# Factoring square roots

#### NeedingWD40

##### New member
I know that one can factor square roots, such as √8 = √(4x2) = √4 x √2 = 2√2 but I don't understand why.

we are saying a1/2.b1/2 = (ab)1/2 which doesn't feel intuitive....

can someone help explain?

#### Subhotosh Khan

##### Super Moderator
Staff member
I know that one can factor square roots, such as √8 = √(4x2) = √4 x √2 = 2√2 but I don't understand why.

we are saying a1/2.b1/2 = (ab)1/2 which doesn't feel intuitive....

can someone help explain?
Law of exponent tells you:

(a * b)m = am * bm

#### NeedingWD40

##### New member
Law of exponent tells you:

(a * b)m = am * bm
Thanks - and I can understand this using non-fractional indices as, say (2*4)3 = (2*4)(2*4)(2*4) , then rearrange to get 24*43 but I don't see this with square roots...

#### Mr. Bland

##### Junior Member
Recall that $$\displaystyle \sqrt[y]{x} = x^{\frac{1}{y}}$$. This works the same as any other exponent.

EDIT:
If you're confused about "what does a non-integer exponent mean", that's understandable because it's not intuitive given the usual description of an exponent. What does it mean to multiply something by itself three and a half times? This can be written as $$\displaystyle x^{3\frac{1}{2}} = x^{\frac{7}{2}} = \sqrt[2]{x^7}$$. Eventually it boils down to a matter of calculating a point on a curve, which has points in between integer values.

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#### JeffM

##### Elite Member
I know that one can factor square roots, such as √8 = √(4x2) = √4 x √2 = 2√2 but I don't understand why.

we are saying a1/2.b1/2 = (ab)1/2 which doesn't feel intuitive....

can someone help explain?
This is a GREAT question.

$$\displaystyle \text {BY DEFINITION, } x * x = y \iff x = \sqrt{y}.$$

OK with that?

$$\displaystyle \text {Given } a = \sqrt{p}, \text { and } b = \sqrt{q}.$$

$$\displaystyle \therefore a * b = \sqrt{p} * \sqrt{q}.$$

$$\displaystyle \therefore a * (a * b) = \sqrt{p} * (\sqrt{p} * \sqrt{q}) = (\sqrt{p} * \sqrt{p}) * \sqrt{x} = p * \sqrt{q} \implies$$

$$\displaystyle (p * \sqrt{q}) * \sqrt{q} = (a * (a * b)) * b \implies p * (\sqrt{q} * \sqrt{q}) = (a * (b * a)) * b \implies$$

$$\displaystyle p * q = ((a * b) * a) * b = (a * b) * (a * b) \implies (a * b) * (a * b) = p * q.$$

$$\displaystyle \therefore a * b = \sqrt{p * q}.$$

$$\displaystyle \text {But } a * b = \sqrt{p} * \sqrt{q}.$$

$$\displaystyle \text {THUS, } \sqrt{p} * \sqrt{q} = \sqrt{p * q}.$$

That is abstract. Let's take a numeric example.

$$\displaystyle \sqrt{25} * \sqrt{49} = 5 * 7 = 35.$$

$$\displaystyle 35^2 = 30^2 + 2(30)(5) + 5^2 = 900 + 300 + 25 = 1225 = 1250 - 50 = 25(50 - 1) = 25 * 49.$$

$$\displaystyle \therefore \sqrt{25 * 49} = \sqrt{35^2} = 35.$$

$$\displaystyle \therefore \sqrt{25} * \sqrt{49} = \sqrt{25 * 49}.$$

This result can be extended to any root.

#### NeedingWD40

##### New member
Yay! Thank you. I could follow it, I don't think I could reproduce it yet, but I could follow the steps so that's a good start and I certainly feel more convinced. Thank you very much.

#### NeedingWD40

##### New member
Recall that $$\displaystyle \sqrt[y]{x} = x^{\frac{1}{y}}$$. This works the same as any other exponent.

EDIT:
If you're confused about "what does a non-integer exponent mean", that's understandable because it's not intuitive given the usual description of an exponent. What does it mean to multiply something by itself three and a half times? This can be written as $$\displaystyle x^{3\frac{1}{2}} = x^{\frac{7}{2}} = \sqrt[2]{x^7}$$. Eventually it boils down to a matter of calculating a point on a curve, which has points in between integer values.
Thanks Mr Bland. I am happy about the definition of non-integer exponents, and I like your link to visualising them as points on a curve - very helpful. I just couldn't 'feel' why what worked for integer exponents would also work for non-integer. JeffM helped too!

