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!