1. ## 2 "basic" proofs

I have to provide direct proofs for the following and I am not sure how to do that.

1. If the square of an integer is odd then the integer must be odd.

Pf: Since the square of an integer is odd then there exists some integer c such that x^2 = 2c+1 where (2c+1) is an integer.

2. If xy is an odd integer then x and y are odd integers.
Pf: Since xy is an odd integer then there exists some integer c so that xy = 2x+1.

Thats as far as I can get on either of those.

2. Originally Posted by algebrapro18
I have to provide direct proofs for the following and I am not sure how to do that.

1. If the square of an integer is odd then the integer must be odd.

Pf: Since the square of an integer is odd then there exists some integer c such that x^2 = 2c+1 where (2c+1) is an integer.
you are begging the question here. that is, you are assuming what you are supposed to be proving. hint, use the contrapositive. assume the integer is even, then what? (you know what "contrapositive" means right?)

2. If xy is an odd integer then x and y are odd integers.
Pf: Since xy is an odd integer then there exists some integer c so that xy = 2x+1.

Thats as far as I can get on either of those.
you made the same mistake here...sort of. again, do the contrapositive. assume that x and y are not odd integers. that is, we have two cases, x and y are both even, or (without loss of generality) x is even and y is odd. what does that tell us about xy?

How are you with implications?

3. These can't be used using the contrapositive, they have to be done using direct proof. There has to be some mathematical work that can be done to arrive at the conclusions of those 2 statements I just can't figure out what.

4. Originally Posted by algebrapro18
These can't be used using the contrapositive, they have to be done using direct proof. There has to be some mathematical work that can be done to arrive at the conclusions of those 2 statements I just can't figure out what.
i doubt that's true. to do these by direct proof (if possible) would be extremely messy i think. i see no reason why you would be restricted to direct proofs. did the text specifically say it wanted direct proofs for these?

5. No but my professor did. These aren't from the text book unfortunately there from my professor who is VERY detail oriented.

6. Originally Posted by algebrapro18
No but my professor did. These aren't from the text book unfortunately there from my professor who is VERY detail oriented.
well, the contrapositive is the most efficient way to do these proofs, like you, i cannot see a way to prove them directly. not yet

7. I am going to prove this using the contrapositive and then take the penilty if there is one.

Thm: If the square of an integer is odd then the integer must be odd.
This contrapositive of this statement is if x is an even integer then the square of an even integer must be even. Now we can do a proof by contrapositive

Pf: Assume that x is even, that is there exists some integer c such that 2c = x. x^2 = (2c)^2 = 4c^2= 2(2c^2) where (2c^2) is an integer. Therefore x^2 is an even number

Is this right?

8. Originally Posted by algebrapro18
I am going to prove this using the contrapositive and then take the penilty if there is one.

Thm: If the square of an integer is odd then the integer must be odd.
This contrapositive of this statement is if x is an even integer then the square of x must be even. Now we can do a proof by contrapositive

Pf: Assume that x is even, that is there exists some integer c such that 2c = x. x^2 = (2c)^2 = 4c^2= 2(2c^2) where (2c^2) is an integer. Therefore x^2 is an even integer

Is this right?
yes, that's the gist of it. and you're done. i'd rephrase somethings though, marked in red, and others, which i'm probably just being a pain in the butt about. but that's good

9. If $z^2$ is odd the we know that $\begin{array}{l} z^2 = 2n + 1 \\ n = \frac{{z^2 - 1}}{2} = \frac{{\left( {z - 1} \right)\left( {z + 1} \right)}}{2} \\ \end{array}$.

This means that one of ${\left( {z - 1} \right) \,\mbox{or}\left({z + 1} \right)}$ must be even. So z must be odd.

10. Originally Posted by Plato
If $z^2$ is odd the we know that $\begin{array}{l} z^2 = 2n + 1 \\ n = \frac{{z^2 - 1}}{2} = \frac{{\left( {z - 1} \right)\left( {z + 1} \right)}}{2} \\ \end{array}$.

This means that one of ${\left( {z - 1} \right) \,\mbox{or}\left({z + 1} \right)}$ must be even. So z must be odd.
brilliant!

11. Maybe this will inspire you (I tried a similar method for the second question, but I could not figure out a way to get it to work).

$z^2=2a+1$ where $a\in\mathbb{W}$

Assume $z=c+1$ for $c\in\mathbb{W}$

Then
$(c+1)^2=2a+1$
$c^2+2c+1=2a+1$
$c^2+2c=2a$
$c(c+2)=2a$

Therefore either c or c+2 is divisible by 2, which implies c must be.
$c=2b$ for $b\in\mathbb{W}$
$z=2b+1$

12. Originally Posted by ecMathGeek
Maybe this will inspire you (I tried a similar method for the second question, but I could not figure out a way to get it to work).

$z^2=2a+1$ where $a\in\mathbb{W}$

Assume $z=c+1$ for $c\in\mathbb{W}$

Then
$(c+1)^2=2a+1$
$c^2+2c+1=2a+1$
$c^2+2c=2a$
$c(c+2)=2a$

Therefore either c or c+2 is divisible by 2, which implies c must be.
$c=2b$ for $b\in\mathbb{W}$
$z=2b+1$
what's $\mathbb{W}$?

13. Whole numbers (unless I'm mistaken)... upon reflection, it has occured to me that these numbers can be positive or negative, so it was not necessary to limit them to whole numbers. Just assume I had used $\mathbb{Z}$ instead. :P