# Thread: Rational Number - Question

1. ## Rational Number - Question

Let p be a rational number.
$p^2 < 2$

Find another rational number q, such that $q>p$ AND $q^2 < 2$

[I am stuck on Rudin pg 1 . I am struggling to find an explanation as how would anyone get (2p+2)/(p+2). It works but I couldn't have imagined it - so is there a way we can do it systematically?]

Thanks

2. Originally Posted by aman_cc
Let p be a rational number. $p^2 < 2$
Find another rational number q, such that $q>p$ AND $q^2 < 2$
I am struggling to find an explanation as how would anyone get (2p+2)/(p+2). It works but I couldn't have imagined it - so is there a way we can do it systematically?]
I answered this very question several times.
The truth is: there is no satisfactory answer. In general, it is a ‘reverse engineering’ job.
Someone, most likely before Rudin, found that one simply worked.
I have seen several others. Bevin Youse uses $\frac{4p}{p^2+2}$ which also works.

3. Originally Posted by Plato
I answered this very question several times.
The truth is: there is no satisfactory answer. In general, it is a ‘reverse engineering’ job.
Someone, most likely before Rudin, found that one simply worked.
I have seen several others. Bevin Youse uses $\frac{4p}{p^2+2}$ which also works.
Thanks. Good to know people have struggled before me as well. I was like little disappointed - It was pg 1 of Rudin after all !!

4. Originally Posted by Plato
I answered this very question several times.
The truth is: there is no satisfactory answer. In general, it is a ‘reverse engineering’ job.
Someone, most likely before Rudin, found that one simply worked.
I have seen several others. Bevin Youse uses $\frac{4p}{p^2+2}$ which also works.
@Plato - I seem to have hit on a way it can be 'worked' out. This is based on a proof in Bartle book.

We know $p^2<2$
we need another rational x>0 such that $(p+x)^2<2$
$x(x+2p)<2-p^2$
Note x<2 => Any $x<\frac{2-p^2}{2+2p}$ will do.