Waste less time on Facebook — follow Brilliant.

How about we think deep about the rationals and irrationals?

Suppose one chooses any of the following types of intervals

  • \([a,b]\),

  • \([a,b)\),

  • \((a,b)\)

  • \((a,b]\)

  • \((-\infty,\infty)\)

in \(\ \mathbb{R}\) where \(a\) and \(b\) can be very arbitrarily close to each other but not equal to each other. Can one always find two irrationals \(p\) and \(q\) within any chosen domain such that \(p+q\) or \(p-q\) is rational?

As an example, let us consider the interval \((-1,3)\), one can find the two irrationals \(\sqrt{2}-1\) and \(\sqrt{2}+1\) within this interval that give a difference of \(2\) which is rational. Also one can find the irrationals \(1- \sqrt{2}\) and \(\sqrt{2}+1\) whose sum gives \(2\), a rational again.

Note by Tapas Mazumdar
3 months, 2 weeks ago

No vote yet
1 vote


Sort by:

Top Newest

For any interval \((a,b)\) of length \((b-a)\) matters where so it is convenient to take \(a,b\gt 0\) . Now choose a rational \(c\) such that \(a<c<b\) we want to find another \(d\in(a,b)\) such that both of \(c+\sqrt{d}\) and \(c-\sqrt{d}\) belong to \((a,b)\). It's clear that by adding a \(\sqrt{d}\) with \(c\) won't make \(c+\sqrt{d}\not\gt b\) which happens only if \(\sqrt{d}\lt b-c\) ,similarly for \(c-\sqrt{d}\gt a\) we must have \(\sqrt{d}\lt c-a\) . It is sufficient to find an \(d\) satisfying \(\sqrt{d}\lt {\rm min} \{b-c,c-a\}=\dfrac{|b-a|-|b+a-2c|}{2}\) so that there are two irrationals \(c\pm\sqrt{d}\) which add upto a rational. SImilar reasoning for \(d\) produces a lower bound for \(d\) such that we can find a \(d\) satisfying \(\dfrac{|a-b|+|a+b-2c|}{2}\lt \sqrt{d}\lt \dfrac{|b-a|-|b+a-2c|}{2}\). Even by choosing \(c\) such that \(\dfrac{a}{2}\lt c\lt \dfrac{b}{2}\) it can be achieved that the sum of the two irrationals also will lie in \((a,b)\) that is \(a\lt (c+\sqrt{d})+(c-\sqrt{d}) \lt b\). Aditya Narayan Sharma · 3 months, 1 week ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...