Proof by induction

\displaystyle u_1 = 5.2, \text{and } u_{n+1} = \frac{6u_n+25}{u_n+6}

.

Prove by induction that

u_n>5

.
How do you deal with these inequality types? It seems so trivial but how to give a sound proper proof. It seems that you rearrange it to

u_{n+1} = \displaystyle 6-\frac{11}{u_n+6}

.
Then it seems trivial.

(edited 8 years ago)

Reply 1

samb1234

13

Original post by Ano123

\displaystyle u_1 = 5.2, \text{and } u_{n+1} = \frac{6u_n+25}{u_n+6}

.

Prove by induction that

u_n>5

.

As a general rule:

1. Show it works for case n=1 or another easy to evaluate one.

2. Assume it works for n=k.

3. Show that it must then work for n=k+1, normally involves relating to the n=k case.

4. Hence holds by induction

Reply 2

Ano123

OP

15

Original post by samb1234

As a general rule:

1. Show it works for case n=1 or another easy to evaluate one.

2. Assume it works for n=k.

3. Show that it must then work for n=k+1, normally involves relating to the n=k case.

4. Hence holds by induction

What would you make the inductive step?

Reply 3

IrrationalRoot

20

Original post by Ano123

What would you make the inductive step?

Showing that the statement holds for

n=k+1

given that it holds for

n=k

.

Reply 4

samb1234

13

Original post by Ano123

What would you make the inductive step?

Assume that Uk>5, then using your rearranged form for Un+1 can easily show that if Uk is greater than 5 than so is Uk+1

Reply 5

Ano123

OP

15

Original post by samb1234

Assume that Uk>5, then using your rearranged form for Un+1 can easily show that if Uk is greater than 5 than so is Uk+1

That's what I've done and demonstrated in the OP, but how can I formally say it. It doesn't seem right saying, it is clearly greater than 5 as long as

u_n > 5

.
But in this case do you think it's so obvious that it is fine to just do that?

Reply 6

samb1234

13

Original post by Ano123

That's what I've done and demonstrated in the OP, but how can I formally say it. It doesn't seem right saying, it is clearly greater than 5 as long as