# Spivak's Calculus Section I Trouble

1. May 15, 2010

### Char. Limit

So, I've been going through Spivak's Calculus, section I, and I'm having some trouble proving 1.1.iii. Also, I'd like to check 1.1.ii and ensure that I did 1.1.i right. So, first, I'm restricted to these nine properties:

a*(b*c)=(a*b)*c <P5, multiplication associative>
a*1=1*a=a <P6, multiplicative identity>
a*a-1=a-1*a=1 <P7, multiplicative inverse>
a*b=b*a <P8, multiplication commutative>
a*(b+c)=a*b+a*c <P9, multiplication distributive>

P7 has the condition that a cannot be 0.

So, the first question asks me to prove that if a*x=a, then x=1. So, I start with a*x=a.

a*x=a
a-1*(a*x)=a-1*a <Multiply by a-1>
(a-1*a)*x=a-1*a <P5>
1*x=1 <P7>
x=1 <P6>

Now that one I'm confident on. The second one I know I got right, but I may have done It the wrong way. The second problem asks me to prove that x2-y2=(x-y)(x+y). So...

x2-y2=(x-y)(x+y)
=(x-y)*x+(x-y)*y <P9>
=(x*x-x*y)+(x*y-y*y) <P9>
=((x2+(-x*y))+(x*y))-y2 <P1>
=(x2+((-x*y)+(x*y)))-y2 <P1>
=(x2+0)-y2 <P3>
=x2-y2 <P2>

So, I got it right, but was I supposed to modify the left side instead? Does it matter?

Now, the third one has me lost. It asks me to prove that if x2=y2, then x=y or x=-y. I tried. Here's how far I got.

x*x=y*y
x-1*(x*x)=x-1*(y*y) <Multiply by x-1>
(x-1*x)*x=(x-1*y)*y <P5>
1*x=(x-1*y)*y <P7>
x=(x-1*y)*y <P6>
y-1*x=((x-1*y)*y)*y-1 <Multiply by y-1>
y-1*x=(x-1*y)*(y*y-1) <P5>
y-1*x=(x-1*y)*1 <P7>
y-1*x=x-1*y <P6>

I don't know where to go from there without recycling the original equation. If only I could just raise both sides of the original equation to the power of 1/2... but I can't...

2. May 16, 2010

### Staff: Mentor

What you have above shows that x2-y2 equals itself, which is obviously true. Start with one side and apply the axioms until you get to the other side. For this problem it doesn't make any difference which side you start from.

You can also do it this way, starting from the left side:
x2 - y2
= x2 -xy + xy - y2 <P2>
= x(x - y) + y(x - y) <P9>
= (x + y)(x - y) <P9>
x*x = y*y
<==> x2 - y2 = 0
Now replace the left side with (x - y)(x + y), which you have already proved you can do in the previous problem. Then, if x + y $\neq$ 0, it has a multiplicative inverse, so you can multiply both sides of the equation to solve for x - y. Do the same to solve for x + y.

3. May 16, 2010

### Char. Limit

Ah. That's going to help wonderfully. The other thing that'll help wonderfully for the later ones is that I can work from either side...

One thing. If I can show that if x^2=y^2, then (x-y)(x+y)=0, then can I not show that either (x-y)=0 or (x+y)=0 by some theorem or another? Or is this just what you're getting at?

4. May 16, 2010

### Staff: Mentor

Yes, that's what I'm getting at. If x + y is not zero, then it has a multiplicative inverse, 1/(x + y), and you can multiply both sides of the equation (x - y)(x + y) = 0 to get a new equation that involves x - y. And you can do a similar thing with x - y (by assuming it's not zero).

5. May 16, 2010

### Char. Limit

All right, I've figured it out now. Thanks a ton, now I can get to problem 2.

6. May 19, 2010

### Unit

This is a very clever move. Is it acceptable then, in proof writing, to move things over from one side of the equals sign to the other? I have this (mis)conception that in proving things, the left side and right side must be treated separately.

7. May 19, 2010

### Staff: Mentor

In this problem, which starts in post #3, Char. Limit isn't trying to prove that x^2 = y^2, but is trying to show that x^2 = y^2 ==> x = y or x = -y using the listed axioms.

The equations x^2 = y^2 and x^2 - y^2 = 0 are equivalent. (Adding -y^2 to both sides yields an equivalent equation.)