## Wednesday, November 22, 2006

### Root Beer. Topic: Real Analysis.

Problem: (Stanford Math 51H) Prove that every positive real number has a positive square root (That is, for any $b > 0$, prove that there is a real number $\alpha$ such that $\alpha^2 = b$). [Usual properties of the integers are assumed.]

Solution: Consider the set $S = \{x \in \mathbb{R}: x > 0, x^2 < b\}$. We can show that $S$ is non-empty by selecting an integer $n$ large enough such that $\frac{1}{n^2} < b$. Since $b$ is a real and the integers are unbounded, there exists a positive integer $k$ such that $k^2 > b$, thus $x^2 < k^2 \Rightarrow x < k$ so $S$ is bounded from above.

Now there must exist $\alpha$ such that $\sup S = \alpha$. We claim that $\alpha^2 = b$. Suppose the contrary; then either $\alpha^2 < b$ or $\alpha^2 > b$.

CASE 1: If $\alpha^2 < b$, then consider $\left(\alpha+\frac{1}{n}\right)^2 = \alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2}$. Choose $n$ large enough such that

$\alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2} < b$

which is definitely true for any $n > \frac{2\alpha+1}{b-\alpha^2}$. But then $\alpha+\frac{1}{n} \in S$ and $\alpha+\frac{1}{n} > \alpha = \sup S$, contradicting the fact that $\alpha$ is the supremum.

CASE 2: If $\alpha^2 > b$, then consider $\left(\alpha-\frac{1}{n}\right)^2 = \alpha^2-\frac{2\alpha}{n}+\frac{1}{n^2}$. Again, choose $n$ large enough such that

$\alpha^2-\frac{2\alpha}{n}+\frac{1}{n^2} > b$

which we know is true for $n > \frac{2\alpha}{\alpha^2-b}$ (furthermore, we impose the restriction $\alpha > \frac{1}{n}$ so our resulting real is positive). Then $x \le \alpha-\frac{1}{n}$ for all $x \in S$ and $\alpha-\frac{1}{n} < \alpha = \sup S$, contradicting the fact that $\alpha$ is the supremum.

Hence we know that $\alpha^2 = b$, or that the square root of any positive real number exists and is a positive real number. QED.

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Comment: This is from the real analysis portion of a freshman honors calculus course, i.e. a rigorous treatment of the real numbers which is the basis for calculus like limits and stuff. Really understanding calculus involves really understanding how the real number system works.

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Practice Problem: (Stanford Math 51H) If $a, b \in \mathbb{R}$ with $a < b$, prove:

(a) There is a rational $r \in (a,b)$.
(b) There is an irrational $c \in (a,b)$.
(c) $(a,b)$ contains infinitely many rationals and infinitely many irrationals.

1. the function f(x) = x^2 is continuous everywhere. since it takes the value f(0) = 0, which is less than b, and since it is unbounded, so it takes a value greater than b, then it takes on the value of b at some point in (0, infty)

2. er okay i guess proving that x^2 is continuous is kind of a stronger result isn't it heh...

a) let m be the smallest integer such that |ma - mb| > 1. then there exists some integer n such that ma < n < mb and hence a < n/m < b

b) buhhh.

c) obvious by induction. now if only i could prove b lol

3. b) If a and b are both real numbers, how can there be a imagionary number in between? where does imagionary numbers fit on the scale anywayz.... since it would be sqrt of -1, would it be inbetween 0 and -1?

still, logically imagionary numbers seem to be a seperate catogory, so it wouldn't even be on the number scale
logically I agree with the problem, i dont see why you need to "prove it" cuz it's already logical, (a) make sense, (b)... doesn't

RootBeer-----sqrt(BEER) = E sqrt(BR)
lol, i wanna tell my joke in math club!

4. b) It says irrational, not imaginary...

And yes, continuity of f(x) = x^2 is a stronger result that the existence of square roots.