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Spivak "root 2 is irrational number" problem

  1. Dec 23, 2015 #1
    Am using Spivak. Spivak elegantly proves that √2 is irrational. The proof is convincing. For that he takes 2 natural numbers, p and q ( p, q> 0)...and proves it.
    He defines irrational number which can't be expressed in m/n form (n is not zero).
    Here he defines m and n as integers.
    But in the earlier proof, he takes p and q as only natural numbers and not integers. Why?
    Is it because of definition, he earlier defined?
    ie....root of a real number a
  2. jcsd
  3. Dec 23, 2015 #2


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    Any rational number m/n (m, n integers, n≠0) can, by multiplying numerator and denominator by -1 if necessary, be written with n>0. If then (m/n)2=2, then also (-m/n)2=2. One of m and -m is positive. Hence, there are postive integers p,q such that (p/q)2=2, and this leads to the contradiction.
  4. Dec 23, 2015 #3


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    There's essentially no difference between looking for natural numbers or integers. An integer may be negative, but that makes no difference to divisibility properties.
  5. Dec 23, 2015 #4
    Yes...got it
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