Assignment 0 : Problem 2

[Aaron Albers]

a^2 - b^2 is odd

Four Cases:

Case 1: a is odd && b is odd -> a^2 - b^2 is odd

Case 2: a is odd && b is even -> a^2 - b^2 is odd

Case 3: a is even && b is odd -> a^2 - b^2 is odd

Case 4: a is even && b is even -> a^2 - b^2 is odd

We need to find a single case that is true.

[Aaron Albers]

Case 1: FALSE

[Aaron Albers]

Case 2: TRUE

A^2 - B^2 = (2a + 1)^2 - (2b)^2 = 4a^2 + 4a - 4b^2 + 1 = 2(2a^2 + 2a - 2b^2) + 1

n is odd iff n = 2k + 1, therefore 

2(2a^2 + 2a - 2b^2) + 1 is odd

[Natalia Duque]

isn't it case 2, a and b both odd. In A^2 - B^2 = (2a + 1)^2 - (2b)^2 = 4a^2 + 4a - 4b^2 + 1 = 2(2a^2 + 2a - 2b^2) + 1, a is odd and b is even, That would be case 4, which is not true (according to me). Since we are looking for a necessary and sufficient condition, we need to prove the implication both ways. if a odd, b even then a^2 -b^2 odd is true but if a^2 -b^2 odd  then a odd, b even is false. I used the contrapositive proof

[Aaron Albers]

Interesting. I didn't know the converse had to be true in order for it to be "necessary and sufficient condition". (Just curious.. where does it state what a "necessary and sufficient condition" is in the book?)

[Anne S]

My question on this one is why would a single case be true? I think
there are two true cases and two false cases.

[Natalia Duque]

yeah, page 11

[Natalia Duque]

All cases are false ... I might be wrong tho. If you find a true case, lmk

[Aaron Albers]

I agree. Two cases will be true and two will be false ( I have not checked the converses)

One variable must be even and the other odd in order for the result to be odd.

[Anne S]

Without writing out my whole proof...
-an even number squared is even
-an odd number squared is odd
-an even number minus an even number and an odd number minus an odd
number are even <-- these are false cases.
- an even number minus an odd number or an odd number minus an even
number are odd. <-- these are true cases.

On Sep 6, 4:48 pm, Natalia Duque <duque...@gmail.com> wrote:
> All cases are false ... I might be wrong tho. If you find a true case, lmk
>

> On Mon, Sep 6, 2010 at 4:46 PM, Natalia Duque <duque...@gmail.com> wrote:
> > yeah, page 11
>

> > On Mon, Sep 6, 2010 at 4:44 PM, Aaron Albers <aaroncalb...@gmail.com>wrote:
>
> >> Interesting. I didn't know the converse had to be true in order for it to
> >> be "necessary and sufficient condition". (Just curious.. where does it state
> >> what a "necessary and sufficient condition" is in the book?)
>

> >> On Mon, Sep 6, 2010 at 4:37 PM, Natalia Duque <duque...@gmail.com> wrote:
>
> >>> isn't it case 2, a and b both odd. In A^2 - B^2 = (2a + 1)^2 - (2b)^2 =
> >>> 4a^2 + 4a - 4b^2 + 1 = 2(2a^2 + 2a - 2b^2) + 1, a is odd and b is even, That
> >>> would be case 4, which is not true (according to me). Since we are looking
> >>> for a necessary and sufficient condition, we need to prove the implication
> >>> both ways. if a odd, b even then a^2 -b^2 odd is true but if a^2
> >>> -b^2 odd  then a odd, b even is false. I used the contrapositive proof
>

> >>> On Mon, Sep 6, 2010 at 4:22 PM, Aaron Albers <aaroncalb...@gmail.com>wrote:
>
> >>>> Case 2: TRUE
>
> >>>> A^2 - B^2 = (2a + 1)^2 - (2b)^2 = 4a^2 + 4a - 4b^2 + 1 = 2(2a^2 + 2a -
> >>>> 2b^2) + 1
>
> >>>> n is odd iff n = 2k + 1, therefore
> >>>> 2(2a^2 + 2a - 2b^2) + 1 is odd
>

> >>>> On Mon, Sep 6, 2010 at 4:17 PM, Aaron Albers <aaroncalb...@gmail.com>wrote:
>
> >>>>> Case 1: FALSE
>

[Natalia Duque]

check the converses. I got true implications for case 3 and 4. The converses were false for both.

[Aaron Albers]

I don't think the converses have to be true. I am not seeing anything stating that on Page 11.

[Natalia Duque]

Below statement 5 , count 4 lines... the sentence starts with "A is necessary and sufficient condition for B...

[Anne S]

So are you saying they're all false because the converse can't say
whether a or b is odd?

On Sep 6, 4:52 pm, Natalia Duque <duque...@gmail.com> wrote:
> check the converses. I got true implications for case 3 and 4. The converses
> were false for both.
>

> On Mon, Sep 6, 2010 at 4:50 PM, Aaron Albers <aaroncalb...@gmail.com> wrote:
> > I agree. Two cases will be true and two will be false ( I have not checked
> > the converses)
>
> > One variable must be even and the other odd in order for the result to be
> > odd.
>

[Aaron Albers]

OOOh. Thanks. Hmmm.. I guess I will check the converse of my solution then.

[Natalia Duque]

Yes to Anne

[Aaron Albers]

The converses of both true statements are also true. They are built into my proof as being so.

[Aaron Albers]

Converse A^2 - B^2 is odd -> A is odd and B is even

A^2 - B^2 = (2a + 1)^2 - (2b)^2

A = 2a + 1 (odd)

B = 2b (even)

[Natalia Duque]

so for case 3, for example, you are saying that a^2-b^2 an odd integer implies that a even, b odd is true ... but a counterexample would be  3^2-2^2 = 5 (odd) where a is odd and b is even

[Aaron Albers]

That is just the reverse.. or the other case that is true. There are two true cases. For me they are Cases 2 & 3 listed above.

[Natalia Duque]

yeah... I guess that if you look at the implication as being a^2-b^2 odd then a even, b odd OR a odd, b even, you are right. If you look at it as 

a^2-b^2 odd then a even, b odd 

a^2-b^2 odd then a odd, b even

then I'm right. Cool. I'll write a note on my HW!