How do you formally show subsequent element in a series is larger?

[golmschenk]

As an example, for the sequence:
$$s_{1} =2$$
$$s_{n+1}=\frac{s^{2}_{n}+1}{2}$$
We see:
When $$s_{n} >1$$ then,
$$s_{n+1}=\frac{s^{2}_{n}+1}{2}>s_{n}$$
But how do I formally show that this last inequality is true? This example is fine or just one similar to it. Thanks!

EDIT: Many of those superscripts are suppose to be subscripts. I don't seem to be able to edit them so there correct though. Sorry, hopefully you can still figure it out. If, it doesn't have any subscript at all, then whatever is the superscript is suppose to be the subscript. However, if there is a subscript then the superscript is correct.

 

[zhentil]

If s_n > 1, is (s_n-1)^2>0?

 

[golmschenk]

Yes, but I'm not quite sure how to use that to formally explain that the inequality is true. Also, again, I know that what you asked is true, but how do you formally show this?

 

[Mark44]

You might try showing this by mathematical induction. Clearly, S₂ > S₁, so that establishes your base case.

Assume that S_n > S_{n - 1}, and see if you can use this assumption and your formula for S_n to show that S_{n + 1} > S_n.

 

[zhentil]

Yes, but I'm not quite sure how to use that to formally explain that the inequality is true. Also, again, I know that what you asked is true, but how do you formally show this?

You're not sure how to show that the square of a non-zero number is greater than zero?

 

[golmschenk]

You're not sure how to show that the square of a non-zero number is greater than zero?

Not formally, no. And that's what this thread was about to begin with. Is it ok just to state that the square of a non-zero, real number is positive?

You might try showing this by mathematical induction. Clearly, S₂ > S₁, so that establishes your base case.

Assume that S_n > S_{n - 1}, and see if you can use this assumption and your formula for S_n to show that S_{n + 1} > S_n.

Thanks, I think I've got it mostly figured out.

 

[HallsofIvy]

Not formally, no. And that's what this thread was about to begin with. Is it ok just to state that the square of a non-zero, real number is positive?

Surely, for any real x, $x^2\ge 0$ which means "either $x^2= 0$ or $x^2> 0$. Further, if $x^2= 0$ then, taking the square root of both sides, x= 0. Therefore, if $x\ne 0$, $x^2> 0$.
That's a "formal" proof.

Thanks, I think I've got it mostly figured out.