- #1
- 519
- 11
Let
[tex]S = \{x | x \in \mathbb{R}, x \ge 0, x^2 < c\}[/tex]
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 [itex]\ge[/itex] 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?
[tex]S = \{x | x \in \mathbb{R}, x \ge 0, x^2 < c\}[/tex]
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 [itex]\ge[/itex] 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?