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Wed 29 Feb, 2012 02:45 pm
I am reading Rudins, Principles of Mathematical Analysis, on page 2 of the text is shown that the equation p^2 = 2 is not satisfied by any rational p.
In the proof...... If there were such a p, we could write p = m/n where m and n are integers THAT ARE NOT BOTH EVEN.
Q) The definition of rational number is that it is any number of the form m/n, where m and n are integers and n neq 0. How can it be assumed that m and n are not both even? Isn't imposing that restriction suggest that we are not dealing with the entire set of rational numbers ?
No. If both numbers in the fraction, which expresses the solution as a rational number, are even then both (numerator and denominator) are divisible by two, and the rartional solution is equivalently expressed by the ratio pf m/2 to n/2, at least one of which is ultimately not divisible by two. Another way of expressing the same idea is by looking for a rational number, expressible as m/n where m and n have no common integer factors.
@umanshan,
m<>n
n<>0
there is only one even prime number (2)
Rap
@georgeob1,
So a more encompassing definition of rational number is that it is any number of the form m/n, where m and n are integers (not both even) and n neq 0?
@umanshan,
No, I think your orignal definition of a rational number is fine. The proof requires the numerator and denominator to be relatively prime but that doesn't mean the definition of rational is incorrect. 3/6 is an acceptable rational number even if it can be reduced to 1/2.
@umanshan,
What it boils down to is that there are an infinite number of ways to represent each rational number. Requiring that the numerator and denominator aren't both multiples of two merely eliminates some of the representations - not some of the rationals.
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