#### NeedingWD40

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*why.*

we are saying a

^{1/2}.b

^{1/2}= (ab)

^{1/2}which doesn't feel intuitive....

can someone help explain?

- Thread starter NeedingWD40
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we are saying a

can someone help explain?

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Law of exponent tells you:why.

we are saying a^{1/2}.b^{1/2}= (ab)^{1/2}which doesn't feel intuitive....

can someone help explain?

(a * b)

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Thanks - and I can understand this using non-fractional indices as, say (2*4)Law of exponent tells you:

(a * b)^{m}= a^{m}* b^{m}

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.

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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This is awhy.

we are saying a^{1/2}.b^{1/2}= (ab)^{1/2}which doesn't feel intuitive....

can someone help explain?

Let's start with definitions and deal with non-negative real numbers.

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

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.

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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!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.

Couple of things.feelmore convinced. Thank you very much.

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

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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Thank you very much for those pointers - and for the encouragement (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)

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

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, 0

0

but

0

(0

but

(\(\displaystyle \sqrt[3]{0^0}\))

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.So that's why zero was excluded?

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

I'm guessing that it would become increasingly difficult to mathematically

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, 0^{0}= 1... so does it still work?

0^{2/3}= \(\displaystyle \sqrt[3]{0^2}\) = 0

but

0^{0/3}= \(\displaystyle \sqrt[3]{0^0}\) = \(\displaystyle \sqrt[3]1\) = 1 and if we cube both sides

(0^{0/3})^{3}= 0^{1}=0

but

(\(\displaystyle \sqrt[3]{0^0}\))^{3}= 0^{0}= 1.

So that's why zero was excluded?

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 mindJeffM'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 mathematicallyprovethat this property holds under all of these different circumstances.

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

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

en.m.wikipedia.org

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.

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

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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 …The non-Platonist has never seen the company dress code.

\(\;\)

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