1. ## Inequality Inductions

Hi,

Could someone please explain how to do this:

Show by Induction that:
n^2 - 2n + 5 > 0 for all n>=1

I have so far:

Show true for n = 1
Assume true for n = k
Show true for n = k+1 (using assumption)
So:
k^2 + 2k + 5 -2k - 1 >0
|---- >0 ---|

??

Thanks.

2. Originally Posted by justmaths
Hi,

Could someone please explain how to do this:

Show by Induction that:
n^2 - 2n + 5 > 0 for all n>=1

I have so far:

Show true for n = 1
Assume true for n = k
Show true for n = k+1 (using assumption)
So:
k^2 + 2k + 5 -2k - 1 >0
|---- >0 ---|

??

Thanks.
True for n = 1.

Assume true for $\displaystyle n = k$, that is, assume $\displaystyle k^2 - 2k + 5 > 0$.

Show that it follows from the assumption that it's true for $\displaystyle n = k + 1$:

$\displaystyle (k + 1)^2 - 2(k + 1) + 5 = k^2 + 2k + 1 - 2k - 2 + 5 = (k^2 - 2k + 5) + 2k - 1$

and it should be clear how to finish things off.

3. But thats the thing, I don't know how to finish things off. I only know how to start these off. Could you please expand on your response, please?

4. Originally Posted by justmaths
But thats the thing, I don't know how to finish things off. I only know how to start these off. Could you please expand on your response, please?
Well, from the inductive assumption you know that k^2 - 2k + 5 > 0. And you also know that 2k - 1 > 0. Therefore ....

And you should have examples to follow (in class notes and/or textbook) for setting out a proof by induction in a formal way ....

5. Yes I understand now. Thanks. I would have notes but I happened to be not in class for that lesson. I understood everything until the part where there are two parts which are bigger than zero. Thanks.