Prove that if a is any integer and a is not 0, then a^2 is any positive integer?

Def: the square of x is defined to be x^2=xx

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  • 9 years ago

    Your question doesn't make sense to me.

    The phrase "then a^2 is any positive integer" is the problem.

    3 is a positive integer but is not the square of any integer.

    The product of two integers is always an integer(another way of saying this is the integers are closed under multiplication) and is always positive if the integer is not zero. Do you mean to show that the product of two negative integers is positive? This follows from the fact that (-1)^2=1 for if a<0 then -a>0 => (-a)*(-a)>0.

    But (-a)(-a)=[(-1)*a]*[(-1)*a]=(-1)^2*a^2=a^2. Since we've shown that (-a)(-a)>0 and (-a)(-a)=a^2,

    it follows that a^2>0.

  • 9 years ago

    Prove both cases:

    if a > 0, then a^2 = a*a > 0

    If a < 0, let a = -b where b > 0, then a^2 = a*a = (-b)*(-b) = b*b > 0

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