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# Proof by induction watch

1. Hello,

Please could you check if my method is correct?

A sequence is defined by the recurrence relation with u1 = 2

Prove by induction that for any natural number n,

Step 1: Check for n=1. We know that u1=2, so 1<u1<5

Step 2: Assume that the inequality holds for all natural n.

Step 3: Check for n+1.
We have

If we sub. un for 1, we get

If we sub. un for 5, we get

Since we had assumed that 1<un<5, we conclude that 1<Un+1<5.

We have shown that the inequality holds for n+1 and thus it holds for all natural n.

Any clarification is much appreciated.
2. (Original post by razzor)
Hello,

Please could you check if my method is correct?

A sequence is defined by the recurrence relation with u1 = 2

Prove by induction that for any natural number n,

Step 1: Check for n=1. We know that u1=2, so 1<u1<5

Step 2: Assume that the inequality holds for all natural n.

Step 3: Check for n+1.
We have

If we sub. un for 1, we get

If we sub. un for 5, we get

Since we had assumed that 1<un<5, we conclude that 1<Un+1<5.

We have shown that the inequality holds for n+1 and thus it holds for all natural n.

Any clarification is much appreciated.
THat is correct but I should check for n=2 because n=1 -> u1 is only a given initial value. We can get for n=2 ->u2 a calculated value through the recurrent relation.
4. (Original post by razzor)
Hello,

Please could you check if my method is correct?

A sequence is defined by the recurrence relation with u1 = 2

Prove by induction that for any natural number n,

Step 1: Check for n=1. We know that u1=2, so 1<u1<5

Step 2: Assume that the inequality holds for all natural n.

Step 3: Check for n+1.
We have

If we sub. un for 1, we get

If we sub. un for 5, we get

Since we had assumed that 1<un<5, we conclude that 1<Un+1<5.

We have shown that the inequality holds for n+1 and thus it holds for all natural n.

Any clarification is much appreciated.
Unfortunately you haven't proved it at all.

You've shown that 1 and 5 map to 1 and 5, but you haven't shown that for some value between 1 and 5 you couldn't output a value that goes outside that range, which is what the question is asking you to do!

Also: just noticed for Step 2 you've said "assume inequality holds for all natural numbers n". That is NOT how you set up the induction hypothesis! You should be saying "assume the inequality holds for a given natural number k". Your task is then to show that it holds for k+1 (that's what you've actually tried to do, but your wording is incorrect and you shouldn't use 'n' because 'n' refers to the general term).
5. (Original post by razzor)
Hello,

Please could you check if my method is correct?

A sequence is defined by the recurrence relation with u1 = 2

Prove by induction that for any natural number n,

Step 1: Check for n=1. We know that u1=2, so 1<u1<5

Step 2: Assume that the inequality holds for all natural n.

Step 3: Check for n+1.
We have

If we sub. un for 1, we get

If we sub. un for 5, we get

Since we had assumed that 1<un<5, we conclude that 1<Un+1<5.
This isn't valid reasoning: substituting two values for u_n (without any extra argument) doesn't prove anything about the behaviour for u_n between 1 and 5.

Your argument should look more like:

Since u_n < 5, we know that 5/u_n is ..., and so -5/u_n + 6 is ...
Since u_n > 1, we know that 5/u_n is ..., and so -5/u_n + 6 is ...

We conclude that as desired.

Step 3: Check for k+1

>
<

We assumed that
which implies that

Since , we conclude that
7. (Original post by razzor)

Step 3: Check for k+1

>
<

We assumed that
which implies that

Since , we conclude that
You have shown that , but you still need to show that . The post by DFranklin highlights the way.
8. Please could you show me how to get there? I don't really understand his explanations
9. You have already done most of it.

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