# Linear algebra proof

## Homework Statement

Let x,y,and z be vectors in a vector space V. Prove that if x+y=x+z
then y=z

I know it cant be as simple as just subtracting vector x from both sides.

What i'm thinking is out goal is to get y=z which means that there much be a zero vector

in the relationship x+y=x+z.

My guess is that x is the zero vector in this case. I'm not that good with proofs yet, but I have been reading "How to prove it" so i split it up into givens and goals.

given is x,y,and z be vectors in a vector space V and x+y=x+z

goal being y=z.

am I allowed to grab one of the axioms as a staring point for the proof?

## The Attempt at a Solution

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper

## Homework Statement

Let x,y,and z be vectors in a vector space V. Prove that if x+y=x+z
then y=z

I know it cant be as simple as just subtracting vector x from both sides.

What i'm thinking is out goal is to get y=z which means that there much be a zero vector

in the relationship x+y=x+z.

My guess is that x is the zero vector in this case. I'm not that good with proofs yet, but I have been reading "How to prove it" so i split it up into givens and goals.

given is x,y,and z be vectors in a vector space V and x+y=x+z

goal being y=z.

am I allowed to grab one of the axioms as a staring point for the proof?

## The Attempt at a Solution

Yes. Exactly. Grab the axiom that says that every vector x has an additive inverse.

I think I got it. Would this be a correct proof.
A1. x+y=y+x
A2. (x+y)+z=x+(y+z)
A3. x+0=x
A4. x+(-x)=0

These are the axioms I used.

Here is the proof: Assume A1. x+y=y+x
using A3 we get the following substitution x-x+y=y+x-x
y=y+(x-x). Using A2 and A1 y=(x+y)-x. Given that x+y=x+z, we can substitute
y=(x+z)-x using A2 y=z+(x-x) now using A3 we get y=z+0
thus y=z

done.

It seems you haven't seen an "If, then" type of question before. Basically, you are given the "if" and you work towards the "then" part. In this case, start with the given: ##x + y = x + z## line and use axiom(s) to get to ##y = z##.

Yea karnage, i'm still a bit rusty on those but if I start with that couldnt I just use one axiom to get to the goal.

E.g x+y=x+z

(x-x)+y=(x-x)+z A3

then y=z

Yup, that is correct. Although you would first add (-x) then use associativity to combine the x and (-x) to make 0, which is two steps.