Help with my proofs

jrh1337 · #1 $Old$ September 17th, 2008, 10:43 AM

If you could check my 2 proofs and help me start on one please.

1a) If $n^3$ is a multiple of 2, then $n$ is a multiple of 2. Domain of n is all integers.

My proof:

An integer n is even if there exists an integer k such that n = 2k.
n = 2k, $n^3 = (2k)^3 = 8k^3 = 2(4k^3)$
$4k^3$ is an integer therefore $n^3$ is even, thus making it a multiple of 2.
By the definition of n = 2k, n is also a multiple of 2.

1b) $\sqrt[3]{2}$ is an irrational number.

My proof(by contradiction):

$\sqrt[3]{2}$ is a positive number such that its cube is 2.

Assume $\sqrt[3]{2}$ is rational.
$\exists$ integers P, Q such that $\sqrt[3]{2}$ = $\frac{P}{Q}$ , fully simplified.

$(\sqrt[3]{2})^3 = (\frac{P}{Q})^3 \rightarrow 2 = \frac{P^3}{Q^3} \rightarrow 2Q^3 = P^3$

(Corollary: If $n^3$ is even, then n is even. (proven in 1a))
An integer n is even if there exists an integer k such that n = 2k.

$2Q^3 = (2k)^3 = 8k^3 \rightarrow Q^3 = 4k^3 = 2(2k^3)$

$2k^3$ is an integer, therefore Q is even, along with P.

Since P and Q are even, they share a common factor. Since we defined that $\frac{P}{Q}$ was fully simplified, we have a contradiction.

Now the one I am having a hard time with is
2) Prove or disprove the following proposition: If x and y are positive integers such that x > y + 1, then $x^2 - y^2$ is not prime.

Thanks in advance
James

Moo · #2 $Old$ September 17th, 2008, 11:10 AM

Hello,

Quote:

Originally Posted by jrh1337 $View Post$

If you could check my 2 proofs and help me start on one please.

1a) If $n^3$ is a multiple of 2, then $n$ is a multiple of 2. Domain of n is all integers.

My proof:

An integer n is even if there exists an integer k such that n = 2k.
n = 2k, $n^3 = (2k)^3 = 8k^3 = 2(4k^3)$
$4k^3$ is an integer therefore $n^3$ is even, thus making it a multiple of 2.
By the definition of n = 2k, n is also a multiple of 2.

Huh

But here you proved that if n is even, then n^3 is even... while you're asked to prove the converse.
Actually, it's more complicated, you proved that if $n^3$ is in a form 2k, with k a very special number, then n is even. Hehe that was unfair

So...what you can do is to prove the contrapositive :
"if n is odd, then n^3 is odd"

Let $n=2k+1$ .
So $n^3=(2k+1)^3= \text{ even or odd ? }$

Quote:

1b) $\sqrt[3]{2}$ is an irrational number.

My proof(by contradiction):

$\sqrt[3]{2}$ is a positive number such that its cube is 2.

Assume $\sqrt[3]{2}$ is rational.
$\exists$ integers P, Q such that $\sqrt[3]{2}$ = $\frac{P}{Q}$ , fully simplified.

$(\sqrt[3]{2})^3 = (\frac{P}{Q})^3 \rightarrow 2 = \frac{P^3}{Q^3} \rightarrow 2Q^3 = P^3$

(Corollary: If $n^3$ is even, then n is even. (proven in 1a))
An integer n is even if there exists an integer k such that n = 2k.

$2Q^3 = (2k)^3 = 8k^3 \rightarrow Q^3 = 4k^3 = 2(2k^3)$

$2k^3$ is an integer, therefore Q is even, along with P.

Since P and Q are even, they share a common factor. Since we defined that $\frac{P}{Q}$ was fully simplified, we have a contradiction.

It's your method, you keep it because it's okay !

Quote:

Now the one I am having a hard time with is
2) Prove or disprove the following proposition: If x and y are positive integers such that x > y + 1, then $x^2 - y^2$ is not prime.

Difference of two squares ^^

x²-y²=(x-y)(x+y)
(note that x-y < x+y since x and y are positive integers)
The only way for it to be a prime is that x-y=1 and x+y is a prime number.
But... It is said that x > y+1; that is to say x-y>1

Got it ?

Quote:

Thanks in advance
James

Blop.

jrh1337 · #3 $Old$ September 17th, 2008, 11:16 AM

Thanks for the reply. I shall work on it some more after class.

But doesn't showing n is even indirectly show it is a multiple of 2?

Moo · #4 $Old$ September 17th, 2008, 11:24 AM

Quote:

Originally Posted by jrh1337 $View Post$

Thanks for the reply. I shall work on it some more after class.

But doesn't showing n is even indirectly show it is a multiple of 2?

Yes it does.
But it's not the problem here.

You found an expression for n^3, starting from n=2k (that is to say even). But you want to prove that n is even, you don't know it yet !

If you want to do the direct proof, you will have to assume that 2 divides n^3. There is a theorem (not sure) that says "if a prime number p divides n^m, then p^m divides n^m". But I don't think you're supposed to use it (and you'll have to prove it).

So do the contrapositive, starting from what I've showed you, and it'll be okay

Nota Bene : the contrapositive of $A \Rightarrow B$ is $\neg B \Rightarrow \neg A$ , and if the contrapositive is true, then the original statement is true too. And conversely.

jrh1337 · #5 $Old$ September 18th, 2008, 09:00 AM

Thanks Moo! I was able to do the problem I couldn't start, and fix my proof today.