Why can I simplify radicals? (eg, $\sqrt{153} = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{17}$)

Question

I know it might sound like a ridiculously easy question to answer, but I just can't put two and two together for some reason.

Say for example you have: $$\sqrt{153}$$ You can break it down to $$\sqrt{3}\cdot \sqrt{3}\cdot \sqrt{17}$$ (and of course you can further simplify it, but we'll just leave it at that for now).

Why can I do that?

Answer 1

We have $\sqrt{x} = x^{\frac{1}{2}}$. Therefore you can use the regular computation rules for powers. For example $6^2 = (2 \cdot 3)^2 = 2^2 \cdot 3^2$. Now we do the same for radicals:

$\sqrt{6} = 6^{\frac{1}{2}} = (2 \cdot 3)^{\frac{1}{2}} = 2^{\frac{1}{2}} \cdot 3^{\frac{1}{2}} = \sqrt{2} \cdot \sqrt{3}$

Answer 2

The $nth$ root of a real number $a$ is simply $a^{1/n}$. Hence, the properties of radicals follow directly from properties of exponents.

Answer 3

Do you know these properties: $\left(\sqrt{x}\right)^2=x$ if $x\ge 0$ and $x^2=y^2\implies x=y$ if $x\ge 0$ and $y\ge 0$?

Assuming $a\ge 0$ and $b\ge0$:

$$ \left(\sqrt{a}\cdot\sqrt{b}\right)^2= \left(\sqrt{a}\right)^2\cdot\left(\sqrt{b}\right)^2= a\cdot b. $$

But $$ \left(\sqrt{a\cdot b}\right)^2=a\cdot b. $$

Thus:

$$ \left(\sqrt{a}\cdot\sqrt{b}\right)^2=\left(\sqrt{a\cdot b}\right)^2\implies \sqrt{a}\cdot\sqrt{b}=\sqrt{a\cdot b}. $$ This can be easily extended to situations where you have more than two terms.

Answer 4

Ignoring some (ahem) complexities, we can say, for non-negative(!) $x$,

$\sqrt{x}\;$ is the non-negative(!) number that multiplies by itself to give $\;x$.

With that in mind, consider what happens when we multiply $\sqrt{x}\cdot\sqrt{y}$ by itself:

$$\begin{align} \left(\;\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\;\right)\cdot\left(\;\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\;\right) &= \color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\cdot\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}} \\ &= \color{red}{\sqrt{x}}\cdot\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\cdot\color{blue}{\sqrt{y}} \\ &= \left(\;\color{red}{\sqrt{x}}\cdot\color{red}{\sqrt{x}}\;\right)\cdot\left(\;\color{blue}{\sqrt{y}}\cdot\color{blue} {\sqrt{y}}\;\right) \\ &= \color{red}{x}\cdot\color{blue}{y} \end{align}$$ Since $\sqrt{x}$ and $\sqrt{y}$ are non-negative, so is their product. Therefore, we have shown that

$\sqrt{x}\cdot\sqrt{y}\;$ is the non-negative(!) number that multiplies by itself to give $\;x\cdot y$.

More simply,

$$\sqrt{x\cdot y} = \sqrt{x}\cdot\sqrt{y}$$

The idea extends to show that $$\sqrt{x\cdot y\cdot z} = \sqrt{x}\cdot\sqrt{y}\cdot\sqrt{z}$$ and so forth. (By a completely-analogous argument, this kind of thing works for cube-roots, and fourth-roots, and $n$-th roots in general.) This is precisely what allows you to simplify radicals.

Answer 5

This is a non-trivial question, because when you take square roots in the complex numbers (where it is possible to take the square root of a negative number), the kind of factorization you have done is no longer valid! But if we stick to real numbers, it always works.

Consider what it means when we say that a number $x$ is the square root of a number $y$, that is, $$ x=\sqrt y.$$ It means two things:

1. $x$ is not negative.
2. $x^2=y.$

So let’s see what happens when you factor the number under the square root and take the square roots of the factors. Take $y=3\cdot3\cdot17$ and $x=\sqrt3\cdot\sqrt3\cdot\sqrt{17},$ and let’s see if $x$ is the square root of $y$.

First, the three factors $\sqrt3,$ $\sqrt3,$ and $\sqrt{17},$ are all positive, so their product $x$ is positive, so $x$ is not negative. That satisfies condition 1 above. Check.

Second, \begin{align} x^2 &= \left(\sqrt3\cdot\sqrt3\cdot\sqrt{17}\right)^2\\ &= \sqrt3\cdot\sqrt3\cdot\sqrt{17}\cdot \sqrt3\cdot\sqrt3\cdot\sqrt{17}\\ &= \sqrt3\cdot\sqrt3\cdot\sqrt3\cdot\sqrt3 \cdot\sqrt{17}\cdot\sqrt{17}\\ &= 3 \cdot 3 \cdot 17, \end{align} so $x^2=y,$ satisfying condition 2. Check.

This is just an example of a more general theorem: $\sqrt{ab}=\sqrt a \sqrt b$ when $a$ and $b$ are non-negative real numbers. The proof is much like the reasons for the specific example already shown.

