Proving c+1 is an Upper Bound of S with Completeness Axiom

[Päällikkö]

Let
$$S = \{x | x \in \mathbb{R}, x \ge 0, x^2 < c\}$$

Show that c + 1 is an upper bound for S and therefore, by the Completeness Axiom, S has a least upper bound that we denote by b.

Pretty much the only tools I've got are the Field Axioms.
I think I'm supposed to do something like:
x² $\ge$ 0. Thus c > 0.
x² < c < c + 1

Thus c + 1 is an upper bound.

By the Completeness axiom, S has a least upper bound that we denote by b.

QEDIt can't be just this, can it? I'm totally lost in maths, these things were dealt with ages ago and I still can't quite grasp the logic.

The part "Thus c + 1 is an upper bound" is where I think my logic fails. If this was the way (which I think is not the case) to prove c + 1 was an upper bound, couldn't we just have concluded that c is an upper bound, and thus b exists?

 

[arildno]

Remember that c is NOT an upper bound of S if c<1!

But in the case of c<1, you'll get that x<1<1+c, i.e, 1+c is an upper bound for S

If c>1, then $x<\sqrt{c}<c<c+1$ and thus, c+1 is an upper bound for S.

 

[HallsofIvy]

Let
$$S = \{x | x \in \mathbb{R}, x \ge 0, x^2 < c\}$$

Show that c + 1 is an upper bound for S and therefore, by the Completeness Axiom, S has a least upper bound that we denote by b.

Pretty much the only tools I've got are the Field Axioms.
I think I'm supposed to do something like:
x² $\ge$ 0. Thus c > 0.
x² < c < c + 1

Thus c + 1 is an upper bound.

No, you just proved that c+ 1 is larger than x², not x.
Since, for some x, x²< x, it does not follow that an upper bound on x² is an upper bound on x.

Try a proof by contradiction. Suppose c+ 1 is NOT an upper bound on S. That is, suppose there is x in S such that x> c+ 1. Now compare
x² and c.

 

[Päällikkö]

Remember that c is NOT an upper bound of S if c<1!
But in the case of c<1, you'll get that x<1<1+c, i.e, 1+c is an upper bound for S
If c>1, then $x<\sqrt{c}<c<c+1$ and thus, c+1 is an upper bound for S.

I don't think I can do this, as I have yet to prove that there exists a c, so that x² = c.

How's this:
x² < c < (c+1)²
Thus x < c+1

Or by contradiction, maybe:
x > c+1
{ x² > (c+1)²
{ x² < c
(c+1)² > c, so x > c+1 cannot hold.

In an unassisted problem, how do I come up with the upper bound (like the c+1 here)? Just make one up?

 

[arildno]

c is either less than one, equal to one, or greater than 1.
You can break up your proof in special cases.