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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
9 months, 2 weeks ago

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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 $$a\lt (c+\sqrt{d})+(c-\sqrt{d}) \lt b$$.

- 9 months, 2 weeks ago