Proving -(-x)=x: Step-by-Step Guide

[em2008]

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

 

[JG89]

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

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

 

[Howers]

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

But 0 = x + -x

so

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

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

x + 0 = x = -(-x)

 

[JG89]

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?

 

[Howers]

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...

 

[JG89]

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.

 

[Howers]

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

 

[JG89]

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

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.

 

[Feldoh]

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)

 

[mistermath]

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.

 

[sutupidmath]

-(-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.

 

[PowerIso]

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]

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).

 

[smuralimohan]

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)

 

[darthalpha]

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

 

[phydis]

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]

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

 

[Jason Morand]

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]

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

 

[SammyS]

You do realize, 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]

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!

 

[EternusVia]

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

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