Proving a^0 = 1 for non-zero values of a | Homework Help

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To prove that a^0 = 1 for non-zero values of a, a precise definition of exponentiation is necessary. The discussion emphasizes that simply substituting numbers does not constitute a proof. Instead, one should utilize the laws of exponents, specifically by setting m=0 in the equation (a^n)*(a^m)=a^(n+m) to derive a^0. Taking the logarithm of both sides can also simplify the proof if using the definition a^x = e^(log(a)*x). A clear understanding of the properties of exponents is essential for a valid proof.
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Homework Statement



Prove that a^0 = 1 if a is not equal to 0.

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The Attempt at a Solution


Well,since a is not equal to 0, I replace it with another number.

(1^0)^0 = 1

(6^0)^0 = 1

ETC

Is this enough to prove that a^0 = 1 if a is not equal to 0.Even thought its a basic and easy question...im doubting.

Thx
 
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hhmmm any1?...
 
You haven't 'proved' anything. You just wrote down some numbers. What's your definition of a^x? If it's e^(log(a)*x) the answer is pretty easy, just take the log of both sides. To prove something you need a precise definition of the thing you are trying to prove. What does a^x mean?
 
Are you just assuming the laws of exponents? Like (a^n)*(a^m)=a^(n+m)? If so, then set m=0 and solve for a^m.
 
Dick said:
Are you just assuming the laws of exponents? Like (a^n)*(a^m)=a^(n+m)? If so, then set m=0 and solve for a^m.

Thanks, i though about doing that before, don't know what came up in mind.
 
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