# Why is the hypotenuse of a triangle with length $\sqrt{2}$ finite even though its decimal expansion continues infinitely?

My answer:

We can represent the decimal part of the number as a sum of fractions. $$\sqrt{2}=1.41421356237...$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.41421356237...=1+\frac{4}{10}+\frac{1}{100}+\frac{4}{1000}+\frac{2}{10000}+\frac{1}{100000}...$$ As you continue to add each term, the value of the each term added will get smaller since the numerator varies between 0 and 9 while the denominator gets exponentially larger. As the denominator approaches infinity, the value of the terms becomes negligible. For this reason, the $$\sqrt{2}$$ is finite in length.

Am I wrong? If so, where at? Also, are there better ways I could have phrased my answer? Thank you so much everyone!

• How could $\sqrt2$ not be finite? Are there numbers which are infinite? Commented Feb 11, 2022 at 9:54
• The summands get smaller quickly enough to ensure that the sum has a finite value. If this were not the case , only the numbers with terminating decimal expansion would have a finite value, not even $\frac{1}{3}$ would be finite. Commented Feb 11, 2022 at 9:54
• $0<\sqrt{2}<2$ since $0<2<4$. Or you can say that if $\sqrt 2$ was not finite, its square, which is $2$, would probably be not finite either.
– Gary
Commented Feb 11, 2022 at 9:55
• I agree your argument, although it could be made more formal. But important is that you understood what is going on and did not fall into a Zeno-like trap. Commented Feb 11, 2022 at 9:59

## 1 Answer

The heart of your question is why the decimal part of any number, represented as an infinite series, converges. Note the generic geometric series converges: $$1+r+r^2+...=\frac{1}{1-r},\quad |r|<1.$$

The decimal part of any number in base 10 as a series satisfies

$$0.a_1a_2a_3...=\frac{a_1}{10}+\frac{a_2}{100}+\frac{a_3}{1000}+...,\quad a_i\in \{0,1,...,9\}\\ \leq \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...\\ = \frac{9}{10}(1+\frac{1}{10}+\frac{1}{100}+...)\\ = \frac{9/10}{1-1/10}=1.$$

• Thank you for your explanation and referencing the geometric series. I didn't know that every decimal part of every number converged until now. Commented Feb 11, 2022 at 10:08
• @anon1212 No prob. I guess it's sort of taught implicitly that the decimal part of a number is never greater than 1. Writing it as a series formalizes this claim. Commented Feb 11, 2022 at 10:10
• @ryang How would you phrase it? Commented Feb 11, 2022 at 10:20
• +1 to this distillation. @anon1212 Your suggested argument (in the Question post) "the individual terms of an infinite series approaches zero, thus the series converges" isn't valid; for example the series $\sum \frac1n$ diverges. [I'm critiquing the argument not the phrasing.] Commented Feb 11, 2022 at 10:21
• Yes, and that's why we're on MSE to learn. Welcome here! @anon1212 Commented Feb 11, 2022 at 10:35