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Can someone prove this basic identity?

by Synopoly
Tags: basic, identity, prove
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Synopoly
#1
Feb16-14, 01:11 AM
P: 1
x^(1/n) = the nth root of x

(I'd use mathematical notation but I don't really know how I'm new sorry)
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Mentallic
#2
Feb16-14, 06:19 AM
HW Helper
P: 3,531
Try raising both sides to the power of n.
lurflurf
#3
Feb16-14, 03:36 PM
HW Helper
P: 2,264
$$x^{\frac{1}{n}}=\sqrt[n]{x}$$
is usually true by definition as those are two notations for the same function.

Mark44
#4
Feb16-14, 06:32 PM
Mentor
P: 21,258
Can someone prove this basic identity?

Quote Quote by lurflurf View Post
$$x^{\frac{1}{n}}=\sqrt[n]{x}$$
is usually true by definition as those are two notations for the same function.
I agree.
rama
#5
Mar13-14, 06:01 AM
P: 9
same as what mentallic said,the n(and root n) gets cancelled and you are left with the original number
lhs=rhs
jing2178
#6
Mar16-14, 12:19 PM
P: 41
Quote Quote by lurflurf View Post
$$x^{\frac{1}{n}}=\sqrt[n]{x}$$
is usually true by definition as those are two notations for the same function.
Not so sure that you can just say they are two notations for the same function.

You can define $$\sqrt(x)$$ as that number when multiplied by itself is $$x$$, ie square root is inverse of squaring.

However when dealing with powers you cannot just say well $$x^{\frac{1}{2}}$$ is a new notation for square root, it is the extension of the notation of powers from positive integers to rationals and its interpretation has yet to be determined.
Using the usual rules for powers then

$$x^{\frac{1}{2}}x^{\frac{1}{2}}=x^{{\frac{1}{2}}+{\frac{1}{2}}} =x^1=x$$

it follows that

$$x^{\frac{1}{2}}=\sqrt(x)$$


In the same way you cannot say $$\frac{8}{12}$$ is just another notation for $$\frac{2}{3}$$ without demonstrating their equality
micromass
#7
Mar16-14, 12:49 PM
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P: 18,240
Quote Quote by jing2178 View Post
Not so sure that you can just say they are two notations for the same function.

You can define $$\sqrt(x)$$ as that number when multiplied by itself is $$x$$, ie square root is inverse of squaring.

However when dealing with powers you cannot just say well $$x^{\frac{1}{2}}$$ is a new notation for square root, it is the extension of the notation of powers from positive integers to rationals and its interpretation has yet to be determined.
Using the usual rules for powers then

$$x^{\frac{1}{2}}x^{\frac{1}{2}}=x^{{\frac{1}{2}}+{\frac{1}{2}}} =x^1=x$$

it follows that

$$x^{\frac{1}{2}}=\sqrt(x)$$


In the same way you cannot say $$\frac{8}{12}$$ is just another notation for $$\frac{2}{3}$$ without demonstrating their equality
Every notation must have a definition. How do you define ##x^{1/n}## and ##\sqrt[n]{x}##? Usually, these are defined to coincide.
DrewD
#8
Mar16-14, 01:54 PM
P: 446
I personally would think of the rules for exponent to be defined so that ##\{a^x|x\in\mathbb{R}\}## with ##a## being some constant that is greater than zero and not 1, is a group under multiplication. From this, like jing2178 shows, it is easy to prove that ##x^{1/n}## is a number such that, when raised to the ##n^{th}## power, one gets ##x##. From there it is clear that the notation ##\sqrt[n]{x}## and ##x^{1/n}## mean the same thing (ignoring discussions of principle roots).

It is equally valid to set the definition that those are the same notation and then show that the exponent rules work out to keep the group structure. To me the other way makes me feel happier inside
micromass
#9
Mar16-14, 02:41 PM
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P: 18,240
Quote Quote by DrewD View Post
I personally would think of the rules for exponent to be defined so that ##\{a^x|x\in\mathbb{R}\}## with ##a## being some constant that is greater than zero and not 1, is a group under multiplication. From this, like jing2178 shows, it is easy to prove that ##x^{1/n}## is a number such that, when raised to the ##n^{th}## power, one gets ##x##. From there it is clear that the notation ##\sqrt[n]{x}## and ##x^{1/n}## mean the same thing (ignoring discussions of principle roots).
That's actually a neat definition
qspeechc
#10
Mar16-14, 03:00 PM
P: 792
$$(x^{1/n})^n = x$$ from which the result follows. E.g. $$\sqrt{4}=2$$ because 2x2=4. $$\sqrt[3]{8}=2$$ because 2x2x2=8.

So if $$(x^{1/n})^n$$ means $$x^{1/n}$$ multiplied by itself n times, and the result gives x, then it must mean $$x^{\frac{1}{n}}=\sqrt[n]{x}$$


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