Question

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

Exponents and radicals
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asked 2021-02-24

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

Answers (1)

2021-02-25

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

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