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As a continuation of Prove that $\sqrt{5}$ exists I came across a different proof of $\sqrt{5}$ exists that is different from the answers on the original post and the proof goes something like this:

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I understand what the person is trying to do in his proof but where I'm having trouble understanding is how he/she came about $\frac{2\alpha +1}{5-\alpha^2}$ and $\frac{2\alpha}{\alpha^2-5}$

I tried to manipulate things that were already given to somehow try to get those values but I couldn't come to a conclusion.

If anyone could show me step-by-step what he/she did to get $\frac{2\alpha +1}{5-\alpha^2}$ and $\frac{2\alpha}{\alpha^2-5}$ that would be really appreciated.

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user3412839
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  • All of the elementary functions (addition, subtraction, multiplication, and division) are closed over the reals (and over the complex numbers). Thus if you only use those operations, the final result must be a real (assuming $\alpha$ was a real to begin with). If $\alpha$ were a rational number then $\frac{2\alpha + 1}{5 - \alpha^2}$ would be a rational number (again, the rationals are closed over those elementary operations). On the other hand, if $\alpha$ had been an integer then the final result would not necessarily be an integer since the integers are not closed over division. – Jared Apr 13 '14 at 02:42
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    "I tried to manipulate things that were already given to somehow try to get those values" <- It works the other way round. You know where you want to end up ($(\alpha+\frac{1}{n})^2 < 5$ resp. $(\alpha-\frac{1}{n})^2 > 5$), and then you find out which conditions $n$ must satisfy. – Daniel Fischer Apr 13 '14 at 02:42
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    You are thinking about this backwards. If $\alpha>0$, and $\alpha^2<5$, we want to show that increasing the value of $\alpha$ a little bit still gives us a number whose square is below $5$. The natural attempt is to look for a number of the form $\alpha+1/n$, and if it works, algebra leads us to the inequality $(2\alpha+1)/(5-\alpha^2)<n$. Same with the other case. Now, once we have identified the relevant numbers, we can proceed as in the proof. But it is not that we thought of those fractions first. – Andrés E. Caicedo Apr 13 '14 at 02:44
  • Thanks guys, so say if we want to prove sqrt(3) instead than we will always do the same thing right (try to get a number in the form α+1/n and α-1/n)? – user3412839 Apr 13 '14 at 03:43

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  1. The case $\dfrac{2\alpha}{\alpha^2-5}\in\mathbb{R}$ is easy. $\alpha^2>5$, and hence, $\alpha^2-5\in\mathbb{R}$. Same thing holds for $2\alpha$. $\mathbb{R}$ is a division algebra, $\dfrac{2\alpha}{\alpha^2-5}\in\mathbb{R}$.
  2. The case $\dfrac{2\alpha-1}{5-\alpha^2}\in\mathbb{R}$ is not hard either. We may write $\dfrac{2\alpha-1}{5-\alpha^2}=\dfrac{2\alpha}{5-\alpha^2}-\dfrac{1}{5-\alpha^2}$. $\alpha^2<5$, and hence, $5-\alpha^2\in\mathbb{R}$. Similarly, since $\alpha$ is positive, $2\alpha\in\mathbb{R}$. Hence, $\dfrac{2\alpha}{5-\alpha^2}-\dfrac{1}{5-\alpha^2}\in\mathbb{R}$ (since $\mathbb{R}$ is a normed division algebra). Therefore, $\dfrac{2\alpha-1}{5-\alpha^2}\in\mathbb{R}$.
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    The question is not really why those fractions are real numbers, but rather why one would think of them. – Andrés E. Caicedo Apr 13 '14 at 02:48
  • @AndresCaicedo The OP has framed the question as "How did he/she know that since $\alpha^2<5$ then ...". Hence, this was my natural interpretation of the question. If this is the wrong interpretation, then I'd be happy to remove my answer. –  Apr 13 '14 at 02:50
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    Yes, the question is phrased that way, but that is an accident resulting of their inexperience with phrasing statements in mathematical language. (The question has now been edited, hopefully removing the ambiguity.) – Andrés E. Caicedo Apr 13 '14 at 02:51