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Proving that -(-a)=a?

  1. May 14, 2008 #1
    proving that -(-a)=a???

    Well, i was trying to prove from the axioms of the field of reals that -(-a)=a, but i am not sure that what i did is valid and follows from the axioms, since i have not seen the actual proof before.

    From the axiom of the existence of the inverse we know that

    a+(-a)=0 , now adding on both sides -(-a) from the left we get

    -(-a)+(a+(-a))=-(-a)+0 now on the left hand side from the commutative axiom of addition we get

    -(-a)+((-a)+a)=-(-a)+0 now on the right hand side, from the axiom of existence of the neutral element we get:

    -(-a)+((-a)+a)=-(-a) now from the associative axiom we get on the left hand side:

    (-(-a)+(-a))+a=-(-a) we know from the axom of the existence of the inverse of addition that the inverse of (-a) is -(-a) so we get:

    0+a=-(-a) now from the axiom of the existence of the neutral element we get

    a=-(-a) what we wanted to prove.

    Is this kind of proof correct, or is there any more rigorous proof to show this. I mean a proof that follows from the axioms of the field of the real numbers, because i think there is another kind of proof usind classes of equivalence or sth.

    Thnx in advance.
  2. jcsd
  3. May 14, 2008 #2
    Wouldn't it be easier to prove first that the inverse is unique, i.e. if b and b' are both inverses of a then b = b'? Then you just say that


    can also be interpreted as stating that a is the inverse of -a.
  4. May 14, 2008 #3

    Well i thought of that too, but i wasn't, as a matter of fact i still am not, sure whether i know how to prove that the inverse is unique. HOwever i will give it a crack to see where i'll get.

    Let's suppose that b and b' are both distinct inverses of a. NOw we want to prove that b=b'???


    Since both b and b' are inverses of a it means, from the axiom of existence of the inverse, that

    a+b=0 and


    now let's start with

    a+b=0, let's add b' from the left on both sides, so we get

    b'+(a+b)=b'+0, now from the associative axiom and also based on the axiom of the existence of the neutral element of addition we get

    (b'+a)+b=b' now from the commutative axiom we get on the right hand side

    (a+b')+b=b', now using the fact that b' is the inverse of a, it means that a+b'=0, so we get

    0+b=b', now also from the axiom of neutral element we get


    Is this some kind of proof about the uniqueness of inverse, or i still need to prove it?

    Now from here we can say that from the fact that

    a+(-a)=0, it means also that a is the inverse of (-a), but since -(-a) is also the inverse of (-a), from the uniqueness of the inverse we can automatically say that a=-(-a), right?
  5. May 14, 2008 #4
    Yes, that proves uniqueness because it shows that if you have 2 inverses of a, then they're actually the same.

    An even easier way to do it is to first prove that you can cancel. I.e., if you have x + y = x + y', then y = y'
    Proof: x + y = x + y', add (-x) to both sides =>
    (-x) + x + y = (-x) + x + y'
    0 + y = 0 + y'
    y = y'

    Now with inverses, you have the additional information that a + b = 0 and a + b' = 0. But we still have that a + b = a + b', so b = b'.
  6. May 14, 2008 #5
    Yeah i did it this way too, but i just wasn't sure whether i would be allowed to use the fact that since

    a+b=0 and a+b'=0 then also a+b=a+b', because the rest i did exactly what you did above. It was only this step that i wasn't sure i could do, i don't know on which axiom is this based?? Since i suppose we are not allowed to use prior information, in other words, everything has to be based strictly on axioms.
  7. May 15, 2008 #6
    Sorry, misunderstood your post: In process of rewriting mine.
  8. May 15, 2008 #7
    This follows from the (often not mentioned in textbooks) assumption that = is an equivalence relation. This means that it has the properties of reflexivity (that a = a), symmetry (that if a = b, then b = a), and transitivity (that if a = b and b = c, then a = c).

    Your question in particular follows from transitivity.

    For your purposes, you can view = as meaning that if a = b, then a and b can be considered indistinguishable and are the exact same object (though I will talk about why in a bit).


    The rest of this post isn't necessary for your question, but if you want to know more about what the author intends when he writes = and if you want to know more about equivalence relations in general, you're welcome to keep reading.

    There are several different symbols that are used for equivalence relations. The most common ones that you will see are [tex]\equiv, \ \cong, \ \sim, \ \text{and} \ =[/tex]. Though there are no set rules as to when you use each, these are how you will typically see them used:

    [tex]\equiv[/tex]: The "equivalence symbol" is typically used in number theory or in certain other algebraic contexts such as when you have that [tex]a \equiv b \ \text{and} \ c \equiv d \ \Rightarrow a + c \equiv b + d[/tex] makes sense and is true.

    [tex]\cong[/tex]: The "congruence symbol" is typically used in geometric contexts such as to say that two angles are congruent or that two shapes are similar.

    [tex]\sim[/tex]: The "similarity symbol" is used in general for any sort equivalence relation. You can be fairly sure that whenever an author defines some sort of relation using [tex]\sim[/tex] that he will prove that it is an equivalence relation.

