1. ## Finish analysis Proof

I have to finish the following proof by showing that with the assumption alpha^2 > 2, this will lead to a contradiction that alpha = sup(T).

THEOREM: There exists a real number alpha "belonging to R" satisfying alpha^2 = 2.

Proof:

Consider the set T = {t "is an element of" R : t^2 <2}

and set alpha = sup(T). We'll prove alpha^2 = 2 by ruling out that alpha^2 < 2 and alpha^2 > 2. Demonstrate that alpha^2 will violate the fact that alpha is an upper bound of T and alpha^2 < 2 will violate that its the least upper bound.

If we assume alpha^2 < 2, in search of an element thats largest than alpha, write:

(alpha + 1/n)^2 = alpha^2 + 2alpha/n + 1/n^2
(alpha + 1/n)^2 < alpha^2 + 2alpha/n + 1/n
(alpha + 1/n)^2 = alpha^2 + (2alpha+1)/n

Choose n_0 "belonging to" N (natural numbers) large enough such that

1/n_0 < (2 - alpha^2)/(2alpha + 1)

This says (2alpha + 1)/n_0 < 2 - alpha^2 and that:

(alpha + 1/n_0)^2 < alpha^2 + (2 - alpha^2) = 2

Therefore, alpha + (1/n_0) "is an element of" T and thus contradcting that alpha is an upperbound of T. Therefore, alpha^2 < 2 can't happen.

What about alpha^2 > 2? Write:

(alpha - 1/n)^2 = alpha^2 - 2alpha/n + 1/n^2
(alpha - 1/n)^2 > alpha^2 - 2alpha/n

FINISH PROOF HERE.

2. I have an extension to this problem and I have to modify the above argument in order to prove the existence of sqrt(b) for any real number b >= 0.

No clue on how to do this...

3. This is completely standard problem. However, it seems that your instructor has a particular way he/she wants the problem done. I have never seen this approach. Therefore, I would not even try to complete someone else’s approach. I really think that you must rely upon the particular instructor.

4. Originally Posted by Plato
This is completely standard problem. However, it seems that your instructor has a particular way he/she wants the problem done. I have never seen this approach. Therefore, I would not even try to complete someone else’s approach. I really think that you must rely upon the particular instructor.
Yup I visited my prof. and it all makes sense now. Thanks!