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# Proof by mathematical induction question Watch

1. Just two question:

1) Am I right in thinking that if for example I've proven 2f(x+1) (or Yf(x+1) where Y is an integer) to be divisible by 5, then that automatically proves that f(x+1) is also divisible by 5?

2) If that is true, then this is my confusion: say if 2f(x+1) is equal to 5, then obviously its divisible by 5. However, how can this also mean that f(x+1) is divisible by 5 because f(x+1) would be 2.5, so clearly 5 would need to be multiplied by something which is NOT an integer (0.5) in order to get 2.5.
2. You've given a counter example to your own question.

Take f(x) = 2.5 for all x. Then 2f(x + 1) is divisible by 5, but f(x + 1) is not. So what you have said in 1) is not true.
3. the proof actually has to be real. you can't pull something random out of a bag. Structuring these questions requires a bit of time and thought
4. Is f(x+1) necessarily an integer?

Also, what if Y is 5.

Suppose you wanted to show that n was divisible by 6. Would showing 3n is divisible by 6 be sufficient? Why?
5. (Original post by CalculusMan)
You've given a counter example to your own question.

Take f(x) = 2.5 for all x. Then 2f(x + 1) is divisible by 5, but f(x + 1) is not. So what you have said in 1) is not true.
I'm struggling with this question I've got here. The solutions page has this:

''As f(k) and 2f(k+1)? f(k) are each divisible by 5, deduce that f(k+1) is also divisible by 5''.

''Use induction to complete your proof''.

what do I have to do? - I've proven 2f(k+1)? f(k) to be divisble by 5 and assumed that f(k) is also divisible by 5 as well.
6. (Original post by ilyking)
the proof actually has to be real. you can't pull something random out of a bag. Structuring these questions requires a bit of time and thought
So am I right or wrong in my first question of my first post?
7. bump
8. (Original post by W.H.T)
I'm struggling with this question I've got here. The solutions page has this:

''As f(k) and 2f(k+1)? f(k) are each divisible by 5, deduce that f(k+1) is also divisible by 5''.

''Use induction to complete your proof''.

what do I have to do? - I've proven 2f(k+1)? f(k) to be divisble by 5 and assumed that f(k) is also divisible by 5 as well.
What are these "?"s supposed to be? Knowing this is crucial for us to be able to help you here. Also bumping your thread after 23 minutes of no reply is a bit lame.
9. (Original post by nuodai)
What are these "?"s supposed to be? Knowing this is crucial for us to be able to help you here. Also bumping your thread after 23 minutes of no reply is a bit lame.

f(n) = 3^(n+2) + [(-1)^n]*2^n

By considering 2f(n+1) - f(n) and using the method of mathematical induction,
prove that 3 ^(n+2) + [(-1)^n]*2^n is divisible by 5.

So the bits you mentioned about were part of the hint (in bold) which was given in the question.
10. (Original post by W.H.T)

f(n) = 3^(n+2) + [(-1)^n]*2^n

By considering 2f(n+1) - f(n) and using the method of mathematical induction,
prove that 3 ^(n+2) + [(-1)^n]*2^n is divisible by 5.

So the bits you mentioned about were part of the hint (in bold) which was given in the question.
I see... it would have been useful for us to know what f(n) is.

As a hint, think factorisation.
11. (Original post by nuodai)
I see... it would have been useful for us to know what f(n) is.

As a hint, think factorisation.
I've completed the majority of that question.

I've shown that the hint given by the question, 2f(n+1) - f(n), is divisible by 5:

2f(n+1) - f(n) = 5[ 3^(k+2) - [(-1)^k]*2^k ]

which is correct based on the solutions page. However, I'm not sure if this is enough to satisfy the question which was to prove that f(n): 3^(n+2) + [(-1)^n]*2^n is divisble by 5.

So I checked again at the solutions page. It mentions this at the end:

''As f(k) and 2f(k+1) - f(k) are each divisible by 5, deduce that f(k+1) is also divisible by 5''

''Use induction to complete your proof''

I'm not sure what I have left to do for the question. Do I still have to prove f(k+1) is divisible by 5?

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Updated: December 27, 2010
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