#### JeffM

##### Elite Member
Yay! Thank you. I could follow it, I don't think I could reproduce it yet, but I could follow the steps so that's a good start and I certainly feel more convinced. Thank you very much.
Couple of things.

A mathematician observes relationships such as

$$\displaystyle \sqrt{1} * \sqrt{169} = 1 * 13 = 13 = \sqrt{169} = \sqrt{1 * 169}.$$

$$\displaystyle \sqrt{4} * \sqrt{9} = 2 * 3 = 6 = \sqrt {36} = \sqrt{4 * 9}$$

$$\displaystyle \sqrt{9} * \sqrt{25} = 3 * 5 = 15 = \sqrt{225} = \sqrt{9 * 15}$$

That gets a mathematician wondering under what conditions this very simple relationship is true. In other words, what frequently happens is that something is true in specific cases, and we try to find out what is the general case such that that something is true. You probably did not notice that I restricted the discussion to non-negative real numbers. The reason is that the relationship is not so simple if we extend it beyond non-negative real numbers. There is a similar but more complex relation that applies more generally.

The point is: before trying to develop or understand a general proof, try some very simple examples.

Number two. Detailed proofs in algebra are frequently hard to understand because they are lengthy strings of blindingly obvious steps. The mind goes: of course, of course, of course ... WHAT how did we get here. If it is hard even to follow the proofs, do not worry that you cannot yet create such proofs.

Finally, rational exponents are fairly easy to understand with this definition

$$\displaystyle \text {Given: } r \in \mathbb R, \ a,\ c. \in \mathbb Z,\ r > 0,\ a \ge 0, \text { and } c \ge 1, \text { then}$$

$$\displaystyle r^a \equiv (r^{(a/c)})^c.$$

If you think about this definition, you will realize that it simply means (under the conditions given)

$$\displaystyle r^{(a/c)} \equiv \sqrt[c]{r^a}.$$

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#### NeedingWD40

##### New member
Finally, rational exponents are fairly easy to understand with this definition

$$\displaystyle \text {Given: } r \in \mathbb R, \ a,\ c. \in \mathbb Z,\ r > 0,\ a \ge 0, \text { and } c \ge 1, \text { then}$$

$$\displaystyle r^a \equiv (r^{(a/c)})^c.$$

If you think about this definition, you will realize that it simply means (under the conditions given)

$$\displaystyle r^{(a/c)} \equiv \sqrt[c]{r^a}.$$
Thank you very much for those pointers - and for the encouragement (yet!). I will keep your advice in mind. I think my big problem is the meeting in the middle - you know where to start, and where you want to finish, it's the meeting in the middle that I can't find!

The rational exponents explanation is delightful. That was fun. I first sat and carefully thought through: Why that condition on that variable? Ok, zero raised to any power is zero, so wouldn't it still work for zero? But then I thought, except for raising to the power zero, 00 = 1... so does it still work?
02/3 = $$\displaystyle \sqrt[3]{0^2}$$ = 0
but
00/3 = $$\displaystyle \sqrt[3]{0^0}$$ = $$\displaystyle \sqrt[3]1$$ = 1 and if we cube both sides
(00/3)3 = 01=0
but
($$\displaystyle \sqrt[3]{0^0}$$)3 = 00 = 1.

So that's why zero was excluded?

#### Cubist

##### Full Member
So that's why zero was excluded?
JeffM's constraints work, logically they are correct, but they don't show the full range of numbers for which the identity holds. It works under other circumstances too.

See this post about "power of power" and contrainsts (click)

I'm guessing that it would become increasingly difficult to mathematically prove that this property holds under all of these different circumstances.

#### JeffM

##### Elite Member
Thank you very much for those pointers - and for the encouragement (yet!). I will keep your advice in mind. I think my big problem is the meeting in the middle - you know where to start, and where you want to finish, it's the meeting in the middle that I can't find!

