Resolved: "Deduce from the Completeness Axiom that there exists..."

High School
Algebra
Algebra I
Exponents and radicals

defazajx

Answered question

2021-02-24

Deduce from the Completeness Axiom that there exists a square root of a real number a if and only if $a \geq 0$

Answer & Explanation

Jozlyn

Skilled2021-02-25Added 85 answers

Suppose that sqrta exists. Then $a = {(\sqrt{a})}^{2}$ , and we know that a square of a real number is alwais nonnegative, so $a \Rightarrow 0$ .
Suppose that $a \Rightarrow 0$ . Define a set $S = {x \in R ∣ x^{2} \leq a}$
Then $0 \in S$ , so $S = 0$ . Furthermore, S has an upper bound, fi $a \leq 1, 1$ can be an upper bound, and if $a > 1$ , a can be an upper bound (because $a^{2} > a$ when a>1).
By the Completeness Axiom, there exists a supremum of S: $s =\supset S$
Now we can see that we must have $s^{2} = a$ (since it is the least upper bound of S, otherwise, if $s^{2} > a$ , we could find a smaller upper bound, which is a contradiction). Thus, $s = \sqrt{a}$
as required.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Algebra I

Didn't find what you were looking for?