# How to prove -(-x)=x

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

x = x
-1(x) = -1(x)
-x = -x
(-x)/-1 = x

and since 1/-1 = -1/1 = -1, then we have
-(-x) = x

0 + -(-x) = -(-x)

But 0 = x + -x

so

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

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

x + 0 = x = -(-x)

0 + -(-x) = -(-x)

But 0 = x + -x

so

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

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

x + 0 = x = -(-x)

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?

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?

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...

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...

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.

y + -y = 0

let y = -x

-x + -(-x) = 0

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

hence, -(-x) = x

QED

Same flaw, I give up y + -y = 0

let y = -x

-x + -(-x) = 0

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

hence, -(-x) = x

QED

Same flaw, I give up 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.

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)

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?

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.

-(-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.

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.

HallsofIvy
Homework Helper
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).

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)

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
= 0

so -(-x) is the additive inverse of -x, by lemma 4
-(-x) = x

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)

lurflurf
Homework Helper

-x+-(-x)=0
I adding x to each side is allowed we have
x+(-x+-(-x))=0+x

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.

PeroK
Homework Helper
Gold Member
2020 Award
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.

You do realise, of course, that the previous post in this thread was over 3 years ago!

• LemmeThink
SammyS
Staff Emeritus
Homework Helper
Gold Member
You do realise, of course, that the previous post in this thread was over 3 years ago!
Right!
Furthermore:
Two are from 2013, one before that is from 2010 and the earlier ones as well as the OP are from 2008 .

PeroK
Homework Helper
Gold Member
2020 Award
Right!
Furthermore:
Two are from 2013, one before that is from 2010 and the earlier ones as well as the OP are from 2008 .

The OP, let's hope, has graduated by now!

• SammyS
JasonMorand, I think that stupidmath's proof is valid.

He proves it essentially the same way you do -- by claiming 0 = 0.