Spivak's Calculus: The Basic Properties of Numbers

[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

 

[dmoravec]

well that's 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.

 

[Tasaio]

Thanks for the reply!

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?

 

[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)

 

[Tasaio]

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

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!

 

[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?

 

[Tasaio]

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?

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

We were asked to show:

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?

 

[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.)

 

[Tasaio]

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

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

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?

 

[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.)

 

[Tasaio]

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.)

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.

 

[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?

 

[nocturnal]

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

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

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

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

 

[matt grime]

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.

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.

 

[Tasaio]

Try breaking into the following 3 cases
...
2) x &lt; 0 \leq y
...

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

Case: x &lt; 0 \leq y

Then
x - y &lt; -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?

 

[matt grime]

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

 

[Tasaio]

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

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

 

[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.

I'll try Proof by Contradiction.

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?

 

[BSMSMSTMSPHD]

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

 

[nocturnal]

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

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.

 

[Tasaio]

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.

Hi,

Here's my attempt so far...

Prove: x^n = y^n and n even --> x = y OR x = -y

My attempt

Suppose x^n = y^n and n even for n = 1, 2, 3,...

Proof by Contradiction: Suppose NOT(x = y OR x = -y)
Then x != y and x != -y (by DeMorgan Theorem)

Cases:
0 < x < y
0 = x < y
x < 0 = y
x < 0 < y
x < y < 0

Which reduces to:

0 <= x < y (1)
x < 0 <= y (2)
x < y <= 0 (3)

And:
0 < y < x
y = 0 < x
y < 0 = x
y < 0 < x
y < x < 0

Which reduces to:
0 <= y < x (4)
y < 0 <= x (5)
y < x <= 0 (6)

Case (1): 0 < x < y

By 6(a), x^n < y^n for n = 1, 2, 3,...

But x^n = y^n : CONTRADICTION!

Case(2): x < 0 <= y

Case: y = 0

Then x < 0 = y
Then - x > 0 = -y = y

So -x > y >= 0

By 6(a), (-x)^n > y^n for n = 1, 2, 3,...
(-1)^n*x^n > y^n for n = 1, 2, 3...

But since n even,
Then (-1)^n = 1

so x^n > y^n

But x^n = y^n: CONTRADICTION!

Case: y > 0

Then x < 0 < y
-x > 0 > -y

Now x^n = (-x)^n since n even
(-x)^n > 0 since (-x) > 0

...

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