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How can one be sure that for every real number $x\in\mathbb R$ with $x\geq0$ there is a unique number $a\geq 0$ such that $a^2 = x$?

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  • $\begingroup$ Btw, you can use the notation $x\in\mathbb R^+$. And there isn't a unique $a$, since $-a$ also satisfies that. Perhaps require $a>0$. $\endgroup$
    – Simply Beautiful Art
    Commented Nov 13, 2016 at 14:52
  • $\begingroup$ Thanks; I edited the answer such that $a\geq 0$ $\endgroup$
    – user384011
    Commented Nov 13, 2016 at 14:53
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    $\begingroup$ we need to know more about your background to give a tailored answer to you. What do you know about real numbers ? $\endgroup$
    – Ewan Delanoy
    Commented Nov 13, 2016 at 14:59
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    $\begingroup$ I think you can prove by contradiction. So suppose $a^2 = x = b^2$ where $a \ne b$. We can assume $a > b$. With simple reasoning you can show that $a<b$, a contradiction. $\endgroup$
    – Jerry
    Commented Nov 13, 2016 at 15:03
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    $\begingroup$ There's a very slick proof in the first few pages of Baby Rudin, using only the least upper bound property, that every non-negative real has an $n$th root for all $n \in \mathbb{N}$. $\endgroup$
    – MathematicsStudent1122
    Commented Nov 13, 2016 at 15:11

3 Answers 3

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The "axiom of completeness" distinguishes the reals from the rationals (where this statement is not true, e.g. $\sqrt 2$). It says every set of real numbers with an upper bound has a least upper bound.

If $x > 1$ then clearly $\sqrt x \lt x$, while if $x \le 1$ then $\sqrt x \le 1$.

So the set of all numbers $a$ such that $a^2 < x $ is bounded above and therefore has a least upper bound. One can then show that the LUB is in fact the square root.

(as a point of interest this is normally ignored in the usual proof that $\sqrt 2$ is an irrational number. The proof normally given uses contradiction to show that $(p/q)^2 $ cannot equal $2$ for any integer $p, q$ but fails to show that it is a real number).

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If we prove $x=a^2$ is bijective for $a\ge0$, we prove that it is one-to-one, which establishes that $a$ is unique for each $x$. Since we have $x'=2a\ge0$ for all $a\ge0$, we know that this is monotone. Since $x=a^2$ is continuous, it is bijective.

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Using the intermediate value theorem for $f(a)=a^2-x$ one finds that $f(0)=-x<0$ and $f(1+\frac{x}2)=1+\frac{x^2}4>0$ so that $f$ has at least one root inside that interval. Strict monotonicity, $0<a<b\implies f(b)-f(a)=(b-a)(b+a)>0$, guarantees that there is exactly one root.

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