Does This Proof Prove the Equation (a^{n})^{m}=a^{nm}?

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Just make sure you are clear in your proof that you are using the fact that the relationship holds for m as well. You could say something like "By the inductive hypothesis, we know that (a^n)^m = a^{nm}, so (a^n)^{m+1} = (a^n)^m * (a^n)^1 = a^{nm} * a^n = a^{nm+n} = a^{n(m+1)}." Great job!In summary, using proof by induction, it can be shown that (a^{n})^{m}=a^{nm}, where m is an integer and a is a fixed number. This is proven by first establishing the relationship for m=1, and
  • #1
a_skier
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Homework Statement



Prove (a[itex]^{n}[/itex])[itex]^{m}[/itex]=a[itex]^{nm}[/itex]

Homework Equations



Proof by induction
a[itex]^{n}[/itex]*a=a[itex]^{n+1}[/itex]
a[itex]^{n}[/itex]*a[itex]^{m}[/itex]=a[itex]^{nm}[/itex]

The Attempt at a Solution



Let a and n be fixed. I will induct on m.

Suppose m=1. Then a[itex]^{(n)(m)}[/itex]=a[itex]^{n(1)}[/itex]=(a[itex]^{n}[/itex])[itex]^{1}[/itex]

Now assume the hypothesis is true for any integer m in P. I will show this is true for m+1.

a[itex]^{n(m+1)}[/itex]=(a[itex]^{n}[/itex])[itex]^{m+1}[/itex]

Thus the hypothesis is true for m+1.

Is this proof sufficient? I am once again struck by a problem that seems almost too simple to be proved.
 
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  • #2
Your two equations in section 2 of your post seem fairly irrelevant to the problem at hand.

Your inductive step seems to have zero content. Why is
[tex] a^{n(m+1)} = \left( a^{n} \right)^{m+1}[/tex]
true at all? The whole point is you're supposed to prove that using your inductive hypothesis. You need to use the fact that [tex] a^{nm} = \left(a^n\right)^m [/tex]
at some point
 
  • #3
Your induction will still only prove that the relationship is true for integer m and n if you succeed. Clearly the relationship is more general than that.
 
  • #4
an(m+1)=an*(m+1)=(by the theorem) (an)(m+1)

How else could I move from the left side of the equality to the right? If someone could give me a little direction I would love to solve it myself.
 
  • #5
Fightfish said:
Your induction will still only prove that the relationship is true for integer m and n if you succeed. Clearly the relationship is more general than that.

Luckily I only need to prove it for integers right now. I don't think I could handle anything more atm.
 
  • #6
You need to show explicitly that the proposition is true for m+1 making use of the assumption that the proposition holds true for m, which you are clearly lacking here.
Start with [itex](a^{n})^{(m+1)}[/itex], rearrange / rewrite it until you manage to obtain a term [itex](a^{n})^{m}[/itex] somewhere. Then you can replace that term with [itex]a^{nm}[/itex]
 
  • #7
a_skier said:
an(m+1)=an*(m+1)=(by the theorem) (an)(m+1)

How else could I move from the left side of the equality to the right? If someone could give me a little direction I would love to solve it myself.

Fightfish gives a good suggestion... alternatively from an*(m+1) do literally the only algebra you are allowed to do and re-write it as anm+n
 
  • #8
Here's the proof for m=1

(an)m=(an)1=(By the definition)an

Now for m+1
Because xn+m=xn*xm

Let x=(an)

(x)(m+1)=(x)m*(x)1 and by the hypothesis and the case m=1 I have already proved and replacing x with (an)
(x)m*(x)1=(an)m*(an)1=anm*an=anm+n=an(m+1)

I'm pretty sure this proves it!? IM SO EXCITED THIS IS AWESOME HAHA!
 
  • #9
Yeah that looks good to me
 

1. What is the proof of (a^n)^m = a^(nm)?

The proof of (a^n)^m = a^(nm) is based on the properties of exponents. It can be proven by using the laws of exponents, specifically the product rule and power rule.

2. Can you provide an example to illustrate the proof of (a^n)^m = a^(nm)?

For example, let's say we have a = 2, n = 3, and m = 2. In this case, (a^n)^m = (2^3)^2 = 8^2 = 64. On the other hand, a^(nm) = 2^(3*2) = 2^6 = 64. Therefore, (a^n)^m = a^(nm) is proven to be true.

3. What are the properties of exponents used in the proof of (a^n)^m = a^(nm)?

The properties of exponents used in this proof are the product rule, which states that a^(m+n) = a^m * a^n, and the power rule, which states that (a^m)^n = a^(m*n).

4. Is the proof of (a^n)^m = a^(nm) applicable for all values of a, n, and m?

Yes, the proof of (a^n)^m = a^(nm) is applicable for all values of a, n, and m. This is because the properties of exponents hold true for all real numbers.

5. Why is understanding the proof of (a^n)^m = a^(nm) important in mathematics?

Understanding the proof of (a^n)^m = a^(nm) is important in mathematics because it helps in solving more complex problems involving exponents. It also provides a basis for understanding other mathematical concepts, such as logarithms and the laws of logarithms.

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