Given the set {x: 0 is less than or equal to x which is less than or equal to square root of 2,?

Question

and x is rational }

Prove that the supremum (i.e. least upper bound) of the set is square root of 2.

Please show your work. thanks.

Answer 1

The only hard part is proving that there are rational numbers arbitrarily close to the apparent supremum.

Well, if you can use the decimal expansion of the square root of two, you're immediately done!

Answer 2

This is difficult -- proving the obvious!
If there is a least upper bound which is less than √2, we need to prove there is a rational number between this bound and √2, which contradicts its being an upper bound. Hence √2 must be the upper bound. However at this stage of the night I can't prove that crucial bit, that "there is a rational number between this bound and √2". Sorry.

Answer 3

This is difficult -- proving the obvious!
If there is a least upper bound which is less than √2, we need to prove there is a rational number between this bound and √2, which contradicts its being an upper bound. Hence √2 must be the upper bound. However at this stage of the night I can't prove that crucial bit, that "there is a rational number between this bound and √2". Sorry.

Answer 4

evaluate a semicircle with diameter AB and a factor on the circumference C From C drop a perpendicular to the circumference to fulfill it at D and enable advert = a and BD = b Now evaluate ?'s ACD and CBD on condition that