advanced calc help!!!

[theverymooon]

I just started adv calc and im having some trouble. i was wondering if anyone could help me with these proofs

1. prove the sqrt of 3 exists

2. prove that every interval contains an irrational number

if anyone could help me with these i would really appreciate it, thanks

 

[mmm4444bot]

… prove the sqrt of 3 exists …

Proofs are not my forte, but how about the following.

By the Pythagorean Theorem, the diagonal of a unit square is sqrt(2).

Therefore, the square root of 2 exists, so we can now construct a right-triangle with legs sqrt(2) and 1.

Again, by the Pythagorean Theroem, the length of the hypotenuse is sqrt(3).

For the second proof, my first thought is a Dedekind Cut, but my memory is fuzzy on that.

 

[daon]

Depends on what you "know."

If you know that the real numbers are complete,

\(\displaystyle a_{n+1} = \frac{1}{2}(a_n + \frac{3}{a_n}); a_0=1\)

\(\displaystyle \sqrt{3} = \lim_{n \to \infty} a_n\)

If you know that \(\displaystyle exp(x)=e^x\) and \(\displaystyle log_e(x) = ln(x)\) exist:

\(\displaystyle \sqrt{3} = e^{\frac{1}{2}ln3}\)



For the second, assume there was an interval \(\displaystyle (a,b)\) which contained no irrationals. By definition, then, \(\displaystyle (a,b)\) contains only rationals. Can you construct a bijection f: (a,b)->R? If so, that would mean the Reals are a countable set (a contradiction).