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Why is it that (-1)^2 = 1?

  1. Aug 19, 2011 #1
    I got confused reading spivak's calculus book and couldn't understand why. They said something like

    "if you add ab to both sides, you get (-a)(-b)=(ab)"

    It's on the 7th page at the bottom. I believe it's the older version of the textbook.
  2. jcsd
  3. Aug 19, 2011 #2
    Well, before that paragraph, Spivak proved that

    [tex](-a)\cdot (-b)+(-(a\cdot b))=0[/tex]

    So adding [itex]a\cdot b[/itex] to both sides gives us

    [tex](-a)\cdot (-b)+(-(a\cdot b))+(a\cdot b)=a\cdot b[/tex]

    and thus

    [tex](-a)\cdot (-b)+0=a\cdot b[/tex]

    and because 0 is neutral:

    [tex](-a)\cdot (-b)=a\cdot b[/tex]
  4. Aug 19, 2011 #3
    Could you also help me understand why a(b+c) = ab+ac?

    I was looking at it and wondering "wait a minute, i don't see what he did in P(8) had any relation to ab+ac=a(b+c)
  5. Aug 19, 2011 #4


    Staff: Mentor

    Textbooks in English usually use the terminology "0 is the additive identity," meaning that you can add 0 to any number without changing the number.
  6. Aug 19, 2011 #5
    That is true by axiom P9. It is (b+c)a=ba+ca that can be proven by P8. Indeed

    & = & a(b+c)\ \ \ \ P8\\
    & = & ab+ac\ \ \ \ P9\\
    & = & ba+ca\ \ \ \ P8\\
  7. Aug 19, 2011 #6
    I was looking at the textbook but P8 seems like it had nothingi to do with why a(b+c)=ab+ac. I don't really understand it.
  8. Aug 19, 2011 #7
    Do you understand my previous post?
  9. Aug 19, 2011 #8
    To be quite honest, no not really.
  10. Aug 19, 2011 #9
    a(b+c)=(b+c)a by P8. Do you see that?
  11. Aug 19, 2011 #10
    I really don't to be honest. Wasn't p8 talking about commutative law?
  12. Aug 19, 2011 #11
    Yes. P8 says that for any x and y it holds that


    Now take x=a and y=b+c. Then

  13. Aug 19, 2011 #12
    Hm ok but then for p9, how'd they get the distributive law? Sorry i'm still kinda not seeing this.
  14. Aug 20, 2011 #13
    I don't have access to your book, so it's entirely possible that I'm misreading the question... but I'll try to help out anyway, and hopefully not create further confusion.

    a(b+c) = ab + ac
    The distributive law isn't as bad as it might look. Sometimes the variables can mess with my head a bit, but to alleviate this I just have to remind myself that they simply represent numbers. Plug in some numbers and try it out. Remember that anything contained in parentheses is multiplied by the number outside it (either before or after, it doesn't matter as per "p8"). [a(b+c) = (b+c)a]

    Anyway, here's a simple example:
    a= 7
    b= 2

    Multiply 7 by 2 (answer: 14), and add it to 7 multiplied by 3 (answer: 21). You get 35. I could have just as easily written 7(2+3) as 7(2)+7(3). In variable form, it would look like a(b+c) = ab+ac. Remember that ab is the same as a(b). This is how you'll see it written once it's converted to number form, because while ab is clear, 72 is not when you actually intend 7(2). So just as a(b) is [a] multiplied by , a(b+c) is [a] multiplied by (added to) [a] multiplied by [c]. You're just juggling numbers. Or variables. (same thing, really!)

    It's easy to see the process when you replace the variables with the numbers they represent; (b+c) is representative of (2+3). Interestingly, if you add (2+3) together to get 5, and multiplying THAT by 7, you still get 35. The distributive law just allows you to move things around to reduce clutter -- it will still give you the same answer. With simple 7s, 2s and 3s it might seem like a waste of time, but when you're using larger numbers, different variables and more complicated equations, it comes in handy.

    So to sum it up.
    7(2+3) = 35
    7(2) + 7(3) = 35
    7(5) = 35
    This is the distributive property. ;) Hope it helps.
  15. Aug 20, 2011 #14
    ^Thanks but i'm wondering how and why it works for all real numbers. For any natural numbers, i know it will work but i'm wondering how it could also work for the rest of the numbers?
  16. Aug 20, 2011 #15


    User Avatar
    Homework Helper

    Well, if you understand why (-1)(-1) = 1, then it automatically follows for all real numbers. Let a be a real number, so:

    (-a)(-a) = (-1)(-1)(a)(a) = a2
    Last edited: Aug 20, 2011
  17. Aug 20, 2011 #16
    Actually now i'm wondering why the distributive law works and what the proof is for all Real number.
  18. Aug 20, 2011 #17
    Multiplication means that you take something y by x times.

    for example 4*5 means you add 4 up 5 times, ie 4+4+4+4+4

    It doesn't matter if you add two fours up now and then fours up later and then add both of them together.


    Multiplication is repeated addition, and as long as you repeat the addition the same amount of times, it doesn't matter if you add some up now and some later.
  19. Aug 20, 2011 #18
    The distributive law is an axiom, there is no proof.
  20. Aug 20, 2011 #19
    Let [itex]a[/itex] and [itex]b[/itex] be an integer number such that [itex]\{\exists\;a, b \mid a,b \in \mathbb{Z}\}[/itex]

    The distributive law is an axiom which states that:

    [itex]a(b + c) = ab + ac[/itex]

    Now suppose we have the expression

    [itex](-a)(b) + (a)(b)[/itex]

    By factoring the common divisor:

    [itex](-a)(b) + (a)(b) = b [(-a) + a] = b(0) = 0[/itex]

    [itex]b(0) = 0[/itex] because

    [itex]b(0) = b(-b + b) = (-b)(b) + (b)(b) = -b^2 + b^2[/itex]

    if [itex]b^2[/itex] is a new number d such that [itex]\sqrt{d} = b[/itex] then

    [itex]-b^2 + b^2 = -d + d = 0[/itex]

    This just shows that a negative number plus a positive number is 0 or that a negative number times a positive number is negative.

    Now suppose that

    [itex](-a)(-a) + (-a)(a) = (-a)[(-a) + a] = (-a)(0) = 0[/itex]

    We have shown that [itex](-a)(b)[/itex] must be negative and therefore [itex](-a)(a)[/itex] must be negative and in order for this to equal zero [itex](-a)(-a)[/itex] must be positive

    [itex](-a)(-a) = (-a)^2 = a^2[/itex]

    [itex]a^2 \in \mathbb{N}[/itex] which is the same thing as saying [itex]a^2 \in \mathbb{Z}^+[/itex]

    Now [itex]\mathbb{N} = \mathbb{Z}^+ \subset \mathbb{Z} \subset \mathbb{R}[/itex] thus the laws must hold for all [itex]\mathbb{R}[/itex].

    I probably made very bad mistakes.
    Last edited: Aug 20, 2011
  21. Aug 20, 2011 #20


    User Avatar
    Science Advisor

    What is an "axiom" and what is a "theorem" depends upon where you start. Yes, we can start with the distributive law and others as axioms but if you start, instead, from Peano's axiom the distributive law is proved as a theorem.
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