Simplifying the surd $\sqrt{3 - \sqrt{8}}$

Recently I was solving a math problem and I came across the following ( Only part of the problem ):

$$\sqrt{3 - \sqrt{8}}$$

Here's is what I did to simplify the above:

$$(a-b)^2=a^2+b^2-2ab$$ $$\sqrt{1+2-2\sqrt{2}}$$ $$\sqrt{(1-\sqrt{2})^2}$$ $$=1-\sqrt{2}$$

When you expand $$(1-\sqrt{2})^2$$ you do get $$3 - \sqrt{8}$$. However, I found that upon the expansion of $$(\sqrt{2}-1)^2$$ I also arrive at the same answer as before $$3 - \sqrt{8}$$.

Yet the in the solution of the problem I was doing it required me to use the latter $$(\sqrt{2}-1)^2$$ to find the correct answer. $$(1-\sqrt{2})^2$$ did not help to solve the problem. Why is this the case? Why does the latter only hold true?

I suspect this might have something to do with the square root function only returning a positive value. Might this be the case? If so, could someone explain why a square root function can only return a positive value. My working above seems to give a negative answer ( As $$\sqrt{2}>1 )$$ in the square root function, but I can't seem to find what I'm doing wrong.

• If $x$ is a real number, $\sqrt{x^2} = |x|$. In particular, $\sqrt{(1 - \sqrt{2})^2} = |1 - \sqrt{2}| = \sqrt{2} - 1$.
– kobe
Commented Feb 6, 2023 at 2:02
• Why is it defined to be like this? Also is there some general method to find the absolute value? How do we work this out |1−√2|=√2-1 ? Commented Feb 6, 2023 at 2:40
• Definition of absolute value.$$1<2\implies\sqrt1=1<\sqrt2\implies|1-\sqrt2|=-(1-\sqrt2)=\sqrt2-1$$ Commented Feb 6, 2023 at 2:46
• It can be confusing: in English, you might hear it said that $5$ is the square root of $25$, but $-5$ is a square root of $25$. To a non-native speaker, it is not immediately obvious what the difference is. But when the symbolic form $\sqrt{25}$ is used, it means, by definition, the non-negative square root of $25$. So $x^2$ is equal to the absolute value $|x|$ of $x$. Hence $\sqrt{3 - \sqrt{8}}=|1-\sqrt 2|=\sqrt 2-1$. Commented Feb 6, 2023 at 15:44
• @HaowenXie it's defined like this to avoid confusion, so if you want to refer to the negative root, you write $-\sqrt a$ or if you want to refer to them at once, you write $\pm \sqrt a$. Suppose you're solving an equation to find the number of students in a class, and you end up with, say $\sqrt{484}$. You're not going to say there are $-22$ students, right? The positive square root is required most of the time, like for calculating distance, the magnitude of vectors, sides of triangles, etc. so to avoid confusion, the square-root function is defined that way.
– D S
Commented Feb 10, 2023 at 5:26

$$(1-\sqrt{2})^2$$

I found that upon the expansion of $$(\sqrt{2}-1)^2$$ I also arrive at the same

This is not surprising, since $$(-a)^2=(-1)^2\times a^2=a^2.$$

$$\sqrt{3 - \sqrt{8}}$$ $$=\sqrt{(1-\sqrt{2})^2}\\\color{red}{=1-\sqrt2}$$ it required me to use $$(\sqrt{2}-1)^2$$ to find the correct answer. $$(1-\sqrt{2})^2$$ did not help, seems to give a negative answer

Not true: the second line that I quoted above is perfectly fine, however, you can replace that erroneous final line with $$=\sqrt{(1-\sqrt{2})^2}\\=\sqrt{(\sqrt{2}-1)^2}\\=\sqrt{2}-1.$$ Here, you were falsely believing that $$\color{red}{\text{regardless of whether a is negative or nonnegative, \sqrt{a^2}=a}}.$$ But $$\sqrt{(-5)^2}=\sqrt{25}=5\ne-5.$$

If so, could someone explain why a square root function can only return a positive value.

We've established that $$\sqrt{\quad}$$ is never negative. It is always just a nonnegative value because it means "the nonnegative square root" instead of "the square roots". This is because the $$n$$th-root (surd) symbol $$\sqrt[n]{\quad}$$ is defined to mean "the principal $$n$$th root" instead of "the $$n$$th roots".

This is just to avoid the untidiness of dealing with multiple-valued expressions. If we really need to consider both square roots of $$25,$$ we can write $$\pm\sqrt{25},$$ like in the quadratic formula. On the other hand, if we were to let $$\sqrt{25}=\pm5,$$ then how do we symbolically indicate that we want to pick just the negative (or just the positive) square root?

Summary: $$\pm5\ne\sqrt{(-5)^2}\ne-5,\\\sqrt{a^2}=|a|.$$

In the complex world, principal root has no universal definition and $$\sqrt[3] {-1}$$ could mean either $$e^{i \frac\pi3}$$ (smallest nonnegative argument) or $$-1$$ (real), so it is common to allow surd symbols only inputs from $$[0,\infty)$$. If you adopt this convention (in which case principal root just means nonnegative root), then $$\sqrt[n]{a^n}\equiv|a|\quad\quad(n\in\mathbb N).$$

• My A was to show the general method find the simplified value, if it exists, mainly for the benefit of other readers. Commented Feb 22, 2023 at 6:28
• Hello @DanielWainfleet. What's the context of this out-of-the-blue announcement? -) Commented Feb 22, 2023 at 6:31

We have $$\sqrt {3-\sqrt 8 }=\sqrt {3-2\sqrt 2 }.$$ We ask whether there exist rational $$a,b$$ such that that $$3-2\sqrt 2=(a+b\sqrt 2)^2=(a^2+2b^2)+2ab\sqrt 2.$$ Since $$a,b\in\Bbb Q$$, this requires $$(\bullet) \quad a^2+2/a^2=3 \land -2ab=2.$$ So $$b=-1/a$$ from the 2nd equation of $$(\bullet).$$ So $$3=a^2+2b^2=a^2+2/a^2.$$

Let $$a^2=x.$$ Then $$3=x+2/x.$$ So $$x^2-3x+2=0.$$ Take the solution $$x=1.$$ Now if $$a=-1$$ then $$a^2=1=x,$$ and with $$b=-1/a=1,$$ we see that $$(\bullet)$$ must hold.