direct proof or contraposition?

**Discrete** · June 24th 2007, 08:41 PM

so what plato means here is it just depends on how in depth you want to proof it right?? so Jhevon answer is not wrong right?? I just don't get it b/a is also a rational number right?

**Jhevon** · June 24th 2007, 08:49 PM

Originally Posted by Discrete

so what plato means here is it just depends on how in depth you want to proof it right??

yes. it is up to you to tell us how in depth you want us to go. as Plato asked you before, what axioms were you given? what have you done in class regarding this topic, how did your professor prove a similar problem? did he just do the simple algebra as i did, or did he take it step by step identifying each axiom as he went along? do you know what axioms are?

so Jhevon answer is not wrong right??

no it is not wrong, it is just incomplete--i basically just did the steps, but you may be required to show how and why those steps work, and why is it valid for me to do what i did.

I just don't get it b/a is also a rational number right?

look at how we defined a rational number. it is a fraction made up of two integers where the bottom integer is not zero, doesn't b/a have that property as we defined it?

**Discrete** · June 25th 2007, 07:32 AM

Originally Posted by Jhevon

yes. it is up to you to tell us how in depth you want us to go. as Plato asked you before, what axioms were you given? what have you done in class regarding this topic, how did your professor prove a similar problem? did he just do the simple algebra as i did, or did he take it step by step identifying each axiom as he went along? do you know what axioms are? no it is not wrong, it is just incomplete--i basically just did the steps, but you may be required to show how and why those steps work, and why is it valid for me to do what i did. look at how we defined a rational number. it is a fraction made up of two integers where the bottom integer is not zero, doesn't b/a have that property as we defined it?

thanks Jhevon for clearing up this confusion, I actually haven't heard anything about axiom in my class and the professor has just explained some simple problems such as proofing if n is an odd number proof that 3n+2 is odd, or something similar like that.. he just mainly use algebra to proof the problem in class..

**Jhevon** · June 25th 2007, 07:59 AM

Originally Posted by Discrete

thanks Jhevon for clearing up this confusion, I actually haven't heard anything about axiom in my class and the professor has just explained some simple problems such as proofing if n is an odd number proof that 3n+2 is odd, or something similar like that.. he just mainly use algebra to proof the problem in class..

well, ok. i'm going to give you the benefit of the doubt and see you as one of those students who actually pays attention in class, so what i did should be acceptable. you made a comment, i think, about how i got b/a, do you get it now? why does putting 1 over a fraction flip it up-side down?

**Discrete** · June 25th 2007, 09:38 AM

Originally Posted by Jhevon

well, ok. i'm going to give you the benefit of the doubt and see you as one of those students who actually pays attention in class, so what i did should be acceptable. you made a comment, i think, about how i got b/a, do you get it now? why does putting 1 over a fraction flip it up-side down?

yeah I get all that, because of x cannot be equal to 0 therefore a cannot be 0 and b also can't be 0 since the first time it is stated. So b/a is still a rational number although it is flipped.

though there is still another question that I am willing to do but I just need some hints:

If n is a perfect square, then n+2 is not a perfect square.

Here's what I've did so far:

The definition of a perfect square is:

An integer a is a perfect square if there is an integer b such that $a = b^2$

Therefore in this question:
$n = t^2 \mbox { and } n+2 = (t^2 + 2)^2$
= $t^4+4t^2+4$
= $t^2(t^2+4) + 4$
= $t^2(....)$ is a perfect square but because you add 4 to it, it doesn't become a perfect square again

therefore it's not a perfect square?

is this right?

**ThePerfectHacker** · June 25th 2007, 10:41 AM

Originally Posted by Discrete

If n is a perfect square, then n+2 is not a perfect square.
?

$n=a^2$ and $n+2=b^2$

That means,
$a^2 + 2 = b^2$

Thus,
$b^2 - a^2 = 2$
Thus,
$(b+a)(b-a) = 2$

By factorization we have that $b+a = 2 \mbox{ and }b - a =1$ . Impossible*

*)The cool

way of saying is that the integers form a unique factorization domain. Which basically is the Fundamental Theorem of Arithmetic is that we can factorize the integers > 1 uniquly.

**Discrete** · June 25th 2007, 10:46 AM

Originally Posted by ThePerfectHacker

$n=a^2$ and $n+2=b^2$

That means,
$a^2 + 2 = b^2$

Thus,
$b^2 - a^2 = 2$
Thus,
$(b+a)(b-a) = 2$

By factorization we have that $b+a = 2 \mbox{ and }b - a =1$ . Impossible*

*)The cool

way of saying is that the integers form a unique factorization domain. Which basically is the Fundamental Theorem of Arithmetic is that we can factorize the integers > 1 uniquly.

why did you say that $n+2 = b^2$
and why is it also b-a = 1? shouldn't it be b-a = 2
n+2 is said to not be a perfect square in the question...

**ThePerfectHacker** · June 25th 2007, 10:50 AM

Originally Posted by Discrete

why did you say that $n+2 = b^2$
and why is it also b-a = 1? shouldn't it be b-a = 2
n+2 is said to not be a perfect square in the question...

I am doing this by contradiction. I assume that n=a^2 and n+2=b^2 and show I arrive at a contradiction.

**Jhevon** · June 25th 2007, 10:50 AM

Originally Posted by Discrete

...

Therefore in this question:
$n = t^2 \mbox { and } n+2 = (t^2 + 2)^2$ ...

is this right?

No, this is wrong.

If $n = t^2$ then $n + 2 = t^2 + 2$ not $(t^2 + 2)^2$

**Discrete** · June 25th 2007, 10:53 AM

Originally Posted by Jhevon

No, this is wrong.

If $n = t^2$ then $n + 2 = t^2 + 2$ not $(t^2 + 2)^2$

oh yeah, just some algebraic issues, sorry bout that.

also one more thing is that I am only asked to do this whether by direct proof or contraposition.

If I use direct proof, thereforeL
$t^2 + 2$ is not a perfect square right? because you can't factor it?

is this right if I am trying to proof this using direct proof?

**Discrete** · June 25th 2007, 02:09 PM

anyone??

**Jhevon** · June 25th 2007, 02:11 PM

Originally Posted by Discrete

oh yeah, just some algebraic issues, sorry bout that.

also one more thing is that I am only asked to do this whether by direct proof or contraposition.

If I use direct proof, thereforeL
$t^2 + 2$ is not a perfect square right? because you can't factor it?

is this right if I am trying to proof this using direct proof?

yes, that will work

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