$\begin{align}\sqrt{2} = \frac{a}{b} \\ 2a^2 = b^2\end{align}$
I have the equation above and was told that since the left land side is divided by $2$, $b^2$ is an even number. But to me, the left hand side is times $2$.
$\begin{align}\sqrt{2} = \frac{a}{b} \\ 2a^2 = b^2\end{align}$
I have the equation above and was told that since the left land side is divided by $2$, $b^2$ is an even number. But to me, the left hand side is times $2$.
I suspect that you were told that the left-hand side is divisible by $2$, which means that it can be divided by $2$ leaving an integer, or equivalently that $2$ is a factor in its prime factorization.
So you've got the equation:
$2a^2 = b^2$
$2a^2$ is obviously even, and as it is equal to $b^2$, $b^2$ is even as well.
EDIT: My mistake, the question was confusion over the phrase "divided by 2".
The LHS is $2a^2$ so it is an even number, since $2a^2=b^2$ it holds that $b^2$ is also an even number.
Note: a number is divisible by $2$ means the number is even
Consider following example:
$a = b$
Now if a is even, that is divisible by 2, then also, since they're equal, b has to be divisible by 2. (Numbers divisible by 2 are 2, 4, 6,etc.)
Let's get back to your example now. You have:
$2a^2 = b^2$
Let's put in following notation, let us instead of $2a^2$ write $x$ and instead of $b^2$ write y.
So we have:
$x = y$
Since $x$ is divisible by 2($x = 2a^2$ so whatever you put instead of $a$ it would be divisible by 2), and therefore using example 1, we have that $y$ is divisible by 2. And since $y$ is the same thing as $b^2$, then $b^2$ is divisible by 2.