Let x, y be an integer. Prove that (x+1)y^2 is even iff x...

[shivers20]

Let x, y be an integer. Prove that (x+1)y^2 is even if and only if x is odd or y is even.


"Theorem: A if and only if B." means you must prove that A and B are true and false at the same time. In other words, you must prove "If A then B" and "If not A then not B". Equivalently, you must prove "If A then B" and "If B then A".

Proof: Assume if (x+1)y^2 is even, x is odd or y is even.
Case1: x=2k+1 and y=(2m) for some integers x and y.
So, (x+1)y^2 = (2k+1+1)= (2k+2) = 2(k+1),
y^2= (2m)^2= 2(1m^2)

Case2: For the converse, assume (x+1)y^2 is odd if x is even or y is odd


Is this correct so far? What exactly do I prove in the first case? Am I proving
that (x+1)y^2 is even or should I prove that x is even and y is odd, then plug into equation?

 

[stapel]

Re: Let x, y be an integer. Prove that (x+1)y^2 is even iff

"Theorem: A if and only if B." means....

Generally (in my experience, anyway), "if and only if" is taken to mean "if-then, but both ways", so "if A, then B; and also if B, then A". So you do "if-then" in two directions.

B->A: If x is odd or y is even, then (x + 1)y² is even.
Proof: If x is odd, then x = 2m + 1 for some integer m. Then x = 2m + 2 = 2(m + 1), so (x + 1)y² = 2[(m + 1)y²], which, by definition, is even. If, on the other hand, y is even, then y = 2n for some integer n, and y² = (2m)² = 4m² = 2(2m²). Then (x + 1)y² = 2[(x + 1)(2m²)}, which, by definition, is even.

A->B: If (x + 1)y² is even, then at least one of the following must hold: x is odd or y is even.
Proof: If (x + 1)y² is even, then (x + 1)y² = 2m for some integer m. Suppose y is not even; y is odd. Then y = 2n + 1 for some integer n, and y² = 2n² + 4n + 1. Then:

. . .(x + 1)y²

. . . . .= (x + 1)(2n² + 4n + 1)

. . . . .= 2n²x + 4nx + 2n² + 4n + x + 1

. . . . .= 2(n²x + 2nx + n² + 2n) + x + 1

If x is even, then this result is, by definition, odd. But we assumed that (x + 1)y² was even. Then, if y is odd, we must have x being odd. Then, for (x + 1)y² even, either y is even, or else x must be odd.

Hope that makes sense....

Eliz.

 

[pka]

Suppose (x+1)y<SUP>2</SUP> is even.
If x is odd then we are done. If x is even then x+1 is odd.
Thus y<SUP>2</SUP> is even so what about y?

If y is even then what about (x+1)y<SUP>2</SUP>?
If x is odd then what about (x+1)y<SUP>2</SUP>?

 

[shivers20]

Suppose (x+1)y<SUP>2</SUP> is even.
If x is odd then we are done. If x is even then x+1 is odd.
Thus y<SUP>2</SUP> is even so what about y?

If y is even then what about (x+1)y<SUP>2</SUP>?
If x is odd then what about (x+1)y<SUP>2</SUP>?

I am very confused. What exactly are you asking.

 

[pka]

I am not asking anything!
That is a proof of the problem. Is it not?
The whole proof turns on a simple fact: if the product of two integers is even at least one of them must be even. Moreover if one is even then the product must be even.