Proving: Vector Equality in a Vector Space

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In proving that if x+y=x+z then y=z in a vector space, the discussion emphasizes using vector axioms, particularly the existence of an additive inverse. Participants suggest starting with the equation x+y=x+z and manipulating it to isolate y and z. The proof involves substituting and applying axioms such as commutativity and associativity of vector addition. It is confirmed that using the additive inverse allows for a straightforward deduction that leads to the conclusion y=z. The conversation highlights the importance of understanding the structure of vector spaces and the properties that govern them.
Mdhiggenz
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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 can't 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?


Homework Equations





The Attempt at a Solution

 
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Mdhiggenz said:

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 can't 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?


Homework Equations





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 couldn't 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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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