Prove that the rationals are dense

[PcumP_Ravenclaw]

Show that there are infinitely many rational numbers $\frac{a +m*c}{b + m*d}$ between the two rational numbers $a/b$ $c/d$. m is any positive integer.

My attempt:

first make common denominator

$\frac{a*d}{b*d}$
$\frac{c*b}{d*b}$
all numbers going from $a*d$ to $c*b$ divided by $b*d$ is in this interval. but these are finite so we have to increase the resolution. say multiply numerator and denominator by 1000 so now $a*d*1000$ to $c*b*1000$ divided by $b*d*1000$ .

It can be 100000 or anything very large say multiplied to infinity. how do I write this mathematically?

 

[RUber]

I am not sure I understand what you are trying to argue. Is ad<cb? How are you going to show the form stated in the problem?
Try looking at values for m from zero to infinity. You should be able to conclude that they are all in the interval [a/b,c/d).

 

[Mark44]

Show that there are infinitely many rational numbers $\frac{a +m*c}{b + m*d}$ between the two rational numbers $a/b$ $c/d$. m is any positive integer.

My attempt:

first make common denominator

$\frac{a*d}{b*d}$
$\frac{c*b}{d*b}$
all numbers going from $a*d$ to $c*b$ divided by $b*d$ is in this interval. but these are finite so we have to increase the resolution. say multiply numerator and denominator by 1000 so now $a*d*1000$ to $c*b*1000$ divided by $b*d*1000$ .

It can be 100000 or anything very large say multiplied to infinity. how do I write this mathematically?

I would do this in steps.
1. Show that $\frac{a + mc}{b + md}$ is less than c/d, for any choice of a positive integer m.
2. Show that $\frac{a + mc}{b + md}$ is greater than a/b, for any choice of a positive integer.
I am assuming that a/b < c/d.
If you can prove the above propositions, you're done, since m can be any integer in the set {1, 2, 3, ...}

 

[PcumP_Ravenclaw]

I am not sure I understand what you are trying to argue. Is ad<cb? How are you going to show the form stated in the problem?
Try looking at values for m from zero to infinity. You should be able to conclude that they are all in the interval [a/b,c/d).

I want to know how to derive the form $\frac{a +m*c}{b+m*d}$

 

[Mark44]

I want to know how to derive the form $\frac{a +m*c}{b+m*d}$

I don't believe that you need to derive it, just show that there are an infinite number of rationals of this form between a/b and c/d.

I think that how it works is that given a/b and c/d, where a/b < c/d, the rational number with m = 1, $\frac{a +1*c}{b+1*d}$, is halfway between a/b and c/d. The same fraction with m = 2 gives a number that is 2/3 of the way between a/b and c/d. When m = 3, the fraction gives a number that is 3/4 of the way between a/b and c/d, and so on. These are the results I get when I choose a/b = 1/4 and c/d = 1/2, for a few selected values of m.

 

[PcumP_Ravenclaw]

Is not halfway between $a/b$ and $c/d$ equal to $\frac{a*d+c*b}{2*b*d}$?

 

[SammyS]

Is not halfway between $a/b$ and $c/d$ equal to $\frac{a*d+c*b}{2*b*d}$?

Yes it is.

Furthermore, the result for $\displaystyle \ \frac{a+m\cdot c}{b+m\cdot d} \$ depends not only on the two rational numbers, a/b and c/d but depends on the particular values used for each of a, b, c, and d .

For Mark's example with a/b = 1/4 and c/d = 1/2:

Using a = 1, b = 4, c = 2, d = 4 the result with m = 1 is: $\displaystyle \ \frac{1+1\cdot 2}{4+1\cdot 4}=\frac38 \ .$

Using a = 1, b = 4, c = 1, d = 2 the result with m = 1 is: $\displaystyle \ \frac{1+1\cdot 1}{4+1\cdot 2}=\frac13 \ .$​

Each of these results is between 1/4 and 1/2. Only the first is halfway between .

 

[PcumP_Ravenclaw]

Can I say that my way to find rational numbers inbetween $a/b$ and $c/d$ is
$\frac{a*d+c*b}{m*b*d}$ where m is an integer from 1 to infinity. I realize that this will miss certain rational numbers but there is no way to encompass all the rational numbers using any formula involving integers like m? because between m = 1 and m = 2 there are infinitely many fractions? so the formula $\frac{a+m\cdot c}{b+m\cdot d}$ is also similar to mine missing some rational numbers so whose formula has higher resolution?

Thank you!

 

[haruspex]

Can I say that my way to find rational numbers inbetween $a/b$ and $c/d$ is
$\frac{a*d+c*b}{m*b*d}$ where m is an integer from 1 to infinity.

that only works for m=2. You need some more references to m in the expression.
Anyway, the way I read the question, you are required to show specifically that there are infinitely many of the form $\frac{a +m*c}{b + m*d}$

 

[RUber]

It seems pretty straightforward that if m = 0, you have a/b and the limit as m goes to infinity will surely be c/d. From my vantage point, the only challenge would be to show explicitly that $\frac ab \leq \frac{a+mc}{b+md} < \frac cd$ for all natural m.
This might be easier if you look at this as some sort of average.
Perhaps a is a total number of something done by b people, and c is the total number done by d people, the $\frac{a+c}{b+d}$ is the average done by all the people.
If every additional group of d people can do c things, than you can continue to average these events out by taking the total number done (a+mc) and dividing by the total number of people (b+md).
From this perspective, it should be clear that your mean can never exceed your largest value (c/d) and it can never be less than your smallest value (a/b).

 

[SammyS]

that only works for m=2. You need some more references to m in the expression.
Anyway, the way I read the question, you are required to show specifically that there are infinitely many of the form $\frac{a +m*c}{b + m*d}$

I agree with haruspex here. (I usually do.)

If you can show that $\displaystyle \ \displaystyle \ \frac{a+c}{b+d} \$ is between $\displaystyle \ \frac ab \$ and $\displaystyle \ \frac cd \,,\$ then that will show that $\displaystyle \ \frac{a+m\cdot c}{b+m\cdot d} \$ is between $\displaystyle \ \frac ab \$ and $\displaystyle \ \frac cd \$ in general. Can you see why?