How Can I Prove There Are c Sequences of Rational Numbers?

[MathematicalPhysicist]

i need to prove that there are c sqequences of rational numbers.
basically, i need to show that |Q^N|=c.
here, are a few attempts from my behalf:
i thought that Q^N is a subset of R^N, so |Q^N|<=c, but this doesn't help here, so i thought perhaps to find a bijection from {0,1}^N to Q^N.
i know that each rational number can be represented in base 2 by the digits 0,1, but I am having difficulty to formalise this idea.

 

[HallsofIvy]

Any real number can be identified with a specific sequence of rational numbers. For example, the real number $\pi$ can be identified with the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, with each term (a rational number because it is a terminating decimal) being one more decimal place in the decimal expansion of $\pi$. This gives a one-to-one function from the set set of real numbers to the set of all sequences of rational numbers and so shows that $c\le |Q^N|$.

 

[Hurkyl]

Grr. If I remember correctly, it's a theorem that for infinite c:

$$a^c = b^c$$

whenever $a \leq b \leq c$. The proof isn't immediately leaping to mind, though.

Oh, bother, I just checked Wikipedia, and it looks like you need the axiom of choice to prove that, so the proof won't be as easy as I had hoped.

Anyways, LQG, remember that |Q| = |N|. It might be easier to try and prove |N|^|N| = 2^|N| or |N|^|N| = |R| instead.

But, we see HoI seems to have proven it directly, so don't listen to me.

 

[HallsofIvy]

Oh, Hurkyl, I'm blushing.


By the way, I had a friend who, on another forum, used the name "Hog on Ice" which was regularly abbreviated "HOI". So whenever I see "HOI" used for "HallsofIvy", I'm surprized!