Confusing proof for $\sqrt{2}$ being irrational $\sqrt{2}$ is irrational using proof by contradiction.
say $\sqrt{2}$ = $\frac{a}{b}$ where $a$ and $b$ are positive integers. 
$b\sqrt{2}$ is an integer. ----[Understood]
Let $b$ denote the smallest such positive integer.----[My understanding of this is 
that were are going to assume b is the smallest possible integer such that $\sqrt{2}$ = $\frac{a}{b}$, ... Understood]
Then $b^{*}$ := $b(\sqrt{2} - 1)$----[I'm not sure I understand the point that's being made here,
from creating a variable  $b^{*} = a - b$ ]
Next, $b^{*}$ := $b(\sqrt{2} - 1)$ is a positive integer such that $b^{*}\sqrt{2}$ is an integer.----[ I get that ($a - b$) has to
be a positive integer, why does it follow that then $b^{*}\sqrt{2}$ is an integer?]
Lastly, $b^{*}<b$, which is a contradiction.----[I can see that given $b^{*}$ := $b(\sqrt{2} - 1)$, we then have   
$b^{*}<b$, I don't get how that creates a contradiction]
Any help is appreciated, thank you. 

Your initial premise is that $b$ is defined to be the smallest positive integer such that $b\sqrt 2$ is a positive integer. This is equivalent to reducing $\frac ab$ to its lowest terms (that is, make $a$ and $b$ coprime).
Obviously $b^\mathrm * = b\sqrt 2 - b = a-b$ will be a positive integer such that $b^*<b$. 
And we now construct a new positive number $b^\mathrm {**}= b^\mathrm *\sqrt 2$. We can show this new number is an integer because $b^\mathrm*\sqrt 2 = (b\sqrt 2 - b) \sqrt 2 = 2b - b\sqrt 2 = 2b-a$.
But now we have a problem. If $b^\mathrm{**} = b^\mathrm*\sqrt 2$ is a positive integer as we've just shown, then $b$ cannot be the smallest positive integer that possesses the defining property of giving an integer when multiplied by $\sqrt 2$. We've just found an even smaller positive integer with the same property. 
We have arrived at a contradiction. Therefore it is not possible to have an integer $b$ and hence $\sqrt 2$ is not rational. 
