**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.

how about this:

ReplyDeletethe 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)

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

ReplyDeletea) 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

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?

ReplyDeletestill, 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!

b) It says irrational, not imaginary...

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