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