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Sciman
Lv 6
Sciman asked in Science & MathematicsMathematics · 1 decade ago

Can you answer this conjecture? I have plenty of evidence to suggest it is correct.?

a, b, and k are positive integers. k is a prime number (not including 1). 0<a<k and 0<b<k. Prove that k is not a factor of ab.

Update:

Tomp gives an interesting answer, which I am studying. It will take me time to go through it.

However, one flaw might be that a or b could be prime numbers, though each less than k. I do not know if the proof assumes that a and or b are composite.

Perhaps, if that is the case, the proof may not work.

2 Answers

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  • Tomp
    Lv 7
    1 decade ago
    Favorite Answer

    Let the prime factorisation of a be

    a = p1^m1 * p2^m2 * ...... * pr^mr

    where pi is one of the prime factors of a raised to the power of mi and pi <= a < k

    Similarly for b

    b = q1^n1 * q2^n2 * ...... * qs^ns

    where qj is a prime factor of b raised to the power of nj and qj <=b < k

    Note that the above two factorisations are unique to a and b.

    Now

    ab = (p1^m1 * p2^m2 * ...... * pr^mr)(q1^n1 * q2^n2 * ...... * qs^ns)

    Since k is prime, none of the prime factors of a or b can divide k.

    Conversely k cannot divide pi (1 <= i <= r) nor qj (1 <= j <= s)

    Hence k cannot be a factor of the product ab

    By the way, 1 is not a prime number. A prime number k is defined by the fact that it can only have two divisors, 1 and k itself.. The number 1 has only one divisor, 1 itself.

  • 1 decade ago

    the only POSITIVE PRIME is 2 Therefore k = 2.

    If a and b are positive and less than 2 then they do not exist

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