1. ## prove

prove that

squar root x^2+y^2 is a irrational if x and y is odd , positiv integer .

thanks

2. Originally Posted by hoger
prove that

squar root x^2+y^2 is a irrational if x and y is odd , positiv integer .

thanks
Suppose that:

$\displaystyle x, y \in \mathbb{Z}$ and that the square root of the sum of their squraes is rational, then there exist $\displaystyle a, b \in \mathbb{N}_+$ with $\displaystyle \gcd(a,b)=1$ such that:

$\displaystyle \sqrt{x^2+y^2} =\frac{a}{b}$

Then:

$\displaystyle b^2(x^2+y^2)=a^2$

Now consider any prime factor of $\displaystyle b$ and the divisibility of $\displaystyle a^2$ by this prime, and in consequence the divisibility of $\displaystyle a$ by the same prime.

This will show that if $\displaystyle \sqrt{x^2+y^2}$ is rational it is an integer and so $\displaystyle x^2+y^2$ ia a square.

CB

3. You can also consider quadratic residues mod 4. It is easy to show that all squares leave a remainder of 1 or 0 when divided by 4.
$\displaystyle x^2 + y^2$ leaves a reminder of 2 if both x and y are odd so it cannot be a perfect square.

Bobak

4. Originally Posted by bobak
You can also consider quadratic residues mod 4. It is easy to show that all squares leave a remainder of 1 or 0 when divided by 4.
$\displaystyle x^2 + y^2$ leaves a reminder of 2 if both x and y are odd so it cannot be a perfect square.

Bobak

hi
can you explain more for me .

thank you

5. Originally Posted by hoger
hi
can you explain more for me .

thank you
You need both of the responses to follow what is going on. Suppose that $\displaystyle x$ and $\displaystyle y$ are integers and that $\displaystyle \sqrt{x^2+y^2}$ is rational, then the first post shows that this is of necessity an integer and that $\displaystyle x^2+y^2$ is a square.

Then bobak shows that any square leaves remainder $\displaystyle 0$ or $\displaystyle 1$ when divided by $\displaystyle 4$, but if $\displaystyle x$ and $\displaystyle y$ are odd integers $\displaystyle x^2$ leaves remainder $\displaystyle 1$ when divided by $\displaystyle 4$ as does $\displaystyle y^2$, so $\displaystyle x^2+y^2$ leaves a remainder of $\displaystyle 2$ and so cannot be a square. but this contradicts our assumption that $\displaystyle \sqrt{x^2+y^2}$ is rational, and si it must be irrational.

CB