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Antineutron
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ice109 said:antineutron:
Show that the solution set of y = 2x+1 fails to be a vector space.
I'm trying to solve a problem ice109 recommended. I'm trying to show how y=2x+1 is not a vector space. Here I go.
Let u=(x,2x+1) v=(x',2x'+1) w=(x",2x"+1)
1. If u and v are objects in V, then u + v is in V.
u+v=(x+x',2x+2x'+2) fails because 2 is not in V?
2. u + v = v + u
passes by inspection
3. u + (v + w) = (u + v) + w
also passes by inspection
4. There is an object 0 in V, called a zero vector for V, such that 0 + u = u+ 0 = u for all u in V
If x=0, the zero vector is (0,1)?
(0,1)+(x,2x+1)=(x,2x+2) which is not u, so it fails this test.
5. For each u in V, there is an object -u in V, called a negative of u, such that u + (-u) = (-u) + u = 0.
It passes obviously
6. If k is any scalar and u is any object in V, then ku is in V.
I don't know about this one... I don't know the meaning of this axiom...
7. k(u + v) =ku + kv
k[(x,2x+1)+(x',2x'+1)]=(kx,k2x+k)+(kx',k2x'+k) passes
8. (k + m)u = ku + mu
(k+m)(x,2x+1)=[(k+m)x,(k+m)2x+(k+m)]
k(x,2x+1) + m(x,2x+1)=(k+m)(x,2x+1)=[(k+m)x,(k+m)2x+(k+m)]
passes
9. k(mu) = (km)(u)
passes
10. 1u=u
passes
I can say for sure it fails axiom 1 and 4, but how do I check to see if it fails 6?