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Meaning of \({ x }^{ 0 }\)

\(\bullet\) What does \(x^0\) signify ?

\(\bullet\) What is the actual meaning of \(0!\) ?

Note by Abhijeet Verma
2 years, 6 months ago

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\(0!\) means the number of permutations of 0 distinct objects, which is nothing!

The way of permuting 0 objects is not doing anything, which is 1 way!

Therefore, \(0! = 1\) Samuraiwarm Tsunayoshi · 2 years, 6 months ago

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Well , \(x^{0}\) means x is raised to the power of 0 .

Now moving on to your second question ,

\[ n! = n\times (n-1)! \\ (n-1)! = \dfrac{n!}{n}\]

Input n=1 to get 0!=1 .

But what you have asked is the significance of 0! right ?

Actually it has no significance of its own , we have just alloted it an arbitrary value of one , so that we may use it for our benefits . One of it's applications is in the Choose function .

\[ \binom{12}{0} = \dfrac{12!}{(12)!(0)!}\]

This way , one can even incorporate 0! into calculations ignoring the fact that \(\dfrac{1}{0}\) is undefined .

If you want , you can read this , though this might be slightly off topic .


@Sandeep Bhardwaj A bit of expert's advice may be needed here since I'm not satisfied with my reply . Pls help him out sir .

Thanks :) Azhaghu Roopesh M · 2 years, 6 months ago

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@Azhaghu Roopesh M

\(0! = 1\) and \(x^0 = 1\) are defined this way. But there are reasons for these definitions, these are not arbitrary.

We can't reason that \(x^0 = 1\) by thinking of the meaning of powers as "repeated multiplications" because you cannot multiply x zero times. Similarly, you cannot reason out \(0!=1\) just in terms of the meaning of factorial because you cannot multiply all the numbers from zero down to 1.

Mathematicians define \(x^0 = 1\) in order to make the laws of exponents work even when the exponents can no longer be thought of as repeated multiplication. For example, \((x^3)(x^5) = x^8\) because you can add exponents. In the same way \((x^0)(x^2)\) should be equal to \(x^2\) by adding exponents which implies that \(x^0\) must be \(1\) because when you multiply \(x^2\) by it, the result is still \(x^2\). This is the reason that \(x^0 = 1\) only makes sense here.

In the same way, when thinking about combinations we can derive a formula for "the number of ways of choosing \(k\)things from a collection of \(n\) things." The formula to count out such problems is \(\dfrac{n!}{k!(n-k)!}\).
For example, the number of handshakes that occur when everybody in a group of \(10\) people shakes hands can be computed using \(n = 5\) and \(k = 2\) (2 people per handshake) in this formula. (So the answer is \(\dfrac{10!}{2! \cdot 8!}=45\)).

Now suppose that there are \(2\) people and "everybody shakes hands with everybody else." Obviously there is only one handshake. But what happens if we put \(n = 2\) and \(k = 2\) in the formula? We get \(\dfrac{2!}{2! 0!}\). This is \(\dfrac{2}{2 \cdot \lambda}\), where \(\lambda\) is the value of \(0!\). The fraction reduces to \(\dfrac{1}{x}\), which must \(1\) since there is only \(1\) handshake. So the only value of \(0!\) that makes sense here is \(1\). That's why we define \(0!=1\)

I hope, it will give you what you're looking for. If you've any doubts, you can ask me here.

Thanks! Sandeep Bhardwaj · 2 years, 6 months ago

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@Sandeep Bhardwaj Thanks you sir. Abhijeet Verma · 2 years, 6 months ago

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@Sandeep Bhardwaj As usual ,your interpretations are the best :D Azhaghu Roopesh M · 2 years, 6 months ago

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@Azhaghu Roopesh M Thank you very much. \(\ddot \smile\). Sandeep Bhardwaj · 2 years, 6 months ago

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@Azhaghu Roopesh M Thanks, but what is the actual meaning of "something is raised to the power of 0" =1 ? (I don't want the proof ) Abhijeet Verma · 2 years, 6 months ago

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@Abhijeet Verma \(\dfrac{x^1}{x^1} = \dfrac{x^{1-1}}{1} = x^0 = 1\)

Provided \(x \neq 0\)


Is this satisfactory explanation? Krishna Sharma · 2 years, 6 months ago

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@Krishna Sharma Actually I don't want the proof but what is meant by "something is raised to the power of 0" . Like we say that x^4=xxxx. x^2=xx. x^0=? Abhijeet Verma · 2 years, 6 months ago

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@Abhijeet Verma I don't think there is a simple explanation for something like that. For example, what is \( x^{\sqrt{2}} \)? or what is \( x^{\frac{3}{2}} \)? The exponents were first defined for natural numbers using your explanation and then the rules were extended to allow for all real numbers.

A simple analogy would be subtraction. At first, we are taught the rules of subtraction like this. \( 7 - 3 \) is like having \(7\) apples and taking \(3\) away. But \( 3-7 \) isn't defined according to that explanation.Or maybe \( (2+ 3i) - (4 + 7i) \). How would you explain \( (2+ 3i) - (4 + 7i) \). By extending the rules of subtraction beyond natural numbers, we lose the ability to explain the equations but at the same time gain the ability to use it in more scenarios.

Another example might be multiplication. \( 2 \times 3 \) is defined as repeated addition,i.e, \( 2 + 2 + 2 \). What about \( \pi \times e \)? You can't explain it like that. Same case as above. We extend the scope of multiplication beyond natural numbers but lose the ability to "explain" it. Siddhartha Srivastava · 2 years, 6 months ago

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@Siddhartha Srivastava OK, Thank you sir. Abhijeet Verma · 2 years, 6 months ago

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@Siddhartha Srivastava As usual your explanations are too good :) Azhaghu Roopesh M · 2 years, 6 months ago

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@Abhijeet Verma You can write \(x^0 = \dfrac{x}{x}\) or any power of x i.e \(\dfrac{x^n}{x^n}\) \(n \in \mathbb R\) Krishna Sharma · 2 years, 6 months ago

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@Abhijeet Verma Actually I might just not be able to convincingly explain it out to you , so can you wait till Sandeep sir replies ? Azhaghu Roopesh M · 2 years, 6 months ago

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@Azhaghu Roopesh M surely ,thanks. Abhijeet Verma · 2 years, 6 months ago

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@Abhijeet Verma No , not at all ! Why thank me ? I didn't help you out at all . Azhaghu Roopesh M · 2 years, 6 months ago

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@Azhaghu Roopesh M Thanks for finding time to add your comment. I think you will be cracking JEE this year , right? Abhijeet Verma · 2 years, 6 months ago

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@Abhijeet Verma Well , I'll try to crack it . It's not that easy , you know . Azhaghu Roopesh M · 2 years, 6 months ago

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@Azhaghu Roopesh M Best of luck for your exam Abhijeet Verma · 2 years, 6 months ago

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@Abhijeet Verma Thanks :D Azhaghu Roopesh M · 2 years, 6 months ago

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