# Are there more rational numbers than irrational numbers?

##### 1 Answer
Mar 24, 2017

No - There are more irrational numbers than rational.

#### Explanation:

The rational numbers are countably infinite, meaning that they can be put in one to one correspondence with the natural numbers.

For example, we can construct a sequence of rational numbers by listing all of the numbers of the form $\frac{p}{q}$ with integers $p , q$, with $q > 0$ and $\left\mid p \right\mid + q = k$ for $k = 1 , 2 , 3 , \ldots$ as follows...

$\frac{0}{1} , \textcolor{b r o w n}{- \frac{1}{1} , \frac{0}{2} , \frac{1}{1}} , \textcolor{red}{- \frac{2}{1} , - \frac{1}{2} , \frac{0}{3} , \frac{1}{2} , \frac{2}{1}} , \textcolor{\mathmr{and} a n \ge}{- \frac{3}{1} , - \frac{2}{2} , - \frac{1}{3} , \frac{0}{4} , \frac{1}{3} , \frac{2}{2} , \frac{3}{1}} , \ldots$

In this sequence, every rational number occurs at least once, showing that there are at least as many natural numbers as rational.

By way of contrast, the irrational numbers are uncountably infinite. Georg Cantor showed that if you tried to list all of the real numbers, then you could construct a real number different from all of the numbers on your list.

For example, if you attempted to list all of the real numbers in $\left[0 , 1\right)$, then your list could be written something like:

$0.676251977726009 \ldots$
$0.570276777031510 \ldots$
$0.390913128003121 \ldots$
$0.583650810431064 \ldots$
$0.435268835716497 \ldots$
$0.630613188572615 \ldots$
$0.111944804042922 \ldots$
$0.368459378995612 \ldots$
$0.437560677845907 \ldots$
$0.227360508016931 \ldots$
...

We can then replace one digit with a different digit in each line. Here I have used a rule that replaces $7$'s with $8$'s and any other digit with a $7$...

$0 \text{.} \textcolor{red}{7} 76251977726009. . .$
$0.5 \textcolor{red}{8} 0276777031510. . .$
$0.39 \textcolor{red}{7} 913128003121. . .$
$0.583 \textcolor{red}{7} 50810431064. . .$
$0.4352 \textcolor{red}{7} 8835716497. . .$
$0.63061 \textcolor{red}{8} 188572615. . .$
$0.111944 \textcolor{red}{7} 04042922. . .$
$0.3684593 \textcolor{red}{8} 8995612. . .$
$0.43756067 \textcolor{red}{8} 845907. . .$
$0.227360508 \textcolor{red}{7} 16931. . .$
...

Then the number formed from the diagonal digits is not in the original list, since it differs from the first number in the first decimal place, the second in the second, etc.

$0 \text{.} \textcolor{red}{7877787887} \ldots$

So any such list will be incomplete and we can deduce that there is no countable infinite sequence that enumerates all of the real numbers. So there are more real numbers in $\left[0 , 1\right)$ than there are rational numbers. This is Cantor's diagonal argument.

If the irrational numbers were countable, then the real numbers would be countable too, since we could simply interlace rational and irrational numbers in one sequence. Since we have found that this cannot be done, the set of irrational numbers must be uncountably infinite too.