Infinite number of rational numbers in an open interval

Walrus Rider

Hi guys

How would I go about proving this theorem -

"In any open interval (a,b) there are an infinite number of rational numbers"Simplest is to provide a set that has an infinite number of rational numbers between a and b.

Hint: First show there's a rational R between a and b.
Then show there's an infinite number of rationals r close enough to R that we can tell r is also between a and b.

Reply 2

Totally Tom

Can you set up a recurrance relation like if you find rational a, rational a+b which is still in the interval (a,b) is also rational?

Like, if you set b to be

\frac{1}{a^n}

Where a and n are both rational numbers. If you made a small enough coud you run through the number system giving you an infinite number of numbers which are similar to a?

Reply 3

Dadeyemi

take a+(b-a)/n where n is a positive integer above 1

where a 0
=> a+(b-a)/n > a
and
a a(n-1) < b(n-1)
=> an-a < bn-b
=> an-a+b <bn
=> a+(b-a)/n a < a+(b-a)/n < b

as it is true that there are and infinite number of positive integer,
there are an infinite amount of number for which a+(b-a)/n is in the interval (a,b) and as a,b and n are rational a+(b-a)/n must be rational.

Thus it follows that there must be an infinite number of rational numbers between the interval (a,b). QED.

Reply 4

Totally Tom

Dadeyemi

where a 0

??

Reply 5

Simplest way I know is:

1) Prove that in any open interval (x,y) there exists a rational.

Spoiler

Reply 6

Dadeyemi

take a+(b-a)/n where n is a positive integer above 1

where a 0
=> a+(b-a)/n > a
and
a a(n-1) < b(n-1)
=> an-a < bn-b
=> an-a+b <bn
=> a+(b-a)/n a < a+(b-a)/n < b

as it is true that there are and infinite number of positive integer,
there are an infinite amount of number for which a+(b-a)/n is in the interval (a,b) and as a,b and n are rational a+(b-a)/n must be rational.

Thus it follows that there must be an infinite number of rational numbers between the interval (a,b). QED.

What if

a=0

and

b=\sqrt 2

?

Reply 7

Dadeyemi

Ancient Runes

What if

a=0

and

b=\sqrt 2

?

I guess you could then need round b down to the nearest rational between a and b and then make that the new b with the lowest number of decimal places?
and the same could be for a if it was irrational but rounding up.

Reply 8

Dadeyemi

I guess you could then need round b down to the nearest rational between a and b and then make that the new b with the lowest number of decimal places?
and the same could be for a if it was irrational but rounding up.

As I'm sure you know, there is no "nearest rational" (less than b).

Reply 9

Walrus Rider

OP

Hint: First show there's a rational R between a and b.

Prove that in any open interval (x,y) there exists a rational.

Yeah, it's this bit that I'm having the problem with.

Thanks for the help, guys, by the way.

Reply 10

DFranklin

Walrus Rider

Yeah, it's this bit that I'm having the problem with.First show we can find an integer n with b-a < 1/n. Then consider multiples of 1/n and show one of them must lie inside (a,b).

Reply 11

There's also this really easy way I just thought of, by using a and b's decimal expansions, and the fact that any terminal decimal expansion is rational. arr.

Reply 12

Dadeyemi

Ancient Runes

As I'm sure you know, there is no "nearest rational" (less than b).

i mean using any rational between a and b as the new b and any rational between a and new a as the new a.

what i meant to describe is that you would cut off the decimals at a value higher or lower depending and whether it is a or b that is irrational.

e.g. (1,pi) =>1 and 3.14159
or (pi,7) => 3.1416 and 7

This is what I meant, I was just not thinking when i typed that explanation :p:

Reply 13