# SpivakProve x^2 - y^2 = (x-y)(x+y)

1. Nov 18, 2009

Ok. So know that I am supposed to use one or more of the 12 basic properties of numbers from chapter 1 of Spivak's Calculus to prove this, but I can't see to figure out which one to start with.

Can someone start me off here?

Just a 'nudge.'

edit: it says to 'prove' it, not to 'show' it. How do you 'prove' this? I can't see how to even start the proof.

Last edited: Nov 18, 2009
2. Nov 18, 2009

### Staff: Mentor

I would start with the right side, (x -y)(x + y) and use the distributive property to get
(x -y)(x + y) = (x - y)(x) + (x - y)(y).
Then use the commutative property of multiplication to rewrite (x - y)(y) as y(x - y), and the same for the other product. And so on...

Eventually you'll get to x^2 - y^2 and will have justified each step along the way by one of the basic properties of real numbers.

3. Nov 18, 2009

See now, that's what I wanted to do and I know that is a perfectly acceptable way to do it. But I have a question about that:

Perhaps it is just some false notion that I have conjured up somewhere along the way, but I always thought that when they (the math gods) said "Prove A = B," you were supposed to start with 'A' and successively apply 'basic rules' until you arrive at 'B.'

And when they say: "Show A = B," you can just manipulate 'A' or 'B' to show that the two are equivalent.

Is this something that I have made up in my head? Or is there some truth to it?

Thank you,
Casey

4. Nov 18, 2009

### VeeEight

'Proving' a statement and 'showing' a statement is true are the same. If you are asked to show that A=B, then you can manipulate both A and B (as long as you're using valid operations, of course). Since A=B and B=C implies that A=C, this strategy is valid.

5. Nov 18, 2009

### Staff: Mentor

No, you can start with either side. To elaborate on what VeeEight said, you can also start with A = B and apply reversible operations to both sides until you arrive at a new equation that has the same solution set. Because all the operations are reversible, that means that your first equation has the same solution set.

For a simple example, suppose we start with x + 2 = 8. I can add -2 to both sides (a reversible operation) to get x = 6. The solution set for the second equation is obviously {6}, and because the operation I applied was one-to-one, and therefore invertible, the first equation's solution set is also {6}.

On the other hand, if the operation is not reversible, then the solution sets of the starting and ending equations do not have to be the same. Starting with x = -2, and squaring both sides, I get x2 = 4. The solution set of this equation is {2, -2}, while that of the original equation is {-2}. The squaring operation is not reversible, for the same reason that the function f(x) = x2 is not 1-to-1.

6. Nov 18, 2009

Okay. Thanks Mark44 and VeeEight

That is very helpful to know! I had a piece of paper in front of me that just had:

x^2 - y^2 =

written on it and I was trying to find a way to apply basic properties of numbers to this expression such that I would naturally 'arrive at' (x - y)(x + y).

7. Nov 18, 2009

### l'Hôpital

I did that problem starting with just that.

x2 - y2 = x2 - xy + xy -y2

x(x - y) + y(x - y) = (x-y)(x+y)

8. Nov 19, 2009

### Staff: Mentor

If you work it through from the side with (x - y)(x + y), it makes the "trick" of subtracting and adding xy more understandable. Obviously, you can work it from either end.

9. Nov 19, 2009

How would you start it from the LHS? What basic property can you apply such that:

x2-y2=x2+xy-xy+y2 ?

I am missing how one logically deduces the RHS from the LHS?

How did you arrive at that l'hopital? Without just knowing that the two are equivalent...

10. Nov 19, 2009

### lanedance

i think you dropped a sign somewhere, but i would go as follows
x2- y2 = x2- y2 + 0 = x2 - y2 + ((xy)+ -(xy)) = x2 + xy - xy - y2

but as I think Mark was pointing out, all the steps you took working from RHS to LHS are perfectly valid in the opposite direction

11. Nov 19, 2009

### l'Hôpital

x2 - y2
x2 - y2 + 0
x2 - y2 + (xy -xy)
... and so on

12. Nov 19, 2009

Oh jeesh....how do you guys come up with this stuff?! Who would ever think to do that? I mean...I guess you would But seriously.....

Thanks!

And yes, I missed a '-' sign.

EDIT: And apparently l'hopital would think to do that. It seems that I missed the memo about randomly adding zero to expressions :rofl:

13. Nov 19, 2009

### lanedance

zero is often pretty good to use in these types of proofs as you an turn it into a sum or multiplication with almost anything to help

14. Nov 19, 2009

Thank you. I will need to hold onto these tricks and learn start thinking a little differently if I am to make it through this text by the time classes start up again.

15. Nov 19, 2009

### Staff: Mentor

Not only is adding zero a good technique to remember, it's one of a very few things you can do to just one side of an equation, with others being multiplying by 1 in some form, simplifying, and expanding.

16. Nov 19, 2009