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Math Help - proposition proofs

  1. #1
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    proposition proofs

    to prove that there is no rational numbers whose square is equal to 2, my book states that:

    lets use the fact that a prime factor of a square number occurs an even number of times. this p^2 = 2q^2. on the LHS, the prime factor 2 must occur an even number of times while RHS, 2 will occur an odd number of times.

    thus this is a contradiction to the fact that prime factorization of a number is unique.

    hi! could someone explain to me what does it mean to the sentence in bold and why it leads to the proposition being true?

    thanks
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    to prove that there is no rational numbers whose square is equal to 2, my book states that:

    lets use the fact that a prime factor of a square number occurs an even number of times. this p^2 = 2q^2. on the LHS, the prime factor 2 must occur an even number of times while RHS, 2 will occur an odd number of times.

    thus this is a contradiction to the fact that prime factorization of a number is unique.

    hi! could someone explain to me what does it mean to the sentence in bold and why it leads to the proposition being true?

    thanks

    Well do you know the fundamental theorem of arithmetic? It states that any integer n \geq 1 factors uniquely into a product of primes.

    Hence if a is such an integer it can be written as a unique product of primes ie. a = p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}. Where e_1, e_2, \ldots, e_r are the exponents of the prime number, these can be any integer greater than or equal to 0. For example 2 = 2^13^0 etc. or 4 = 2^2, so no matter what we can assign a variable to the value exponent of any prime factor. Hence in the proof in your textbook here is what they did.

    Claim: \sqrt{2} is irrational
    Proof: Suppose it is rational, if supposing this leads to a contradiction, then \sqrt{2} must be irrational.
    So if \sqrt{2} was rational we could write \sqrt{2} = \frac{p}{q}. This imples 2 = \frac{p^2}{q^2} or that 2q^2 = p^2. Now both p, q have unique prime factorizations, so each will have a certain exponent of 2 in their prime factorization. ie. p = 2^ep_2^{e_2}\cdots p_r^{e^r} and q = 2^fp_2^{f_2}\cdots p_r^{f^r}. So when we square these values we get p^2 = 2^{2e}p_2^{2e_2}\cdots p_r^{2e_r} and q^2 = 2^{2f}p_2^{2f_2}\cdots p_r^{2f_r} .

    Now we have that 2q^2 = p^2, if we replace them by the factorizations we get 2*2^{2f}p_2^{2f_2}\cdots p_r^{2f^r} = 2^{2e}p_2^{2e_2}\cdots p_r^{2e^r}, now combine the exponent of the 2 on the LHS, to get 2^{2f + 1}p_2^{2f_2}\cdots p_r^{2f_r} = 2^{2e}p_2^{2e_2}\cdots p_r^{2e_r}.

    So now see what what has happened, we have a 2 numbers being equal to each other but for one the prime factor 2 has an exponent of 2f + 1 and for the other it's 2e. Since the factorization is unique we must have the two numbers be equal or that 2f + 1 = 2e, thus we would have an odd number equaling an even number(contradiction). Or we could say the factorization is not unique and these are 2 different factorizations, which is a contradiction again. And it was this contradiction your book mentioned.
    \blacksquare
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