Spivak: Is his how to approach these proofs?

[Saladsamurai]

Hello all

I have started the problem set for Chapter one (basic properties of numbers) in Spivak's Calculus (self study). I think I am doing these right, but I have some questions.

As a solid example, problem 1-(iv) says to prove the following:

x^3 - y^3 = (x-y)(x^2+xy+y^2)\qquad(1)

My approach was to add and subtract terms to the left hand side of (1) until it could be factored into the desired form:

\begin{array}{l}<br /> x^3 - y^3 &amp;= x^3 - y^3 + (x^2y - x^2y) + (xy^2 - xy^2) \\ <br /> &amp;= x^3 +x^2y+xy^2 - y^3 - x^2y - xy^2 \\<br /> &amp;=x(x^2+xy+y^2)-y(x^2+xy+y^2) \\<br /> &amp;= (x-y)(x^2+xy+y^2)<br /> \end{array}<br />

Now this seems correct to me, but I feel a little guilty because I only knew what to add and subtract to the LHS of (1) because I knew what I was trying to achieve- the RHS of (1).

My second question is similar, but requires some clarifying of Spivak's intent. In problem 1-(vi) he says: Prove the following:

x^3 + y^3 = (x+y)(x^2-xy+y^2)\qquad(2)

Then he says, "There is a particularly easy way to do this, using (iv), and it will show you how to find a factorization of x^n+y^n whenever n is odd."

Well, I solved this one the exact same way I did with (iv) above: I used the right hand side of (2) to infer what terms to add/subtract and obtained the solution. However, the fact that I do not see what he means by the quoted text above leads me to believe that there was some other approach. That is, I have not discovered a way to factorize x^n+y^n whenever n is odd in my procedure.

Any thoughts are appreciated

 

[LCKurtz]

Certainly, if you know (1), you can get (2) by simply substituting -y in for the y.

 

[Saladsamurai]

Certainly, if you know (1), you can get (2) by simply substituting -y in for the y.

Hi LC Yes. I arrived at the solution, but in particular, I am concerned about my approach as I do not see what he means by

...and it will show you how to find a factorization of xⁿ+yⁿ whenever n is odd

 

[LCKurtz]

OK, I see what you mean now. Apparently you need a more clever approach to (1) that easily generalizes to higher odd exponents. I don't know -- I'll think about it if this football game doesn't keep me too distracted. But don't count on me for this one

 

[LCKurtz]

Also, I guess that observing that y = x is a root for odd exponents and factoring (y-x) by long division doesn't count?

 

[Saladsamurai]

Also, I guess that observing that y = x is a root for odd exponents and factoring (y-x) by long division doesn't count?

Maybe he made a typo? He says "this can easily be done using (iv)". Maybe he meant (v). In problem (v) we proved that

<br /> x^n-y^n = (x-y)(x^{n-1} + x^{n-2} y+\dots+xy^{n-2} +y^{n-1} )<br />

Anyway, no big deal! Get back to your game!

 

[micromass]

Maybe he made a typo? He says "this can easily be done using (iv)". Maybe he meant (v). In problem (v) we proved that

<br /> x^n-y^n = (x-y)(x^{n-1} + x^{n-2} y+\dots+xy^{n-2} +y^{n-1} )<br />

Anyway, no big deal! Get back to your game!

My thoughts:

Spivak is correct, (vi) can be easily done by (iv) by just substituting -y for y.
This gives you a hint how to obtain the factorization of x^n+y^n. Indeed, take (v) and substitute -y for y again...

 

[206PiruBlood]

My thoughts:

Spivak is correct, (vi) can be easily done by (iv) by just substituting -y for y.
This gives you a hint how to obtain the factorization of x^n+y^n. Indeed, take (v) and substitute -y for y again...

I am pretty sure this is what Spivak intended.

Also note that you don't have to start from the left side for the first one..

(x-y)(x^{2}+xy+y^{2})=x(x^{2}+xy+y^{2})-y(x^{2}+xy+y^{2})=(x^{3}+x^{2}y+xy^{2})-(x^{2}y+xy^{2}+y^{3})=x^{3}-y^{3}

This approach will come in handy for at least one problem in chapter two, but I will let you figure out which one!