    [tex]=[/tex]: The "equality symbol" is typically used in two different but very specific contexts. One of which is that if [tex]a = b[/tex] then a and b are actually the exact same object. The other context is that if [tex]a = b[/tex], then you can treat them as the exact same object and you cannot tell them apart with anything you will typically do with the objects. Often, with equality, the reason why you cannot tell a and b apart is because everything else in your mathematical theory is defined in terms of =. For instance, one part of the definition of a function or formula on the members of a field is that if [tex]a = b[/tex], then [tex]f(a) = f(b)[/tex].
    Also, every other equivalence relation is defined in terms of =. For instance, if [tex]\sim[/tex] is another equivalence relation, then if [tex]a = b[/tex], you know that [tex]a \sim b[/tex] (In general, if you have two equivalence relations, say [tex]\sim[/tex] and [tex]\equiv[/tex], then you do not know that if [tex]a \sim b[/tex] then [tex]a \equiv b[/tex]. For instance, I can define [tex]a \sim b[/tex] as meaning that a and b have the same remainder when divided by 2 and I can defined [tex]a \equiv b[/tex] as meaning that a and b have the same remainder when divided by 3)

    What this means is that if you are starting from the axioms and reason entirely from the axioms, using nothing else at all, then if [tex]a = b[/tex], then there is no way at all to distinguish between a and b as objects, so you can use them entirely interchangeably.

    However, you have probably heard of set theory and know that the typical way of looking at math is by looking at every object as being a set. (For instance, a typical construction of the natural numbers (including 0) makes 0 the empty set, 1 as the set including the empty set, and so on). A typical construction of the real numbers will sometimes have many different sets all being considered the number 1/2 (though often, you will instead take the entire set of all of these numbers and call that 1/2 instead so that there is only a single set called 1/2). What this means is that depending on your construction, you might not be guaranteed that a = b means that a and b are the exact same object. However, since, as I said, everything in your theory is based on the = symbol, as long as you reason from the axioms alone, they can be considered the same object.
  9. May 15, 2008 #8
    Sorry about the above post, I was tired and in a very odd state of mind when I wrote it.

    The first part was to answer your question. The second was to motivate something else that you did that does not follow from the axioms of field theory but is usually assumed to be true anyway.

    That is to say, that in field theory (as well as most other theories of mathematics), if A(x) is any formula involving x that can be constructed with the assumed logic of field theory (i.e., that it uses only things that can be discussed within field theory, or some extension, and nothing else), then if a = b, A(a) is true if and only if A(b) is true.

    This is usually taken as an axiom of either field theory or of the background logic. On the other hand, if you are constructing the field in some other theory such as in set theory, the field is constructed in a way so that it is true (sorry, my post was more a description of how one constructs a field in set theory so that this is true).

    You used this when you wrote that
    a+b' = 0
    and (a+b')+b = b'
    Therefore, we can replace a+b' with 0 to get 0+b = b'

    Note that the ability to replace a+b' with 0 is not something that follows from the field axioms (as they are usually stated). Along with the property of equality being an equivalence relation, this is usually an assumed property of your background logic.
  10. May 15, 2008 #9
    How about multiplying by -1 on both sides?

    -(-a)=a <=> -a=-a

  11. May 15, 2008 #10
    That won't help because this depends on the fact that minus 1 times x equals minus x. So, you are then using the property you want to use. The fact that minus 1 times minus 1 equals plus 1 actually follows from the fact that -(-1) = 1
  12. May 21, 2008 #11
    Sorry for taking so long to get back to your replies.

    Well i honestly thank you very very much, for you hve been of a great help. I knew before that = is an equivlence relation, but i just didn't think to relate things this way.

    However, i have anoter question about these issues.

    say that a is a real number with the property that

    a>0, so now what i am concerned is why -a<0?

    Here it is how i have tried to prove that if a>0 then -a<0.

    a>0 we add -a on both sides so we get

    a+(-a)>0+(-a) now from the axiom of the existence of the additive inverse and also from the axiom of the existence of the neutral element of addition we get:

    0>-a, now from a definition we know that this is the same as: -a<0.

    I understand this part, however, it seems that my questio should be phrased like this:

    why when we multiply an inequality by -1 it changes its sign, i.e

    a>0 /*(-1) => -a<0

    Does this follow from the fact that i just reasoned above, and after that we just take by deffinition that when we multiply an inequality by -1 it changes sign. SO, what i want to know is whether this fact is taken by definition, but that follows from a similar reasoning as i did before?

    Thnx a lot!
  13. May 21, 2008 #12
    Well, you can show that -1*a = -a (also, 1/(-1) = -1, so this is the same as dividing by -1)

    So this gives us that
    a > 0 implies that -1*a < 0

    By the way, it is an axiom that multiplying by a positive number does not change the direction of the inequality. I.e. if a < b and c > 0, then a*c < b*c. This cannot be derived from any of the other axioms of an ordered field.

    From these facts, you can easily show that multiplying by c < 0 flips the inequality. As you've already seen, this isn't actually taken as a definition. Instead it is derived from the axioms.
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