Why is the Factorial of Zero Defined as One?

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Discussion Overview

The discussion revolves around the definition of the factorial of zero, specifically why it is defined as one. Participants explore various methods and definitions related to factorials, including mathematical reasoning and alternative interpretations.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • Some participants provide the definition of factorial and mention that 0! is defined as 1, referencing the Gamma function.
  • One participant presents a mathematical derivation using the Gamma function to show that 0! equals 1.
  • Another participant expresses skepticism about the clarity of the mathematical explanation provided, suggesting it may be too complex for the original poster.
  • There is a discussion about the recursive definition of factorial, where n! = n(n-1)!, leading to the conclusion that if 1! = 1, then 0! must also equal 1.
  • One participant questions the implications of defining factorial for negative integers, specifically asking about (-1)!.

Areas of Agreement / Disagreement

Participants generally agree that 0! is defined as 1, but there are differing views on the clarity and implications of the definitions and methods presented. The discussion includes some skepticism and questions regarding the understanding of these concepts.

Contextual Notes

Some participants note that the recursive definition of factorial and the relationship to the Gamma function may involve assumptions that are not fully explored, particularly regarding the factorial of negative integers.

Who May Find This Useful

This discussion may be useful for individuals interested in mathematical definitions, particularly in combinatorics or calculus, as well as those exploring the foundations of factorials and their applications.

ananthu017
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can u give me the method to find the factorial of zero ?
 
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What is the definition of the factorial?
 
0!=\Gamma(1)=\intop_{0}^{\infty} t^{1-1}\exp(-t)\mathrm{d}t=-[\exp(-t)]_{0}^{\infty}=-(0-1)=1.

Alternatively, there's exactly one way to arrange an empty set.
 
DeIdeal said:
0!=\Gamma(1)=\intop_{0}^{\infty} t^{1-1}\exp(-t)\mathrm{d}t=-[\exp(-t)]_{0}^{\infty}=-(0-1)=1.

I highly doubt the OP would be able to make heads or tails of this.

n! = n(n-1)!

Work with this definition, it's all you need.
 
So your answer to the original question is that 0!= 0(-1)!? But what is (-1)!?

(Yes, you can write that 1!= 1(0!) and, if you know that 1!= 1, then it follows that 0!= 1. But better is just to state the basic definition of n! that asserts 0!= 1 for as part of the definition.)
 

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