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easy math problem...I cannot solve

S

Semidevil

Diamond Member
prove that -(-v) = v for all v in V.

ok, this is the very first problem given in my abstract alg class...and I dont know how to solve this. How am I suppose to approach this?

so, -(-v) = -1(-v). Now what?

I mean, does this have anything to do with it having an identity, or inverse?

I dont know the right frame of mind to go about solving this...
 
Well Ive never taken any type of algebra proof class but maybe its like

-(-v) = v -> -1* (-1*v) = v, then you can use associative property so (-1*-1)*v = v and then (1)*v = v

Just a guess though...so I could be way off.
 
You could divide each side by -v, which gives you -1=-1. I don't know if that is a valid proof though.
 
You could do it by contradiction (but you shouldn't):

Assume -(-v) doesn't equal v. Then you have (-1)(-v) doesn't equal v. Divide both sides by -1 to get -v doesn't equal -v. That's obviously false since, by defintion, -v = -v. Thus, you have a contradiction proving -(-v) equals v.

There is absolutely no reason you can't (and shouldn't because this is so elementary) do a direct proof. A direct proof shows exactly what you want to show using equations, manipulations, algebra, etc. So:

-(-v) = (-1)(-v) = (-1)(-1)(v) = (-1)^2 * v = 1 * v = v. Thus, -(-v) = v.

Make sense?
 
Originally posted by: hellfreeze
gotta love the contradiction method
Click to expand...

I cast my vote for prove it by exhaustion... 🙂 Show it's true for every number.
😉

The proof would depend on what you have to work with.. do you have the associative property? If so, use it. Do you have -v = -1(v)? Can't use it if it isn't proven or given...
 
I dont' think any of those proofs given above works. Abstract algebra proofs require rigor and you have to use the given axioms. I don't think that -v = (-1)v is an axiom, so you can't assume that.
 
Originally posted by: raptor13
You could do it by contradiction (but you shouldn't):

Assume -(-v) doesn't equal v. Then you have (-1)(-v) doesn't equal v. Divide both sides by -1 to get -v doesn't equal -v. That's obviously false since, by defintion, -v = -v. Thus, you have a contradiction proving -(-v) equals v.

There is absolutely no reason you can't (and shouldn't because this is so elementary) do a direct proof. A direct proof shows exactly what you want to show using equations, manipulations, algebra, etc. So:

-(-v) = (-1)(-v) = (-1)(-1)(v) = (-1)^2 * v = 1 * v = v. Thus, -(-v) = v.

Make sense?
Click to expand...

Can you use negative 1 that is an integer and may not be in his group.
 
I'll use some of the axioms given here.

If v is in V, then there exists a -v in V such that v + (-v) = 0 (Additive inverse property)

Since -v is in V, then there exists a -(-v) in V such that -(-v) + (-v) = 0 (Additive inverse property)

Since the additive identity is unique, we have v + (-v) = -(-v) + (-v). (Otherwise known as the transitive property)

Therefore, v = -(-v) (Cancellation property of Addition, since v, -v, -(-v) are all in V)
 
Originally posted by: eigen
Originally posted by: raptor13
You could do it by contradiction (but you shouldn't):

Assume -(-v) doesn't equal v. Then you have (-1)(-v) doesn't equal v. Divide both sides by -1 to get -v doesn't equal -v. That's obviously false since, by defintion, -v = -v. Thus, you have a contradiction proving -(-v) equals v.

There is absolutely no reason you can't (and shouldn't because this is so elementary) do a direct proof. A direct proof shows exactly what you want to show using equations, manipulations, algebra, etc. So:

-(-v) = (-1)(-v) = (-1)(-1)(v) = (-1)^2 * v = 1 * v = v. Thus, -(-v) = v.

Make sense?
Click to expand...

Can you use negative 1 that is an integer and may not be in his group.
Click to expand...

No, you can't. One thing about math is....make NO assumptions.
 
Originally posted by: chuckywang
I'll use some of the axioms given here.

If v is in V, then there exists a -v in V such that v + (-v) = 0 (Additive inverse property)

Since -v is in V, then there exists a -(-v) in V such that -(-v) + (-v) = 0 (Additive inverse property)

Since the additive identity is unique, we have v + (-v) = -(-v) + (-v). (Otherwise known as the transitive property)

Therefore, v = -(-v) (Cancellation property of Addition, since v, -v, -(-v) are all in V)
Click to expand...

Good show Chucky,Thats what i was hinting at but did not want to give away.
 
sorry if you were offended by me just giving away the answer. however, the OP now knows how these kinds of proofs go, maybe the rest of them will be easier. 🙂
 
Originally posted by: chuckywang
I'll use some of the axioms given here.

If v is in V, then there exists a -v in V such that v + (-v) = 0 (Additive inverse property)

Since -v is in V, then there exists a -(-v) in V such that -(-v) + (-v) = 0 (Additive inverse property)

Since the additive identity is unique, we have v + (-v) = -(-v) + (-v). (Otherwise known as the transitive property)

Therefore, v = -(-v) (Cancellation property of Addition, since v, -v, -(-v) are all in V)
Click to expand...

Chucky, I agree with you completely. However, since we have absolutely no information as to the type of data the OP's set V contains, I didn't feel too bad incorporating -1 into the fold. Also, if you're going to start busting out properties of the integers (and I can't fault you for that) and assuming the set V contains the integers, I don't feel too bad using one of them in my proof(s). 😀

Of course, considered in the context of the real math world, the correct answer is probably "The excercise is trivial and left to the reader." I did always love that proof. :beer::beer::beer:

NOTE: Edited for more beer. :beer:
 
Perhaps my link was a little bit confusing, since the axioms dealt with integers. Having taken abstract algebra, there are only three real axioms. They are:

Associativity: if x, y, z are in group G, then (xy)z=x(yz)
Identity: there exists an x in G such that xy=y=yx.
Inverses: for every x in G, there exists a y such that xy = yx =1 where 1 is the identity element in the group.

From these three axioms, one can proof the cancellation property, which states:
For x,y, z in G, if xy=xz, then y=z.

I think the proof goes something like:

Let a be the inverse of x in G, then

a(xy) = a(xz)
(ax)y = (ax)z by associative property
1y = 1z by property of inverses
y=z by the identity property.

I think this completes my original proof to the OP, since the cancellation property was not an axiom, but is now proven. 🙂

QED.
 
Theorems, axioms... whatever. The point is we're all assuming the set V is the set of integers. Of course, the axioms for integers hold for rational, real, and irrational numbers as well. That's not important here.
 
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