MHB Is Adding Even Numbers to Fractions Enough to Prove Countable Infinity?

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The discussion revolves around proving that the set of fractions q = a/b, where a is even and b is odd, is countably infinite. One participant suggests starting with the general proof that the set of rational numbers, Q, is countably infinite, which implies that any subset, including the specified set, is also countable. The "swan-walk" method is introduced as a way to enumerate fractions in a countable manner. Another participant questions whether this approach applies given the specific conditions of a and b. To demonstrate that the set is infinite, it is suggested to show that there are infinitely many distinct elements, noting that the condition of b being odd allows for an infinite number of even integers in the numerator.
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I am trying to prove how this set is countably infinite:

q∈Q:q=a/b where a is even and b is odd

a needs to be even and b needs to be odd, so I thought this would prove that it would be countably infinite:

q = a/b + x/x, where x is any even number.

a always needs to be even and b always needs to be odd, so if they have the value of x added to them with x being any even number, they would always be positive or negative.

My question is: is this enough to prove that this set is countably infinite? If not, what do I need to do?
 
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JProgrammer said:
I am trying to prove how this set is countably infinite:

q∈Q:q=a/b where a is even and b is odd

a needs to be even and b needs to be odd, so I thought this would prove that it would be countably infinite:

q = a/b + x/x, where x is any even number.

a always needs to be even and b always needs to be odd, so if they have the value of x added to them with x being any even number, they would always be positive or negative.

My question is: is this enough to prove that this set is countably infinite? If not, what do I need to do?

Hey JProgrammer! ;)

How about proving that $\mathbb Q$ is countably infinite to begin with?
So any subset would be countably infinite as well.

Have you heard of the "swan-walk"?
That is, we run through all fractions in the following order:
$$\frac 11, \frac 21, \frac 12, \frac 13, \frac 22, \frac 31, \frac 41, \frac 32, \frac 23, \frac 14, \frac 15, ...$$
This is the swan walk. Its meaning will be apparent if we visualize it in X-Y coordinates.
It shows that we can iterate through all fractions in a countable fashion.
Put otherwise, we have a bijective function $\mathbb Z \to \mathbb Q$, proving that $\mathbb Q$ is countable... (Thinking)
 
I like Serena said:
Hey JProgrammer! ;)

How about proving that $\mathbb Q$ is countably infinite to begin with?
So any subset would be countably infinite as well.

Have you heard of the "swan-walk"?
That is, we run through all fractions in the following order:
$$\frac 11, \frac 21, \frac 12, \frac 13, \frac 22, \frac 31, \frac 41, \frac 32, \frac 23, \frac 14, \frac 15, ...$$
This is the swan walk. Its meaning will be apparent if we visualize it in X-Y coordinates.
It shows that we can iterate through all fractions in a countable fashion.
Put otherwise, we have a bijective function $\mathbb Z \to \mathbb Q$, proving that $\mathbb Q$ is countable... (Thinking)

Thanks for your reply.

I have heard of this before, but I wasn't sure if it would work for this problem because a needs to be even and b needs to be odd. This would work for this problem?

Since this set has been proven to be countable, it needs to be proven to be infinite. How would I prove that?
 
I like Serena said:
How about proving that $\mathbb Q$ is countably infinite to begin with?
So any subset would be countably infinite as well.
Well, this is an overstatement. (Smile)

JProgrammer said:
Since this set has been proven to be countable, it needs to be proven to be infinite. How would I prove that?
You need to find infinite number of different elements. This is not hard.
 
"b odd" includes b= 1 so the set of all even integers is a subset.
 
There is a nice little variation of the problem. The host says, after you have chosen the door, that you can change your guess, but to sweeten the deal, he says you can choose the two other doors, if you wish. This proposition is a no brainer, however before you are quick enough to accept it, the host opens one of the two doors and it is empty. In this version you really want to change your pick, but at the same time ask yourself is the host impartial and does that change anything. The host...

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