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How to prove -(-x)=x

  1. Sep 18, 2008 #1
    I realize this is really simple but I'm stuck, I've worked on it for awhile All that needs to be done is proving that -(-x)=x. We are allowed to use:
    1) (x+y)+z=x+(y+z)
    2) (xy)z=x(yz)
    3) x(y+z)=xy+xz
    4)There is a unique element 0 element of real numbers s.t. 0+x=x for all x
    5) For each x an element of real numbers there is a unique y element of real numbers s.t. x+y=0 and we write y=-x
    6) There is a unique element 1 an element of the real numbers s.t. x(1)=x for all x element of r and 0 does not equal 1
    7) For each x an element of the real numbers with x not equal to 0 there is a unique element y an element of the real numbers s.t. x(y)=1 and or x(1/x)=1
    8) x is greater than y implies x+z greater than y+z
    9) x less than y and y less than z implies x less than z
    10) For x,y an element of the real numbers exactly one of the following is true: x greater than y, y greater than x, or x=y
    11) x greater than y and z greater than 0 implies xz less than yz
     
  2. jcsd
  3. Sep 18, 2008 #2
    x = x
    -1(x) = -1(x)
    -x = -x
    (-x)/-1 = x

    and since 1/-1 = -1/1 = -1, then we have
    -(-x) = x
     
  4. Sep 18, 2008 #3
    0 + -(-x) = -(-x)

    But 0 = x + -x

    so

    [x +-x] + -(-x) = -(-x)

    x + [-x + -(-x)] = -(-x)

    x + 0 = x = -(-x)
     
  5. Sep 18, 2008 #4
    In this step:

    x + [-x + -(-x)] = -(-x)

    Aren't you assuming that -x + -(-x) = 0, which is the same as assuming that -(-x) = x?

    Or am I missing something?
     
  6. Sep 18, 2008 #5

    Doesn't it have to be true though? Let -x = y

    Then y + -(y) = 0. You bring up a good point though, so I'm wondering if my logic is correct...
     
  7. Sep 18, 2008 #6
    It has to be true, but it's what we're trying to prove, so I don't think you can assume it. But what you just posted, I think that would be a good proof.
     
  8. Sep 18, 2008 #7
    y + -y = 0

    let y = -x

    -x + -(-x) = 0

    Add x to both sides,

    LS:x -x -(-x) = 0 -(-x) = -(-x)
    RS: x + 0 = x

    hence, -(-x) = x

    QED

    Same flaw, I give up

    :cry:
     
  9. Sep 18, 2008 #8
    I don't see any flaw. If you let y = -x then you have y + -y = 0, which is -x + -(-x) = 0, which implies that -(-x) = x.

    Looks good to me.
     
  10. Sep 18, 2008 #9
    We have x + (-x) = 0 by the definition of -x (from thm. 5). But this equation tells us that x is the negative of -x. That is, x = -(-x)
     
  11. Sep 19, 2008 #10
    No, he is using rule #5, -x is a real number, so there exists another number y such that -x + y = 0, and we write y = -(-x).

    You might be able to see it clearer if you use different variables for your axioms than you do for what you're trying to prove. Rule 5 says if I have:

    <triangle> in Reals, then there exists y in Reals such that <triangle> + y = 0; we say y = -<triangle>.

    Stuff like this is confusing because the axioms are not proveable. You start out with a set you like, and derive the rest from it. You could have, instead, started with -(-x) = x and then proved #5 (probably).. just boils down to w/e axioms you want to assume.
     
  12. Sep 19, 2008 #11
    -(-x)+(-x)=0,

    also x+(-x)=0 so

    -(-x)+(-x)=x+(-x) we add -(-x) on both sides

    [-(-x)+(-x)]+[-(-x)]=[x+(-x)]+[-(x)]

    -(-x)+[(-x)+(-(-x))]=x+[(-x)+(-(-x)]

    -(-x)+0=x+0

    -(-x)=x

    Or, we could have concluded right away from the uniqueness of the inverse element, that since

    -(-x)+(-x)=0. and since x+(-x)=0, it means that x is the inverse of (-x) but also -(-x) is the inverse of x, so the only way this can happen is if

    -(-x)=x.
     
  13. Sep 19, 2008 #12
    Let a and b be two real numbers.

    Let x = ab + (-a)b + (-a)(-b)
    x = ab + (-a)[ b + -b]
    x = ab + (-a)*0
    x = ab

    Now once again let x = ab + (-a)b+ (-a)(-b)
    x = (a - a)b + (-a)(-b)
    x = (-a)(-b)

    By the transitivity
    ab = (-a)(-b)

    Note ab = (-1)(-1)ab, let x = ab and you get x = (-1)(-1)x which is x = -1(-1x) or simply x = -(-x).

