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# P2 Proof watch

1. Anyone know how to prove there is an infinite number of rational numbers between 0 and 1?
2. (Original post by Mikan18)
Anyone know how to prove there is an infinite number of rational numbers between 0 and 1?
Rationa numbers can be written as a fractions.
So:
Create a fraction by x/(x+1). (x>0). This is greater than 0, but less than one as x<x+1.
You can create another fraction by adding one to boththe denominator and numeriator, resulting in (x+1)/(x+2)
You can keep adding 1's to top and bottom, ad infinitum. Hence there are infinite rational numbers between 0 and 1.
3. you *can* label each rational number 1, 2, 3, .... because there are an infinite amount of labels available, there are an infinite amount of rational numbers.
4. (Original post by elpaw)
you *can* label each rational number 1, 2, 3, .... because there are an infinite amount of labels available, there are an infinite amount of rational numbers.
Erm, between 0 and 1 Paw.
5. (Original post by Mikan18)
Anyone know how to prove there is an infinite number of rational numbers between 0 and 1?
Interesting. Here is an idea,

Assume there are a finite number of rationals between 0 and 1. Then there must be a smallest reational number between 0 and 1. Let the smallest raional be 1/n, where n is rational. But, 1/(n+1) is also a rational between 0 and 1 which is smaller than 1/n. That is a contradiction, so the original assumption is false.

There may be a nicer way, but I think the above may do.
6. Damn, I hate it when you have been on the post reply page for a while and then by the time you have posted loads of other people have replied.
7. here is one way
consider the closed intervals [0.2,0.4] and [0.6,0.8].
We now have 2 disjoint closed intervals between 0 and 1.
The process can be repeated between the upper limit of our first interval and the lower limit of our second interval. Thus we have an infinite number of closed disjoint intervals between 0 and 1. Every closed interval contains at least one rational and irrational number [I will show the proof if you like].
We therefore have an infinite number of sub disjoint closed intervals each containing a rational and irrational number. Thus we have an infinite number of rationals and irrationals over the closed interval [0,1]
8. (Original post by mikesgt2)
Damn, I hate it when you have been on the post reply page for a while and then by the time you have posted loads of other people have replied.

Yeh, but I think your proof is definitely the most elegant.
9. In general you could say
Given [a,b] where a<b and a,b are rational and a,b>0

then a+b is rational. sum of rationals is rational

(a+b)/2 is rational. Division of a rational by a rational is rational

now a < (a+b)/2 < b

repeating this process on [(a+b)/2 , b] and repeating gives

(a+b)/2<(a+2b)/4 < b

etc

thus

a< (a+nb)/(2^n) < b

for all integer n

(a+nb)/(2^n) is rational and thus there are an infinite number of rationals on the closed interval.

max
10. sorry i made a mistake it should be

(a+b)/2<(a+3b)/4 < b

and

(a+((2^n)-1)b)/(2^n)

max
11. (Original post by mikesgt2)
Damn, I hate it when you have been on the post reply page for a while and then by the time you have posted loads of other people have replied.
Ditto
Nice proof!
12. Given any two rationals, their average will be rational and a number between the two, so there are infinitely many.
13. (Original post by Ralfskini)
Yeh, but I think your proof is definitely the most elegant.
Hmmm...

Here are infinitely many rational numbers between 0 and 1:

1/2, 1/3, 1/4, ... .
14. (Original post by Jonny W)
Hmmm...

Here are infinitely many rational numbers between 0 and 1:

1/2, 1/3, 1/4, ... .
That is what I was thinking first, the fact that there are infinitely many natural numbers is sufficient to show that there are infinitely many rationals between 0 and 1. I just thought that sometimes questions force you to arrange a proof in nice steps.

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