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.

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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 - 4 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$