    Essentially that's the more general idea for what you are trying to prove.
     
  14. Sep 19, 2008 #13

    HallsofIvy

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    But what if you are not working in the real numbers? The OP never did say what x was nor what course this is. -x can be defined for x a member of any group (with additive notation). The idea given in the first response, that we can multiply on both sides by "-1", also doesn't work is we have a group (and so no "multiplication").

    What is being asked here is really a proof that the "negative" or inverse for the group operation is "dual"- that is that the "negative of the negative" is the original element.

    For any group, using additive notation and with 0 as the group identity, -x is defined as the member of the group such that x+ (-x)= 0 and (-x)+ x= 0. Letting y= -x, we have y+ x= 0 and x+ y= 0. From the definition, it follows that x is the inverse of y- and if you have already proved that negatives are unique in a group, that x is the inverse of -x or x= -(-x).
     
  15. Oct 16, 2008 #14
    This can be proved using field axioms of real numbers

    take y = -x
    from additive identity field axiom, x + 0 = x
    -> 0 = x - x (eq 1)
    -> 0 = x + (-x)
    -> 0 = x + y (eq 2)

    -x is short form of 0-x

    Now -(-x) = 0-(-x)

    = 0-(0-x)
    = x+y -(x+y-x) from eq 2
    = x+y-(y+x-x) from commutativity field axiom
    = x+y-(y+(x-x)) from associativity field axiom
    = x+y-(y + 0) using eq 1
    = x+y-(y) from additive identity field axiom
    = x+y-y = x+(y-y) from associativity field axiom
    = x + 0 (since y-y = 0 as a consequence of additive identity field axiom as shown in eq 1, for any real y)
    = x (from additive identity field axiom)
     
  16. Nov 30, 2010 #15
    Using Axioms of real numbers:
    Lemma 1: 0 is unique.
    Suppose it is not, suppose b + a = a
    so b = 0 + b = 0, b = 0

    Lemma 2: 0x = 0 for all x
    0x + x = (0 + 1) x = 1x = x.
    0x + x = x, by lemma 1, 0x = 0

    Lemma 3: Additive Inverse is unique
    suppose a + b = 0 and a + c = 0
    then c = c + 0 = c + (a + b) = (c + a) + b = 0 + b = b
    c = b

    Lemma 4: -x = -1x
    -1x + x = (-1 + 1) x = 0x
    by lemma 2, 0x = 0
    so -1x + x = 0, so -1x is the additive inverse of x
    by lemma 3, -1x = -x

    Now Claim -(-x) = x
    Proof: -(-x) + -x lemma 4 and distributive law
    = -1 (-x + x) additive inverse
    = -1(0) additive neutral
    = 0

    so -(-x) is the additive inverse of -x, by lemma 4
    -(-x) = x
     
  17. Jun 21, 2013 #16
    x = x
    x + (-x) = x + (-x) = 0
    -1[x + (-x)] = -1.x + {-1(-x)}
    -1.0 = -x + {-(-x)}

    by adding x to boh sides

    x + (-1.0) = x + [(-x) + {-(-x)}]
    x + 0 = [x + (-x)] + {-(-x)}
    x = 0 + {-(-x)}
    x = -(-x)
     
  18. Jun 21, 2013 #17

    lurflurf

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    I would start with by the definition of additive inverse

    -x+-(-x)=0
    I adding x to each side is allowed we have
    x+(-x+-(-x))=0+x
     
  19. Oct 14, 2016 #18
    These are all wrong, unfortuntely. You can't use what you are trying to prove in your proof.

    Everyone has said basically, let x + -(-x) = 0 but that is exactly what you are trying to prove!!

    Here's the proof (using only field axioms):

    0 = 0
    -(-x) + [-(x) + x] = [(-(-x) + (-x))] + x
    -(-x) + 0 = [(-1)(-x) + (1)(-x))] + x
    -(-x) = (-x)[(1-1)] + x (factor out -x)
    -(-x) = (-x)(0) + x
    -(-x) = x

    Q.E.D.
     
  20. Oct 14, 2016 #19

    PeroK

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    You do realise, of course, that the previous post in this thread was over 3 years ago!
     
  21. Oct 14, 2016 #20

    SammyS

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    Right!
    Furthermore:
    Two are from 2013, one before that is from 2010 and the earlier ones as well as the OP are from 2008 .
     
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