Mathematical induction help please (1 Viewer)

Drongoski

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Prove by induction where n is a postive integer
5^n + 3 divisible by 4
Let P(n) = 5^n + 3 for convenience

1) For n=1: P(1) = 5+3 = 4 x 2 (.: divisible by 4)

2) Assume P(n) is divisible by 4 for n=k > 0

i.e. 5^k + 3 = 4 x M for some integer M

3) .: P(k+1) = 5^{k+1} + 3 = 5 (5^k) + 3 = [4+1]5^k + 3 = 4 x 5^k + 5^k + 3
= 4 x 5^k + 4M = 4 x (5^k + M) = 4 x (an integer)

i.e. P(k+1) is divisible by 4 if P(k) is divisible by 4

4) .: by the Principle of Mathematical Induction, P(n) is divisible by 4 for all positive integer n.
 
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HSC2014

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Let P(n) = 5^n + 3 for convenience

1) For n=1: P(1) = 5+3 = 4 x 2 (.: divisible by 4)

2) Assume P(n) is divisible by 4 for n=k > 0

i.e. 5^k + 3 = 4 x M for some integer M

3) .: P(k+1) = 5^{k+1} + 3 = 5 (5^k) + 3 = [4+1]5^k + 3 = 4 x 5^k + 5^k + 3
= 4 x 5^k + 4M = 4 x (5^k + M) = 4 x (an integer)

i.e. P(k+1) is divisible by 4 if P(k) is divisible by 4

4) .: by the Principle of Mathematical Induction, P(n) is divisible by 4 for all positive integer n.
Just being pedantic, but is the restriction on k necessary? We defined n to be a positive integer; thus if n = k, then k would be a positive integer.
 

Drongoski

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Just being pedantic, but is the restriction on k necessary? We defined n to be a positive integer; thus if n = k, then k would be a positive integer.
In this case, this qualifier may be redundant. But I have adopted the practice of specifying, say: for n = k >=3, if the 1st value of n I have established is 3. Most people don't bother to do this.
 
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