• September 14th 2008, 09:00 AM
antrewry
ok I have some problems that I need help with:
if a^2 is even, then prove that a is even too
if a^2 is uneven, then prove that a is uneven too

And another problem:
prove that:
rational + irrational number = irrational number
PLS HELP ME!!! (Doh)
• September 14th 2008, 09:09 AM
Jhevon
Quote:

Originally Posted by antrewry
ok I have some problems that I need help with:
if a^2 is even, then prove that a is even too

use the contrapostive

it suffices to show if $a$ is not even, then $a^2$ is not even.

so assume $a$ is not even, so that $a = 2n + 1$ for some integer n, then .....

Quote:

if a^2 is uneven, then prove that a is uneven too
same approach as above

by the way, "odd" means "uneven" and it sounds nicer.

Quote:

And another problem:
prove that:
rational + irrational number = irrational number
PLS HELP ME!!! (Doh)
Let $\frac ab$ be a rational number and $c$ be an irrational number. what can you say about their sum?
• September 14th 2008, 09:35 AM
reagan3nc
Proof:
Let x be an even interger. Then by definition of an even number we can write:
x=2m
where m is an interger.
Therefore x^2= (2m)^2
=4m^2
=2(m^2)
Let P =2m^2. Since the product of integers is an interger, p is an itnerger. Thus x^2 =2p and x^2 is an even integer.
• September 14th 2008, 09:40 AM
Moo
Quote:

And another problem:
prove that:
rational + irrational number = irrational number
PLS HELP ME!!!
Quote:

Let $\frac ab$ be a rational number and c be an irrational number. what can you say about their sum?
I think it would be funny to assume that $\frac ab+c=\frac de$, where $\frac de$ is a rationnal. (a,b,d,e are non-zero integers)

Now, is it logical to say : $\underbrace{c}_{\text{irrational}}=\frac de-\frac ab=\frac{bd-ae}{be}$ ?
:D
• September 14th 2008, 09:56 AM
Jhevon
Quote:

Originally Posted by reagan3nc
Proof:
Let x be an even interger. Then by definition of an even number we can write:
x=2m
where m is an interger.
Therefore x^2= (2m)^2
=4m^2
=2(m^2)
Let P =2m^2. Since the product of integers is an interger, p is an itnerger. Thus x^2 =2p and x^2 is an even integer.

you should state what question you are answering

and this approach was already given. give the poster some time to see if they can figure things out
• September 15th 2008, 08:34 AM
antrewry
OK not exactly
Quote:

Originally Posted by reagan3nc
Proof:
Let x be an even interger. Then by definition of an even number we can write:
x=2m
where m is an interger.
Therefore x^2= (2m)^2
=4m^2
=2(m^2)
Let P =2m^2. Since the product of integers is an interger, p is an itnerger. Thus x^2 =2p and x^2 is an even integer.

Not exactly what I asked, I want to prove that IF a^2 is even THEN a is even too, not the opposite. I approciate the help and efford though. Thank you very much!
• September 15th 2008, 08:38 AM
antrewry
Quote:

Originally Posted by Jhevon
use the contrapostive

it suffices to show if $a$ is not even, then $a^2$ is not even.

so assume $a$ is not even, so that $a = 2n + 1$ for some integer n, then .....

same approach as above

by the way, "odd" means "uneven" and it sounds nicer.

Let $\frac ab$ be a rational number and $c$ be an irrational number. what can you say about their sum?

Very helpful, I found the solutions through your guidance. Thank you very much. And I will use "odd" next time if you say so (Rofl)
• September 15th 2008, 08:43 AM
antrewry
Thank you
Quote:

Originally Posted by Moo
I think it would be funny to assume that $\frac ab+c=\frac de$, where $\frac de$ is a rationnal. (a,b,d,e are non-zero integers)

Now, is it logical to say : $\underbrace{c}_{\text{irrational}}=\frac de-\frac ab=\frac{bd-ae}{be}$ ?
:D