# How can we be completely sure that a number is really infinite and does not end in millions of additional digits?

187 views

How can we be completely sure that a number is really infinite and does not end in millions of additional digits?

In: Mathematics

Do you mean like irrational numbers? There are mathematical proofs showing them to be irrational. Any rational number can be expressed as a/b, where a is an integer and b is a non-zero integer. Both a and b cannot be even at once because then they simplify

Square Root of two is proven not to be rational [This page explains it better than I can type](https://www.math.utah.edu/~pa/math/q1.html) The proofs Pi is irrational are a bit too complicated to simplify [Just look at them](https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational#Laczkovich%27s_proof) If someone else can post an ELI5 answer for Pi, more power to them.

I presume you’re asking how we decide whether a number’s decimal expansion is infinitely long, not whether the number itself is infinite (which is impossible, since all real numbers are finitely large).

The general answer is that we can’t prove that the decimal digits will go on forever simply by computing those digits, because, as you note, they could always end at some point beyond our calculations. So we resort to using other techniques to prove that a number’s decimal expansion is infinite. One way is to prove that the number is irrational, which is usually done by contradiction: you assume it’s rational and show that this assumption leads to a contradiction with itself or other known facts, which then means that the assumption is false, i.e. the number is irrational. But there are also rational numbers with infinitely long decimal expansions, eg 1/3 = 0.3333…. To prove that the decimal expansion of 1/3 is infinitely long, it’s enough to show that the decimal expansion 0.333… equals 1/3 (which is done by considering the infinitely long decimal expansion as the limit of a sequence of finitely long decimal expansions and showing that the sequence converges to 1/3).

It’s also worth noting that decimal expansions aren’t necessarily unique, so some numbers can have both finite and infinitely long decimal expansions; for instance 1 = 0.999… (which can proven using the same technique mentioned for the case of 1/3 = 1, that is, considering the infinitely long decimal expansion as a limit and proving that the limit is 1). This non-uniqueness of decimal expansions is yet another reason mathematicians generally don’t find decimals to be particularly useful in theoretical work.

[removed]

[removed]

Based on the proofs in another response, there is absolutely no way this can be explained like you’re 5.

I have a university degree and I didn’t even begin to scratch the surface of understanding even a little of it. Math like that is basically its own language.

We can’t be and some mathematicians are skeptical of infinite numbers. In fact some are skeptical of just very large numbers that nobody could possibly count to in a lifetime. What does that mean really?

But in the end most people feel the idea of infinity is just too useful and beautiful to pass up. It’s an assumption we make it and it seems to work out pretty well.

I’m assuming you mean irrational numbers, so I’m just going to prove that a single number with infinite decimals exist. Let’s use the square root of two. Any number with a limited amount of decimals can be represented with a fraction of intergers, a quick way to think about it is the number 1,23 is 123/100, 1,234 is 1234/1000 and so on.

So now since we know that any number with a finite numbers of decimals is a rational number and we are claiming square root of two *(sqrt(2) from now on)* is a number with a finite number of decimals then this means square root of two can be represented as a fraction.

So we say that

sqrt(2)= P/Q

where P and Q are positive intergers. We can make sure that P and Q are one even and the other odd by reducing the fraction (the same way we can say 6/4=3/2, 6 and 4 are both even but 3/2 one is even and the other one is odd)

EDIT/CORRECTION: we can’t make sure P and Q are an even and an odd number (since 3/5 doesn’t have an even and an odd) but we can make sure that they don’t have common factors, meaning both numbers can’t be even since if both are even we can reduce the common factor of 2. So either both are odd or one is odd and the other is even.

So now we square both sides of the equation and get that

2=P^2 /Q^2

2* Q^2 = P^ 2

So if P^2 can be written by an interger times 2 then it means that P has to be even. So we know Q has to be the one that is odd.

Since P is even then we can write is as 2*M where M is an interger and therefore

P^2 = 2^2 * M^2 = 4* M^2

Ang going back to Q we get

2* Q^2 = 4* M^2 and moving the two we get

Q^2 = 2* M^2 which now tells us Q is an even number since is is the result of 2 times a positive integer for the same reason P was. This is an inconsistency since we specifically selected P and Q from one being even and the other odd.

So from assuming sqrt(2) could have a finite amount of decimals we got to an absurd of an odd number being even. Therefore sqrt(2) must have an infinte amount of decimals.

Similar demonstrations can be done with other numbers but this is just to prove that numbers with infinite decimals do exist.

Quick explanation:

If a number’s irrational, then it _must_ have an unending decimal expansion.

Proof: if it didn’t, and the expansion ended at some point, you could write that as (the digits up to the end) / 10^n , where n is the decimal place it ended at. But this is a rational number, oops.

If a number’s rational, but has ANY prime factors other than 2 or 5 in its denominator? It also has an unending decimal expansion.

Proof: Close to the same one. But now we look at what the factors of that denominator, 10^n , are: n factors of 2, and n factors of 5. No others.

So: if all you have are factors of 2 and 5, you can multiply by 2/2 or 5/5 (which both = 1) until the denominator has equal amounts of them … and then you have a terminating expansion, just convert that number back into a decimal that ends at the m-th place, where the denominator with balanced numbers of 2s and 5s had m of each.

And: if you have ANY other factors on the bottom, then you can never get the fraction into a form where 10^n is on the bottom – there’s always gonna be other factors multiplying it, and so you can’t turn it back into a terminating decimal.

–Dave, hope that’s relatively clear