Prove $ \sqrt{2k}$ is irrational where $ k$ is an odd integer.

Question

Question:

Prove $ \sqrt{2k}$ is irrational where $ k$ is an odd integer.

My attempt:

Proof by contradiction:

Now, assume $ \sqrt{2k}$ is rational. Then, $ \sqrt{2k} = \frac{a}{b}$ where $a,b \in Z$, $b$ not equal $0$ and $a,b$ have no common factors.

$ \sqrt{2k} = \frac{a}{b} \implies 2k = \frac{a^{2}}{b^{2}} \implies (b^{2})(2k) = a^{2} \implies 2|a^{2} \implies 2|a, \ since \ 2 \ is\ prime \implies \exists c \in Z$ such that $ a = 2c$

Then, $ (2k)(b^{2}) = a^{2} \implies (2k)(b^{2}) = 4c^{2} \implies kb^{2} = 2c^{2} \implies 2|kb^{2} \implies 2 | b^{2} \implies 2|b$, since $2$ is prime.

So, $ 2|a$ and $ 2|b$ , a contradiction.

Answer 1

The general result is that $\sqrt x$ is Rational iff $x$ is a perfect square. But, if we have $\sqrt {2y}$ , in order for $2y$ to be a perfect square, we must have an even number of $2$s. But this cannot happen when $y$ is odd. So $2y$ is not a perfect square and so $\sqrt {2y}$ is irrational.

Answer 2

You are almost there, suppose that $a$ and $b$ are relatively prime,

write $a^2=b^2(4m+2)=2b^2(2m+1)$, you deduce that $2$ divides $a^2$, and $a$, write $a=2a_1$, you have $2b^2(2m+1)=4a_1^2$, you deduce that $b^2(2m+1)=2a_1^2$. This implies that $2$ divides $b$, contradiction since $a$ and $b$ are relatively prime.

Answer 3

Suppose rational, then $\sqrt{2k} = a$. We can suppose $a \in \mathbb{Z}$. We have: $2k = a^{2}$ If a is odd, $a^{2}$ too, contradction. If $a$ is even, then $a = 2m$, thus $2k = 4m^{2}$, like this $ k = 2m^{2}$, but $k$ is odd.

Answer 4

Proof by contradiction: suppose that $\sqrt{2k}=\frac{n}m$ where $k,n,m\in\Bbb N_{>0}$ and $\gcd(m,n)=1$. Then taking squares both sides we get

$$2k m^2=n^2\tag1$$

Now observe that $2km^2$ is even regardless the parity of $m$, thus $n^2$ must be even if $(1)$ holds, what imply that $n$ is even and $m$ must be odd, because $\gcd(m,n)=1$.

Because $n$ is even we knows that $n^2=4j$ for some $j\in\Bbb N_{>0}$, thus we can rewrite $(1)$ as

$$km^2=2j\tag2$$

but $km^2$ is odd and $2j$ even.$\Box$

Answer 5

Hint: If $k$ is odd then $k^2$ is odd. (You can prove that).

If $k$ even then $k^2$ is not only even but $4|k^2$. (You can prove that).

And if $k$ is odd, then $2k$ is even but $4\not \mid 2k$. (You can prove that--- and you and use $2k = 4m + 2$ to prove it.)

So you have $b^2*(4m + 2) = a^2$. If $b$ is odd is $a$ even or odd? Is $a^2$ divisible by $4$? If $b$ is even is $a$ even or odd? Is it divisible by $4$? Are any of these possible? Is it possible for $a$ and $b$ to both be even?

Answer 6

Hint: Use fundamental theorem of arithmetic.

The thing inside square root has an odd power of two hence it cannot be a perfect square.