Help with induction question
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# Help with induction question

[From: ] [author: ] [Date: 11-08-31] [Hit: ]
So it is true for n=1.Now lets show that if it is true for the integer k then it is also true for k+1.so it is true for k+1.So, by induction, if it is true for 1,......
Use mathematical induction to prove the property for all positive integers n.

For the equation: [a^n]^5=a^(5n)

This makes no sense to me at all.

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First show that it is true for n=1.
[a^1]^5 = a^5
a^(5*1) = a^5
So it is true for n=1.

Now let's show that if it is true for the integer k then it is also true for k+1.
[a^(k+1)]^5
= [a^k * a]^5
= [a^k]^5 * a^5
= a^(5k) * a^5 because we said it is true for k
= a^(5k+5)
= a^[5(k+1)]
so it is true for k+1.

So, by induction, if it is true for 1, and also if it is true for a certain integer then it is also true for the next integer, it must be true for all positive integers.

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Induction involves two steps. First you have to prove the statement is true for the base. Since n is in the positive integers, we start with n=1.

(a^1)^5 = a^5 = a^((5)(1)) = a^(5n)

So it is true for the base.

Step 2 is to assume the statement is true for 'n', and from that show that it is true for 'n+1'.

(a^n)^5 = a^(5n) <----- Assume that's true.

Now stick in 'n+1' and see what happens.

[a^(n+1)]^5 =
[(a^n)(a^1)]^5 =
[(a^n)(a)]^5
[(a^n)^5](a^5) =
[a^(5n)](a^5) = <----- Substituting what we assumed was true for (a^n)^5
a^(5n+5) =
a^(5(n+1))

And thus we have proven by mathematical induction that (a^n)^5) = a^(5n) is true for all n.

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i guess:

first n=1

[a^1]^5=a^(5*1) true a=a

then n=k

[a^k]^5=a^(5k)

when n = k+1

[a^ (k+1)]^5=a^(5(k+1))

(a^k * a^1)^5 = (a^k)^5 * (a^1)^5 from previous expression we know that [a^k]^5=a^(5k)