prove that a is any integer and x is not 0, then a^2 is any positive integer..help please!?

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

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  • 9 years ago
    Favorite Answer

    All integers squared (except zero) are positive integers. This is because when you square a number, you multiply it by itself.

    If you multiply a positive number by itself, you are multiplying a positive by a positive and the answer is going to be positive.

    If you multiply a negative number by itself, you are multiplying a negative by a negative and the answer will also be positive.

    The only time it wouldn't be is if the number is zero, because 0^2 = 0, which is neither positive or negative.

  • Irv S
    Lv 7
    9 years ago

    Proof read your question.

    As you give it , X doesn't enter the equstion.

    If you mean 'a is any positive integer' the proof fails because while the square if any integer

    will be a positive integer, the value of a ^2 is the set of squares of positive integers which does not

    include all integers. (Try to find an integral Sqrt for 5 or 7 eg)

  • 9 years ago

    That can't be what you're supposed to prove. For one thing, it's missing the word "if". For another, it doesn't say anything meaningful. It sounds like you're trying to prove that all positive integers are perfect squares which of course is nonsense. And the statement about a^2 doesn't involve x, so what does the "x is not 0" have to do with it?

    What was the real statement?

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