The rational exponents explanation is delightful. That was fun. I first sat and carefully thought through: Why that condition on that variable? Ok, zero raised to any power is zero, so wouldn't it still work for zero? But then I thought, except for raising to the power zero, 00 = 1... so does it still work?
02/3 = $$\displaystyle \sqrt[3]{0^2}$$ = 0
but
00/3 = $$\displaystyle \sqrt[3]{0^0}$$ = $$\displaystyle \sqrt[3]1$$ = 1 and if we cube both sides
(00/3)3 = 01=0
but
($$\displaystyle \sqrt[3]{0^0}$$)3 = 00 = 1.

So that's why zero was excluded?
JeffM's constraints work, logically they are correct, but they don't show the full range of numbers for which the identity holds. It works under other circumstances too.

See this post about "power of power" and contrainsts (click)

I'm guessing that it would become increasingly difficult to mathematically prove that this property holds under all of these different circumstances.
It is tempting to adopt, and many mathematicians feel no difficulty in adopting, a Platonist view of mathematics. (Platonism taken to its ultimate conclusion is the philosophy that your mother has never really existed but is merely the shadow of the one real mother that no living human has ever known.) In that view, the human mind discovers mathematics. The other view, the one that I prefer, is that the human mind creates mathematics, in part by assuming that observable regularities in the physical world are invariably true in the idealized world of the intellect. The Platonists assume that the invisible realm is logically consistent and thus discoverable by reason. The non-Platonist requires that what we create be logically consistent. In other words, both agree that valid mathematics is logically consistent, and thus the two schools can live in reasonable amicability. For non-Platonists like me, definitions are free exercises of the mind and judged solely on utility and consistency with other definitions.

Notice that I did not define $$\displaystyle r^a.$$ Obviously my definition of a rational exponent depends on my definition of an integer exponent so what I said before was incomplete. Here is a set of definitions that I like for integer exponents

$$\displaystyle \text {Given } r \in \mathbb R, \ a \in \mathbb Z, \text { and } r > 0, \text { then}$$

$$\displaystyle a = 0 \implies r^a \equiv 1, \text { and } r^a \equiv r * r^{(a - 1)}.$$

Now why did I pick that definition?

Mainly because it has the very nice consequence that any integer power of r is a real number. If I allowed r to be 0, then negative powers would not be real numbers. Moreover if I allowed r to be less than 0, then some rational exponents would not be real numbers.

Thus, the reason that I stipulated that r > 0 for rational exponents is that I had implicitly stipulated that for integer exponents.

In fact, if you are willing to restrict exponents to non-negatice rational numbers, there is no reason to restrict r to positive real numbers; it will work just fine to restrict r to non-negative real numbers. In particular, if r is a real number, q is a rational number, and both are non-negative, then no problem arises from a definition of r^q that entails that 0^0 = 1. However, mathematicians may want to work with real functions to a power that is a real function. In that case, allowing both functions to be zero simultaneously can cause problems in the field of mathematics known as analysis. Thus, some mathematicians define things so 0^0 is not defined, and some mathematicians define things so that 0^0 = 1. For more details, see

As cubist correctly said, we can define exponents so that they make sense and are useful with fewer restrictions than I gave. If we do so, however, the laws of exponents become more complex. If we define exponents in a way that restricts r to the positive reals, we get a set of laws for exponents with no exceptions. In practice, I use exponents with numbers that are not positive reals only after checking that I am not about to be bitten by the exceptions.

#### Mr. Bland

##### Junior Member
The non-Platonist requires that what we create be logically consistent.
The non-Platonist has never seen the company dress code.

#### Otis

##### Elite Member
The non-Platonist has never seen the company dress code.
Is that because the company's dress doesn't really exist, but is merely a reflection of the one, true dress that no human has ever worn? Philosophy is so confusing …

$$\;$$

#### pka

##### Elite Member
When I am confronted with questions about the reality of mathematics, I ask the person if s/he has read THE NUMBER SENSE how he mind creates mathematics. by Stanislas Dehaene? Dehaene is a well known brain scientist who divides his time between University of Paris and various universities in North America.