem

...

I'm stuck. Where do I go from here?

19. Aug 31, 2006

### BSMSMSTMSPHD

Can't you just take the nth root of both sides and get +/- x = +/- y?

20. Aug 31, 2006

### nocturnal

That would be nice, the problem is that one (referring to anyone working through the text) has not yet proven that nth roots exist.

What this problem shows is that for even n, if nth roots exist for a positive number then they have the same absolute value. Likewise, the previous problem showed that for odd n, if nth roots exist for some number (positive or negative) then they are unique.

Tasaio: The proof is similar to the previous problems. Try breaking the problem into cases and make use of work you have already done, also note that (x)^n = (-x)^n for even n.

Last edited: Aug 31, 2006