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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 $\frac12$?

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Factor out an $a$ on top, multiply top and bottom by $\sqrt 2$. It's really not complicated. – Matt Samuel Jan 16 at 22:23
Just multiply everthing by $a$. Also $\frac 2{\sqrt 2} = \sqrt 2$ – Ant Jan 16 at 22:23
Despite it not being complicated I don't really understand the downvotes. It doesn't necessarily mean the OP didn't try. – Matt Samuel Jan 16 at 22:30
@MattSamuel Post that as an answer and I'll accept it. – user305676 Jan 16 at 22:38
Of course there is the implicit assumption that $a\ne 0$ ... – Hagen von Eitzen Jan 17 at 12:05
up vote 6 down vote accepted

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 $\frac12$.

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$$ \frac{(2(\frac {a}{\sqrt{2}})+a)a} {2(1+\sqrt{2})a^2}= \frac{(\sqrt{2}a+a)a} {2(1+\sqrt{2})a^2}= \frac{(\sqrt{2}+1)a^2} {2(1+\sqrt{2})a^2}={1\over2}. $$

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I think you may have missed that by definition,


And thus,


This simplification issue is quite common. Of course using this and multiplying out/ factoring terms may get your desired result:


In which the $\frac{(\sqrt{2}+1)a^2}{(\sqrt{2}+1)a^2}$ reduces to one in the case $a \neq 0$ .

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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.

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Whie I don't see any errors, there are a lot of redundant steps here. For example, you distribute the 2 in the denominator just to factor it back out a few steps later. Also, you don't need to multiply and divide by $\sqrt 2$ since $2/\sqrt 2=\sqrt 2. – Sean English Jan 16 at 23:13
@SE318 I'm well aware of all of that. Since we don't know the particular level of OP's understanding, I decided to write out as much detail as possible. To you it's redundant, but clearly that was not redundant to OP. – probablyme Jan 16 at 23:17
I suppose you are right. Even then though, I would still not distribute the $2$ in the denominator. I think it would be more instructive to rewrite the $2$ in the denominator as a $1/2$ multiplied by the whole thing. Since we want to show it is $1/2$, that seems like a natural strategy because then you just need to figure out why the rest is 1. In my opinion, one way suggests attacking the problem with a strategy, the other way seems to be doing whatever can be legally done which I only advise when you've exhausted other strategies and are hoping to see something you didn't before. – Sean English Jan 16 at 23:28
Or another way to put it, if a student asked you why you distributed the 2, what would your explanation be? – Sean English Jan 16 at 23:30
@SE318 Yup, that's the goal. I never claimed it was the fastest, most efficient or instructive way to do. I just needed to get to the right answer in some manner reasonable manner. And I did not feel like using multiplying by any fractions I felt like OP would have noticed that already. It is clear that OP did since otherwise OP would have noticed and accepted an answer using $2/\sqrt 2 = \sqrt 2$. I got the checkmark initially but then OP gave it to the summary of my answer. – probablyme Jan 16 at 23:50

You could try setting $a$ to some arbitrary nonzero value and seeing what happens. I try $a = 3$:

$$\frac{\left(2 \left(\frac {3}{\sqrt{2}}\right) + 3 \right) 3} {2(1 + \sqrt{2})3^2} = \frac{9 + 9 \sqrt{2}}{18 + 18 \sqrt{2}}.$$

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9+$9\sqrt2$ in the numerator? – Nikunj Jan 16 at 22:43
The upvote is curious because this doesn't solve the problem and also contains an obvious error. – Matt Samuel Jan 16 at 22:44
A statement being true for some $a$ does not imply it's true for all $a$. – MathematicsStudent1122 Jan 16 at 22:45
@Matt You've never mistaken a 9 for a 6? – Robert Soupe Jan 17 at 2:37
@MathematicsStudent1122 That's a much more valid point. Still, if the asker tried it, he might learn something. – Robert Soupe Jan 17 at 2:39

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