How to show this fraction is equal to 1/2?
Mia Lopez 
I have the fraction: $$\frac{\left(2 \left(\frac {a}{\sqrt{2}}\right) + a \right) a} {2(1 + \sqrt{2})a^2}$$
Using Mathematica, I've found that this simplifies to $\frac{1}{2}$, but how did it achieve the result? How can I simplify that fraction to $\frac12$?
$\endgroup$ 45 Answers
$\begingroup$$$ \frac{(2(\frac {a}{\sqrt{2}})+a)a} {2(1+\sqrt{2})a^2}= \frac{(\sqrt{2}a+a)a} {2(1+\sqrt{2})a^2}= \frac{(\sqrt{2}+1)a^2} {2(1+\sqrt{2})a^2}={1\over2}. $$
$\endgroup$ $\begingroup$I think you may have missed that by definition,
$\sqrt{2}\sqrt{2}=2$
And thus,
$\frac{2}{\sqrt{2}}=\sqrt{2}$
This simplification issue is quite common. Of course using this and multiplying out/ factoring terms may get your desired result:
$$=\frac{(\sqrt{2}+1)a^2}{2(\sqrt{2}+1)a^2}$$
In which the $\frac{(\sqrt{2}+1)a^2}{(\sqrt{2}+1)a^2}$ reduces to one in the case $a \neq 0$ .
$\endgroup$ $\begingroup$Assume $a\neq 0$, we have \begin{align*} \frac{\left(2\left(\frac {a}{\sqrt{2}}\right)+a\right)a} {2(1+\sqrt{2})a^2}&=\frac{\left(\frac{2a}{\sqrt 2}+a\right)}{2(1+\sqrt 2)a}\tag 1\\ &=\frac{\left(\frac{2}{\sqrt2}+1\right)a}{2(1+\sqrt 2)a}\tag2\\ &=\frac{\left(\frac{2}{\sqrt2}+1\right)}{(2+2\sqrt 2)}\\ &=\frac{\sqrt 2}{\sqrt 2}\cdot \frac{\left(\frac{2}{\sqrt2}+1\right)}{(2+2\sqrt 2)}\\ &=\frac{2+\sqrt 2}{2\sqrt 2+2\cdot 2}\\ &=\frac{2+\sqrt 2}{2\sqrt 2+4}\\ &=\frac{(2+\sqrt 2)}{2(2+\sqrt 2)}\\ &=\frac{1}{2} \end{align*} where in $(1)$ I cancelled the first $a$ and in $(2)$ I factored out the second $a$ to cancel it also.
$\endgroup$ 5 $\begingroup$Factor out an $a$ on top, multiply top and bottom by $\sqrt 2$. You're left with $$\frac{(2+\sqrt2)a^2}{2(2+\sqrt2)a^2}$$ This is clearly equal to $\frac12$.
$\endgroup$ 0 $\begingroup$You could try setting $a$ to some arbitrary nonzero value and seeing what happens. I try $a = 3$:
$$\frac{\left(2 \left(\frac {3}{\sqrt{2}}\right) + 3 \right) 3} {2(1 + \sqrt{2})3^2} = \frac{9 + 9 \sqrt{2}}{18 + 18 \sqrt{2}}.$$
$\endgroup$ 9
