1. About even and uneven numbers and about rational and irrational numbers

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!!!

2. 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 $\displaystyle a$ is not even, then $\displaystyle a^2$ is not even.

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

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.

And another problem:
prove that:
rational + irrational number = irrational number
PLS HELP ME!!!
Let $\displaystyle \frac ab$ be a rational number and $\displaystyle c$ be an irrational number. what can you say about their sum?

3. 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.

4. And another problem:
prove that:
rational + irrational number = irrational number
PLS HELP ME!!!
Let $\displaystyle \frac ab$ 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 $\displaystyle \frac ab+c=\frac de$, where $\displaystyle \frac de$ is a rationnal. (a,b,d,e are non-zero integers)

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

5. 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

6. OK not exactly

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!

7. Originally Posted by Jhevon
use the contrapostive

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

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

same approach as above

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

Let $\displaystyle \frac ab$ be a rational number and $\displaystyle c$ 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

8. Thank you

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

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