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Prove that $\sqrt{5}$ exists; in other words prove that there exists a positive number $x\in \mathbb R$ satisfying $x^2=5$

Here's what I've done:

I let $A= \{x>0:x^2\leq 5\}$

We know that $A$ is not empty because clearly $2$ is in it: $2^2<5$ and we also know that $A$ is bounded by $3$ because

$$x>3\implies x^2>9$$ so by completeness, $\sup A$ exists. I let $α = \sup A$

So now I'm trying to prove that $α^2=5$ (that there exists a positive number $α\in \mathbb R$ satisfying $α^2=5$) and to do this I'm going to try to show that $α^2<5$ and $α^2>5$ are impossible.

Case 1: $α^2<5$ to do this I'm assuming a proof by contradiction will work.

So, suppose $α^2<5$ then ...

Here's where I'm a bit lost, I don't know how to proceed with a proof by contradiction here. If anyone can give me hints or explain what I should do next that would be really appreciated.

AlexR
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user3412839
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3 Answers3

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If we use your strategy, we need to show that we cannot have $\alpha^2\lt 5$, and we cannot have $\alpha^2\gt 5$. We do this with minimal machinery. That makes the argument much longer than if there is some already developed theory to appeal to.

Suppose that $\alpha^2\lt 5$. Let $\alpha^2+\delta=5$ for some positive $\delta$. Note that $\delta \le 1$ and $\alpha\ge 2$. Then $$\left(\alpha +\frac{\delta}{4\alpha}\right)^2=\alpha^2+\frac{\delta}{2}+\frac{\delta^2}{16\alpha^2}=\alpha^2+\delta\left(\frac{1}{2}+\frac{1}{64}\right)\lt 5.$$ It follows that $\alpha +\frac{\delta}{4\alpha}\in A$, contradicting the fact that $\alpha$ is an upper bound for $A$.

Suppose now that $\alpha^2\gt 5$. Let $\alpha^2=5+\delta$ for some positive $\delta$. We leave it to you to find a positive $x$ smaller than $\alpha$ such that $x^2\gt 5$, contradicting the fact that $\alpha$ is the least upper bound of $A$. Hint: Something like what we did will work, except that we must subtract some small number from $\alpha$, instead of adding a small number, as we did earlier.

André Nicolas
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  • You are welcome. I used a "crude" adjustment that does the job. For a delicate adjustment, one can use the Newton Method for square root. – André Nicolas Apr 12 '14 at 21:46
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I don't think that using the intermediate value theorem we are going to obtain a valid proof. The theorem heavily relies on the completeness of the real line, here we are proving such completeness. So I fear we are producing a circular argument. To support my thesis, if you look in the wikipedia page of the theorem you'll see a counter example if the domain doesn't contain a point ($\sqrt{2} $ in that case)

As you said, there are $2$ cases: IF $\alpha^2 <5$ then consider $k=\dfrac{5(\alpha+1)}{\alpha + 5}$. you have that $\alpha < k$ but $k^2 <5$.

IF $5 < \alpha^2$ then consider again $k$. in this case, $k < \alpha$, but $5 < k^2$ and again an absurd

Riccardo
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    Yes, the point is that the completeness of the real line is equivalent to the least upper bound axiom, and you need the completeness of the real line to prove the intermediate value theorem. As you illustrate,, the least upper bound axiom is enough to answer this question directly. – Geoff Robinson Apr 12 '14 at 21:18
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Hint
Show that $f(x) := x^2-5$ is continuous on $\mathbb R$ and use the intermediate value thm. on $(2, -1) \in {\rm graph}(f)$ and $(3,4) \in {\rm graph}(f)$

AlexR
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    IVT is a little brutal to do an existence proof of $\sqrt5$, I think. – chubakueno Apr 12 '14 at 20:40
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    @chubakueno This allows you to prove (slightly modified), that any square root for positive $x$ exists. Agreed that it's overkill if not familiar with the applied theory, though. – AlexR Apr 12 '14 at 21:38
  • Could the downvoter express himself? – AlexR Apr 14 '14 at 07:40