Question

What kinds of numbers are the difference of two unique rational numbers, two unique irrational numbers, and a rational and irrational number?

Answer 1

Rational and Rational

We say a number is rational if it can be written as the ratio of two integers - $\frac{2}{3} ,-\frac{7}{9} ,$ and $\frac{101}{37}$ are all examples of rational numbers. Now, let's assume we have two unique rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, where $a,b,c,d$ are all integers. Their difference would then be

$\frac{a}{b} -\frac{c}{d}$

Converting these fractions to the common denominator bd:

$\frac{a}{b}\big(\frac{d}{d}\big)-\frac{c}{d}\big(\frac{b}{b}\big)=\frac{ad}{bd}-\frac{cb}{bd} =\frac{ad-cb}{bd}$

Since the product of any two integers is also an integer, we know that ad, cb and bd must be integers, and since the difference of any two integers is also an integer, ad - cb must also be an integer. The means that $\frac{ad-cb}{bd}$ is a ratio of two integers, and by definition a rational number.

Rational and Irrational

We say that a number is irrational if it cannot be written as a ratio of two integers. To find the kind of number a difference between a rational and irrational number produces, we're going to make some assumptions that'll lead us to a contradiction, proving what we want in the process.

Let's assume that we have some rational number $\frac{a}{b}$ and some irrational number $x$. We're going to assume that their difference is rational. What this means is that

$\frac{a}{b}-x=\frac{c}{d}$, where $\frac{c}{d}$ is the rational number difference. Adding x to either side, this implies

$\frac{a}{b}=\frac{c}{d}+x$

and subtracting $\frac{c}{d}$ from either side gets us

$\frac{a}{b}-\frac{c}{d}=x$

But wait - we already concluded earlier that the difference of two rational numbers is rational, which means that $\frac{a}{b} -\frac{c}{d}$ must be rational. And since we assumed from the beginning that x was irrational, this leads us to the ridiculous conclusion that

rational = irrational

which is clearly false. Since we assumed the difference was rational and that assumption led to a contradiction, the difference must actually be irrational.

Irrational and Irrational

This one's a little tricky, because there's no one answer to this question - sometimes the difference is rational, and sometimes it's irrational. I'll provide an example of each:

Ex 1: $\sqrt{3}-\sqrt{2}$ is irrational. If it were rational, that would mean $(\sqrt{3}-\sqrt{2})^{2}$ would also be rational, but $(\sqrt{3}-\sqrt{2})^{2} =5-2\sqrt{6}$, and since we've already shown that the difference between a rational number (5) and an irrational number (2√6) is irrational, $\sqrt{3} -\sqrt{2}$ must be rational.

Ex 2: If we take the two unique irrational numbers $2-\sqrt{3}$ and $-\sqrt{3}$, their difference is $(2-\sqrt{3})-(-\sqrt{3})=2-\sqrt{3}+\sqrt{3}=2$, which is a rational number.