If the square root of two is irrational, why can it be created by dividing two numbers?

Question

I'm doing pre-calculus and got a bit caught on this question… I looked online and it said that a number is rational if it can be the quotient of two integers. So I did this:

I know that it is irrational because if this was actually rational, mathematicians would have figured out. So my question is, why is the square root of two irrational?

Answer 1

Squaring the number you posted:

$$1.41421356237 \cdot 1.41421356237 = 1.9999999999912458800169$$

so the result is not $2$, so the number is not $\sqrt{2}$ (but rather a rational approximation thereof).

Answer 2

The square root of two cannot exactly be written out on a computer screen in decimal notation – the digits extend forever. You're looking at an approximation accurate to a lot of digits. The approximation itself, however, is rational, as you have demonstrated. If you expanded it out to more digits, however, you'd realize that your quantity has zeros extending out to infinity, while $ \sqrt{2} $ has more digits. Therefore your rational number does not contradict the proof given by the mathematical community that $ \sqrt{2} $ is irrational.

Note: just because a number is irrational doesn't mean it can't be arbitrarily well approximated by rational numbers – every number, rational and irrational can, and that property is known as "the rational numbers are dense in the reals." For instance, for $ \sqrt{2} $: \begin{align} \frac{1}{1}, \frac{14}{10}, \frac{141}{100}, \frac{1414}{1000}, \frac{14142}{10000}, ... \end{align} get arbitrarily close to $ \sqrt{2} $, but none of them are equal to it.

Related topic: if you're interested in rational numbers which quickly converge to irrational numbers, check out continued fractions. They're really cool. Interesting numbers like $ \sqrt{2} $ often have continued fraction expansions with closed-form coefficients – $ \sqrt{2} $ can be represented by a continued fraction with first coefficient 1 and all following coefficients 2. You can do all sorts of weird things with this, like show that the Golden Ratio (which has a continued fraction expansion of all 1's) is (one of) the irrational number(s) which takes the longest to converge to, or in more plain concepts, the rational numbers used to approximate it require a really large denominator to do a good job.

Answer 3

This is the usual proof by contradiction that $\sqrt2$ is irrational (often misattributed to Euclid):

Suppose $\sqrt2$ is rational.
Then $\sqrt2=\dfrac mn$ where $m,n\in\Bbb N$ and $\gcd(m,n)=1$.
Squaring, $2=m^2/n^2\implies2n^2=m^2\implies m=2k$ for some $k\in\Bbb N$ ($m$ is even).
Then $2n^2=4k^2\implies n^2=2k^2$, so $n$ must be even.
But now we see that 2 divides both $m$ and $n$, which contradicts the initial assumption that $\gcd(m,n)=1$.
Hence $\sqrt2$ is irrational.

Your attempt at showing the rationality of $\sqrt2$ fails because we can take your fraction through this argument and arrive at absurdities (e.g. $2n^2=m^2$, which is false by direct calculation).

However, we can get arbitrarily close to $\sqrt2$ with rational numbers, and good approximations have been known for centuries, such as $99/70$ and $577/408$. These stem from the continued fraction expansion of $\sqrt2$ as $[1;2,2,2,2,\dots]$ and are better than your decimal digit-based expansions in the sense that they have the smallest denominator for a given accuracy.

Answer 4

"Does that mean all irrational numbers are infinite?"

The irrational numbers are exactly those numbers that require infinitely many digits right of the radix point, no matter which base you are using. (Thanks Beanluc: Normally we say "decimal point" but that would obviously wrong in bases other than base 10, so the generic term is "radix point". )

A number like 1/3 needs infinitely many digits in base 2, 10, 16 but not in base 3, 6, 9, 12. $2^{1/2}$ needs infinitely many digits in every base.

Answer 5

There's nothing so mysterious: $$ \left(\frac{141421356237}{10^{11}}\right)^{\!2}= \frac{19999999999912458800169}{10^{22}}\ne 2 $$ Thus $\dfrac{141421356237}{10^{11}}$ is not $\sqrt{2}$.

Without computations: $141421356237$ is odd, so its square is odd too. Hence the fraction cannot simplify to an even integer.