Answer 6

First, you must remember this: most rules involving square roots are valid for positive base only. Subject to that, we have the rule: $$\sqrt{ab}=\sqrt a\sqrt b\,.$$ Why is this valid? Remember also that for positive numbers $x$ and $y$, we have $x=y$ if and only if $x^2=y^2$. This means that to check the displayed formula, all you need to do is square both sides and see whether you get an equality. So to check that $\sqrt{ab}=\sqrt a\sqrt b$, we must verify the claim that $(\sqrt{ab})^2=\bigl[\sqrt a\sqrt b\,\bigr]^2$.

Why’s that true? Look at the right-hand member: $\bigl[\sqrt a\sqrt b\,\bigr]^2=[\sqrt a]^2[\sqrt b]^2$ because $(xy)^2=x^2y^2$. So we need only verify that $\bigl[\sqrt{ab}\,\bigr]^2=(\sqrt a)^2(\sqrt b)^2$. But by the definition of what the square root does, $\bigl[\sqrt{ab}\,\bigr]^2=ab$, and on the right, $(\sqrt a)^2(\sqrt b)^2=(a)(b)$. That does it.

But: Maybe you believed all along that $\sqrt{ab}=\sqrt a\sqrt b$. In that case, your answer is easy. Since $153=9\cdot17$, you get $\sqrt{153}=\sqrt 9\sqrt{17}=3\sqrt{17}$.

Answer 7

First you must define what $\sqrt{b}; b \ge 0$ means.

It means that $\sqrt{b} = c$ where $c \ge 0$ and $c^2 = b$.

This makes a few assumptions. 1) that there actually exists a $c\ge 0$ so that $c^2 = b$ (how do we know that?) and 2) there is only one such number $c$ (how do we know there are not two such numbers?)

If we didn't know either of statements were true it wouldn't make any sense to talk about $\sqrt{b}$ at all.

But if we could convince ourselves that those two statements are true then:

$\sqrt{ab} = c$ so that $c \ge 0;c^2 = ab$; and we know there is exactly one such number and one only; $(\sqrt{a}\sqrt{b})^2 = (\sqrt{a}\sqrt{b})(\sqrt{a}\sqrt{b}) = (\sqrt{a}\sqrt{a})(\sqrt b\sqrt b)=(\sqrt{a})^2(\sqrt{b})^2 = ab$. So $c$ is the only non-negative number which that is true and $\sqrt{a}\sqrt{b}$ is a number for which that is true so.... $c = \sqrt{a}\sqrt{b}$.

That's all.

......

Okay why do we know 1) and 2)?

[If you don't care, stop reading. You are done. The proof was above. $(\sqrt{a}\sqrt{b})^2 = ab$ and so by definition $\sqrt{a}\sqrt{b} = \sqrt{ab}$. End of story.]

[But if you want to know why we know $\sqrt{b}$ exists at all and why it is distinct...read on...]

2) Is easy: Suppose $0 \le c_1 < c_2$ so $c_1 \ne c_2$. Then $c_1^2 = c_1\cdot c_1 < c_1\cdot c_2 < c_2\cdot c_2 = c_2^2$. So $c_1^2$ and $c_2^2$ are not equal and $c_1$ and $c_2$ can not both be square roots of the same number.

1) is ... more abstract.

What is the definition of a real number? Well, Spoiler alert: $\mathbb R$ is a set of numbers so that for every set of $\mathbb R$ that is bounded above, there is a real number that is the least upper bound of the set. That's actually part of the definition.

We showed above that if $0 \le c_1 < c_2$ then $c_1^2 < c_2^2$

Let $S = \{$ all nonnegative numbers that when squared or less than $b\}$.

As there are positive numbers that when squared are bigger than $b$ and those numbers are bigger than all nonegative numbers that when squared are smaller than $b$ that $S$ is bounded above.

(Note [a]: this means if $c > 0$ and $c^2 > b$ then $c$ is an upper bound of $S$.)

Let $k =$ the least upper bound of $S$.

What is $k^2$? Either $k^2 < b$ or $k^2 = b$ or $k^2 > b$.

If $k^2 < b$ the $b-k^2 > 0$. Let $e$ be so that $0 < e < \min(\frac {b-k^2}{2k+1}, 1)$.

$0< e < 1$ so $e^2 < e$.

And $e < \frac {b-k^2}{2k+1}$ so

$(k+e)^2 = k^2 + 2ke + e^2 < k^2 + 2ke + e = k^2 + e(2k+1)< k^2 + (b-k^2) = b$.

So $k+e \in S$. But $k < k+e \in S$ which contradicts that $k$ is an upper bound of $S$.

So $k^2 \ge b$.

We do similar thing if $k^2 > b$. Then $b - k^2 > 0$ and if $e$ is so that $0 < e < \min \frac {b-k^2}{2k}$ then

$(k-e)^2 = k^2 -2ke + e^2 > k^2 - 2ke > k^2 -(k^2 -b) = b$.

So $k-e$ is an upper bound of $S$ (See note [a]). But that contradicts that $k$ is the least upper bound.

So $k^2 \le b$ and $k^2 \ge b$ so $k^2 = b$.

And so $k =\sqrt b$ does exist.