Why is $\sqrt{a^2+b^2}\neq a+b$, and is there another rule to simplify the square root? So I have $\sqrt{a^2+b^2}$. I thought that this was equal to $a^2+b^2$ but it is not. However, even if I convert the square root to powers, I get (based on the power rule $(a^m)^n = a^{mn}$), I get $(a^2+b^2)^{0.5} = a^1 + b^1$ but this is still not true...
Why is this, and is there any other rule for simplifying $\sqrt{a^2+b^2}$?
 A: Actually, $\sqrt{a^2+b^2}$ is the simplest form already.
A: It is true that (with some restriction):
$$ (a^m)^n = a^{mn}$$
It is also true that:
$$ [(ab)^{m}]^n = [a^{m}.b^m]^n=[ab]^{mn}$$
You say:

$(a^2+b^2)^{0.5} = a^1 + b^1$

However This is not a general rule when you have "addition" operation raised to a power. In this specific case it is true at least when $a=b=0$
In case you have $(x+y)^m$, where $m$ is an positive integer, there is an expansion for this using the Binomial Theorem.
In case you have $(x+y)^r$, where $r$ is not an integer, there is an infinite series for this case using several approximation methods such as Taylor Expansion. There is also a binomial expansion for Fractional Exponents. 
A: In power rule that you mentioned, namely $(a^m)^n=a^{mn}$, $a^m$ is a single number, whereas in $(a^2+b^2)^{0.5}$, the 0.5'th power is applied to a sum, so this is a different case.
To see that $\sqrt{a^2+b^2}=a+b$ is really false, find a counterexample. Take a=3 and b=4 for example.
A: Your first attempt amounts to
$$\sqrt s=s$$ which is obviously wrong.
Your second attempt does not fit with the power rule.
$$\sqrt{s^2}=s^{1/2\cdot2}=s$$ would be right, but is not what you considered.

Now have a look at
$$\sqrt{1+t^2}$$ and try to somehow relate it to $t$.

A: It is very tempting to assume that $(a+b)^{2}$ is equal to $a^{2} + b^{2}$, when, in fact, it is not. 
Thus $\sqrt{a^2+b^2}$ is not equal to $a+b$. $\sqrt{a^2 + b^2}$ is about as simplified as you can go.
A: I would like to start with a principle that should be taught in all schools:
In Mathematics, Nothing Is True unless there’s a proof that it’s true.
There are a lot of formulas that look very pretty and seem very reasonable, and are true besides, like $(ab)^n=a^nb^n$, but you should have been shown in school why that formula is true.
You were undoubtedly hoping that the equally pretty and reasonable formula $(a+b)^n=a^n+b^n$ would be true, but there’s no proof for this. In fact, it’s false, but we have something much better, a formula with a stern and crystalline beauty of its own, called the Binomial Theorem:
$$
(a+b)^n=a^n+na^{n-1}b + \frac{n(n-1)}2a^{n-2}b^2+\cdots+\frac{n!}{(n-j)!j!}a^{n-j}b^j+\cdots+nab^{n-1}+b^n\,,
$$
valid when $n$ is a positive whole number.
The moral of my sermon? Don’t ask why an equation or formula isn’t true, because most are not true. Rather, ask what is true.
A: Answering the titular questions, which are very clear:


*

*The reason that $\sqrt{a^2+b^2}\ne a +b$ is that when you square $a+b$ you don't get back $a^2+b^2$ as you should have if indeed that was the square root. In fact, you have instead $(a+b)^2=a^2+2ab+ b^2,$ which is off by the term $2ab.$

*Well, there's no other way given all we know that we can simplify $\sqrt{a^2+b^2}$ further. It's the square root of a sum of two squares, and if $a$ and $b$ are positive it represents the length of the hypotenuse of a right triangle with legs of lengths $a$ and $b.$
