# Thread: Show by mathematical induction:

1. ## Show by mathematical induction:

We have:

a1 = 2, a(n+1) = (an^2 +2) / 2an

Then comes the induction proof:
" Use mathematical induction to show that: an>= 0 for all n"

This should also mean that a(n+1) is greater than, or equal to 0. So that:
a(n+1) >= 0
(an^2 +2) / 2an >= 0
an^2 +2 >= 0

Where do I go from here?
Or could I just say that an has to be greater than 0, because if an=0 then the denominator for a(n+1) would be 0?

Are you saying

$a_1 = 2$ and $a_{n + 1} = \frac{a_n^2 + 2}{2a_n}$?

3. Look at this outline of every induction proof. That post is about a different problem, but every induction proof must follow those steps.

4. Yes Don't know how to write sub/sup text...

5. I don't think induction is necessary. To show that this is always positive, show that the numerator and denominator are always the same sign (I can't see any reason why either of them would ever be negative...)

6. I agree, but the task says I have to prove it by induction

7. Strictly speaking, induction is necessary. The numerator of $(a_n^2 +2) / 2a_n$ is always positive, so one has to know that $a_n$ is positive. This can only come from the induction hypothesis.

8. Maybe I could say:

The only way that a(n+1) < 0, is if an < 0, since the numerator is always greater than 0 because an^2 > 0.
Therefor all I have to prove is that a(n+1) always is greater than zero.

So if:
a(n+1) >= 0

Then:
(an^2 +2) / 2an >= 0
an^2 +2 >= 0

Which is always greater than zero.

Does that sound correct?

9. Unfortunately, no.
all I have to prove is that a(n+1) always is greater than zero.
This is equivalent to the original problem that asks to show that a(n) >= 0 for all n. So you have not simplified anything.

So if:
a(n+1) >= 0
You can't assume what you have to prove. It's like expecting that the cashier in a grocery will give you cash for buying stuff.

There is no way around the steps that I mentioned above. The only way to gloss over them is knowing them so well that it is boring to write every detail.

10. Ok.

Obviously the statement is true for n=1. But how do I proceed from here?

11. Assume that $a_{k + 1}$ is nonnegative.

That would mean that $\frac{a_k^2 + 2}{2a_k} \geq 0$.

We need to show that $a_{k+ 2}$ is nonnegative.

$a_{k + 2} = \frac{a_{k + 1}^2 + 2}{2a_{k + 1}}$

Since $a_{k + 1} \geq 0$ that means $2a_{k + 1} \geq 0$.

And the numerator is always nonnegative.

Since the numerator and denominator are both nonnegative, that means $a_{k + 2}$ is nonnegative.

Q.E.D.

12. Always using n --> n+1, made me forget that n+1 --> n+2 is also allowed. Thanks

13. Originally Posted by jenkki
Always using n --> n+1, made me forget that n+1 --> n+2 is also allowed. Thanks
You can use any two consecutive general terms. The point is to show by the base step that you knock over the first domino, and by the inductive step that knocking over any domino will knock over the next one.

14. There is also a b) part to this task, which I can't seem to solve

It goes:
Show that $a_{n}^2 -2 \geq 0$ for all values of n

This is what I'm thinking:
If we assume that: $a_{n}^2 \geq 2$, we have to show that this is also the case for $a_{n+1}^2 \geq 2$

From before we have that: $a_1 = 2$ and $a_{n + 1} = \frac{a_n^2 + 2}{2a_n}$. And we also know that: $a_{n} \geq 0$ for all n.

I have tried putting $a_{n+1}^2 = (\frac{a_n^2 + 2}{2a_n})^2$.

But from here I'm not able to show that $a_{n+1}^2 \geq 2$
Any ideas?

15. Originally Posted by jenkki
There is also a b) part to this task, which I can't seem to solve

It goes:
Show that $a_{n}^2 -2 \geq 0$ for all values of n

This is what I'm thinking:
If we assume that: $a_{n}^2 \geq 2$, we have to show that this is also the case for $a_{n+1}^2 \geq 2$

From before we have that: $a_1 = 2$ and $a_{n + 1} = \frac{a_n^2 + 2}{2a_n}$. And we also know that: $a_{n} \geq 0$ for all n.

I have tried putting $a_{n+1}^2 = (\frac{a_n^2 + 2}{2a_n})^2$.

But from here I'm not able to show that $a_{n+1}^2 \geq 2$
Any ideas?
P(k)

$\left(a_k\right)^2-2\ \ge\ 0$

P(k+1)

$\left(a_{k+1}\right)^2-2\ \ge\ 0$

Proof

We are trying to show

$\displaystyle\left(\frac{\left(a_k\right)^2+2}{2a_ k}\right)^2-2\ \ge\ 0$

if P(k) is valid

$\displaystyle\frac{\left(a_k\right)^4+4\left(a_k\r ight)^2+4}{4\left(a_k\right)^2}-2\ \ge\ 0$ ?

$\displaystyle\frac{\left(a_k\right)^4+4\left(a_k\r ight)^2+4}{4\left(a_k\right)^2}\ \ge\ 2$ ?

$\left(a_k\right)^4+4\left(a_k\right)^2+4\ \ge\ 8\left(a_k\right)^2$ ?

$\left(a_k\right)^4-4\left(a_k\right)^2+4\ \ge\ 0$ ?

$\left(\left(a_k\right)^2-2\right)^2\ \ge\ 0$ ?

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