# Showing a function forms a vector space.

1. Sep 4, 2014

### schlynn

1. The problem statement, all variables and given/known data
Does the function: 4x-y=7 constitute a vector space?

2. Relevant equations
All axioms relating to vector spaces.

3. The attempt at a solution

x_n for example means x with the subscript n

The book says that the function isn't closed under addition. So it continues by showing that given 2 points, (x_1,y_1) and (x_2,y_2) that when you add 4x_1-y_1=7 and 4x_2-y_2=7 you get 4(x_1+x_2)-(y_1+y_2)=14, how did they get the values for the problem to see that it sums to 14 and not 7? The case for multiplication show: 4x_1-y_1=7, they used 3 as the scalar to show: 3(4x_1-y_1)=12x_1-3y_1, that part makes sense, but then again they say that the right side is 3x7, and I don't see how they got those values.

2. Sep 4, 2014

### vela

Staff Emeritus
You have $4x_1-y_1 = 7$ and $4x_2-y_2=7$. What do you get when you add those two equations?

3. Sep 4, 2014

### schlynn

You get 4(x_1+x_2)-(y_1+y_2)=14, which obviously isn't equal to 7, but does that then mean that all linear function ax+by=c is not a vector space for all c not equal to 0? Because c+c is always not equal to zero except when c=0 right?

4. Sep 4, 2014

### vela

Staff Emeritus
Yup.

5. Sep 4, 2014

### schlynn

Oh, ok, thanks, was just over-thinking it I guess then, it's apparent now, thank you.

6. Sep 4, 2014

### LCKurtz

You certainly weren't overthinking when you phrased your question. $4x-y=7$ is an equation, not a function. And a function doesn't constitute a vector space. As you proceed in your course you are going to have to be more careful with definitions so you know what you are trying to prove or disprove. Perhaps you meant to ask something like: Does the set $S=\{(x,y)|4x-y=7\}$ consitute a vector space with the usual operations?

7. Sep 4, 2014