1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Square Of A Number Is Non-negative

  1. Jan 3, 2014 #1
    Hello, I am seeking some aid in proving that the square of a number is always non-negative. Here is some of my proof:

    A number, call it a, is either positive, negative, or zero. A number squared is produced when you take the number and multiply it by itself. So, we have three cases to consider.

    When a < 0:

    If a < 0, then 0 - a > 0; and so, 0 - a = -a must be in the set of positive numbers. If -a is in the set of positive numbers, then, by property P12 (Closure under multiplication: If a and b are in P, then a • b is in P), (-a)(-a) must be in the set of positive numbers, implying that it is a positive number, and the negative signs may vanish. (-a)(-a) = a x a = a^2 > 0.

    Initially, I thought that this proof would be valid; but as I went along further along, I developed an inkling of a feeling that it was not so.

    What is wrong with the proof of the case a > 0?
     
  2. jcsd
  3. Jan 3, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The only thing "wrong with the proof of the case a> 0" is that the "proof" says nothing about a> 0! You say "when a< 0" but give no other case.

    Of course, if a> 0 then "property P12" itself immediately tells you that a*a is positive.
     
  4. Jan 3, 2014 #3
    No, I intentionally did not include the case when a > 0, because I had ran into trouble with the case when a < 0. Just to be clear, my proof for the case a < 0 is valid?
     
  5. Jan 3, 2014 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    That depends on what properties you are permitted to use. You say that (-a)*(-a) = a^2 (true!), but have you already been given this, or is this yet again another thing that needs to be proved first?
     
  6. Jan 3, 2014 #5

    phinds

    User Avatar
    Gold Member
    2016 Award

    OR ... imaginary. Are you leaving those out on purpose? You didn't say so.
     
  7. Jan 6, 2014 #6
    The difficulty I am having with my proof is when I state, "If a < 0, then 0 - a > 0; and so, 0 - a = -a must be in the set of positive numbers." Doesn't that mean that -a is a positive number, and clearly not a negative number, implying the next step I proceed is simply the multiplication of two POSITIVE numbers, (-a)(-a), which doesn't prove anything about the multiplication of two NEGATIVE numbers? Or am I wrong, and my initial proof is correct?
     
  8. Jan 6, 2014 #7

    rcgldr

    User Avatar
    Homework Helper

    I'm not sure what mathematical rules you're allowed for the proof, but perhaps these statements could be used:

    (+1)(a) = a
    (-1)(a) = -a
    (+1)(+1) = 1
    (-1)(-1) = 1
    ((a)(b))(c) = (a)((b)(c)) ; multiplication is associative
     
  9. Jan 6, 2014 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No, it proves exactly what you want. It all hinges on whether or not you know or are allowed to use the property a*a = (-a)*(-a). Remember me asking about that in my first response? You never gave an answer.
     
  10. Jan 6, 2014 #9
    I am terribly Sorry, Ray, for not having answered your question. Yes, I am able to use the fact that
    a*a = (-a)*(-a)
     
  11. Jan 6, 2014 #10

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No need to apologize: the question was for your benefit, not mine.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Square Of A Number Is Non-negative
  1. Squares and Numbers (Replies: 4)

Loading...