Can someone prove this basic identity?

  • Thread starter Synopoly
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In summary, x^(1/n) is equivalent to the nth root of x, as both notations represent the same function by definition. This can be demonstrated by raising both sides to the power of n, which simplifies to x on both sides. Therefore, x^(1/n) and the nth root of x are interchangeable notations for the same function.
  • #1
Synopoly
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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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  • #2
Try raising both sides to the power of n.
 
  • #3
$$x^{\frac{1}{n}}=\sqrt[n]{x}$$
is usually true by definition as those are two notations for the same function.
 
  • #4
lurflurf said:
$$x^{\frac{1}{n}}=\sqrt[n]{x}$$
is usually true by definition as those are two notations for the same function.
I agree.
 
  • #5
same as what mentallic said,the n(and root n) gets canceled and you are left with the original number
lhs=rhs
 
  • #6
lurflurf said:
$$x^{\frac{1}{n}}=\sqrt[n]{x}$$
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
 
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  • #7
jing2178 said:
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.
 
  • #8
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 :smile:
 
  • #9
DrewD said:
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 :tongue:
 
  • #10
$$(x^{1/n})^n = x$$ 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 $$x^{\frac{1}{n}}=\sqrt[n]{x}$$
 

1. What is a basic identity in mathematics?

A basic identity in mathematics is an equation that is always true, regardless of the values of the variables involved. It is a fundamental concept in algebra and serves as the foundation for more complex mathematical expressions.

2. Why is it important to prove a basic identity?

Proving a basic identity is important because it helps to establish the validity and accuracy of mathematical concepts. It also allows for a deeper understanding of the relationships between different mathematical expressions.

3. How do you go about proving a basic identity?

Proving a basic identity often involves using algebraic manipulation and properties, such as the commutative and associative properties, to show that both sides of the equation are equivalent. This is typically done step by step, providing a logical and systematic approach to the proof.

4. Can anyone prove a basic identity?

While anyone with a solid understanding of algebra and mathematical principles can attempt to prove a basic identity, it often requires a strong mathematical background and critical thinking skills to successfully complete the proof.

5. What is the significance of a proven basic identity in mathematics?

A proven basic identity serves as a building block for more complex mathematical concepts. It also helps to establish the validity and reliability of mathematical theories and principles. Additionally, a proven basic identity can be used as a tool for solving more complex equations and problems in various fields of science and engineering.

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