Why is (n^0=1)? where n is any positive number

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Discussion Overview

The discussion revolves around the mathematical expression \( n^0 = 1 \) for any positive number \( n \). Participants explore the reasoning behind this convention, including algebraic manipulations and definitions of powers.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the justification for \( n^0 = 1 \), indicating it was accepted without understanding in earlier education.
  • Another participant provides an algebraic justification using the property of exponents: \( n^0 = n^{a-a} = n^a(n^{-a}) = (n^a)/(n^a) = 1 \).
  • A participant notes that the convention of \( n^0 = 1 \) allows for a coherent extension of the definition of powers.
  • Further algebraic reasoning is presented, suggesting that since \( n^{a+b} = n^a n^b \), it follows that \( n^a = n^{a+0} = n^a n^0 \), leading to \( n^0 = 1 \), with a caveat regarding \( n = 0 \).
  • Another participant reiterates the algebraic justification with a specific example, stating \( 1 = (5^7)/(5^7) = 5^{7-7} = 1 \).

Areas of Agreement / Disagreement

Participants present various justifications for \( n^0 = 1 \), but there is no explicit consensus on a singular explanation. Some participants acknowledge the convention while others focus on algebraic proofs.

Contextual Notes

Some assumptions regarding the properties of exponents and the definitions involved are not fully explored, particularly the implications for \( n = 0 \), which is noted to have a separate discussion.

Mozart
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I just can't justify this in my simple mind. I just always accepted it because I was told that it is equal to 1 throughout high school, and now in cegep. :confused:
 
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For an integer a:

n^0 = n^(a-a) = n^a(n^-a) = (n^a)/(n^a) = 1
 
Hehe math is so cool. Thanks.
 
Have a search for lots of posts on this topic on these forums; it is essentially a convention that allows us to coherently extend a definitions of powers.
 
Alternatively, since na+b=nanb, then it must be that na=na+0=nan0, so n0 = 1.

Except for n = 0, of course. There's a whole thread on that.
 
gnomedt said:
Alternatively, since na+b=nanb, then it must be that


If we want to extend the definition from its natural domain consistently

na=na+0=nan0, so n0 = 1.

Except for n = 0, of course. There's a whole thread on that.
 
1 = (5^7)/(5^7)= 5^(7-7) = 1 Easy!
 

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