Proving $\,2\sqrt 2 + 1\,$ is irrational by contradiction

I am working on some review questions for my Discrete Structures final and I needed some assistance for the problem "Prove by contradiction that $\,2\sqrt 2 +1$ is irrational".

Now I know how prove that $\sqrt 2$ is irrational on its own, by showing that $\sqrt 2 = a/b$ and showing they are not in lowest terms. My issue is that other stuff is messing up my proof and I am not sure how to solve it.

Here is what I have so far. Assume that $\,2\sqrt 2 +1$ is rational, therefore $\,2\sqrt 2 +1 =$ $\frac {a}{b}$. Then: $$2 \sqrt 2 +1 = \frac {a}{b}$$ $$\sqrt 2 = \frac {a-b}{2b}$$ $$a = 2(b \sqrt 2 + \frac{b}{2})$$ $$a= 2k$$ $$\sqrt 2 = \frac{2k-b}{2b}$$

And since they are not in lowest terms therefore $\sqrt 2$ is irrational.

Any help would greatly appreciated. Thanks.

• Your expression $\sqrt{2}=\frac{a-b}{2b}$ shows that then it would be rational, so that gives your contradiction. – user84413 Apr 5 '15 at 23:37
• Well, if you already know that $\sqrt{2}$ is irrational, aren't you done on line 2 ... ? – Neal Apr 5 '15 at 23:38
• Why do you think that $\ k = b\sqrt 2 + b/2\$ is an integer? – Bill Dubuque Apr 6 '15 at 0:08
• I dont know that $\sqrt \2$ is irrational in the problem. I was just saying that I know how to prove it is when it is by itself. Also I just made 2(b $\sqrt \2$ + $\frac{b}{2}$) = k so that I could show they aren't in lowest terms in the next step. – stalindroid Apr 6 '15 at 0:34

If you already know that $\sqrt{2}$ is irrational, because you know how to prove this fact, then having written $$\sqrt{2}=\frac{a-b}{2b}$$ is sufficient to end the proof, because the right hand side is rational by construction.
There's nothing wrong in showing something by reducing it to the proof of another fact that you already know how to do. Indeed, this is a normal way of doing things in mathematics. Since proving $\sqrt{2}$ is irrational is easy, you can use this for proving that, whenever $x$ and $y$ are rational and $x\ne0$, then $x\sqrt{2}+y$ is irrational too: from $x\sqrt{2}+y=r$ you get $$\sqrt{2}=\frac{r-y}{x}$$ and, if $r$ is rational, also $(r-y)/x$ is rational.
You can prove it in other ways, though. If $u=2\sqrt{2}+1$, then $u-1=2\sqrt{2}$, so $u^2-2u+1=8$ and so $$u^2-2u-7=0$$ A rational root of the polynomial $X^2-2X-7$ has to be among $1$, $-1$, $7$ and $-7$, but none of these numbers is a root: $$1^2-2\cdot1-7=-8\\ (-1)^2-2\cdot(-1)-7=-4\\ 7^2-2\cdot 7-7=28\\ (-7)^2-2\cdot(-7)-7=56$$