#Are a, b and c subspaces of V?
ashen quarry
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ashen quarry
What do
ashen quarry
Check the axioms. Like a is evidently not a vector space, + isn't commutative
I understood that
But idk how to apply the axioms
ashen quarry
I dont really understand the steps
red sinew
(a,b)+(c,d)=(a,b)
(c,d)+(a,b)=(c,d)
Hence it's not commutative since (a,b) and (c,d) can be different points
ashen quarry
(a,b)+(c,d)=(a,b)
(c,d)+(a,b)=(c,d)
Oh a+b=b+a? Do i start with this axiom?
red sinew
That's commutative, yes...
ashen quarry
Ok do i just write it as is?
red sinew
Again, if all the axioms hold its a vector space
If one fails, it's not a vector space
red sinew
Ok do i just write it as is?
I've written out the argument for why a isn't a vector space
u+v != v+u for all u,v in V. Specifically u+v=v+u iff u=v
ashen quarry
@red sinew like dis?
red sinew
You have extremely wrong notation present but yes
Like (a1,b1) isn't a real number clearly
You also introduce those indexed points then not use them
Also for counterexamples you can just use specific points
ashen quarry
You also introduce those indexed points then not use them
Oh yeah that was from trying to do it myself
Forgot to erase it
In the top just for all (a, b, c, d) belongs to R?
red sinew
Does R contain 4-tuples?
red sinew
You can write $a,b,c,d\in\mathbb{R}$ or $(a,b),(c,d)\in V$, but you cant write $(a,b,c,d)\in\mathbb{R}$
true coralBOT
Omegabet_
ashen quarry
Oh i see
red sinew
cause (a,b,c,d) isnt a real number
ashen quarry
Is dis good
red sinew
no
ashen quarry
Omg
red sinew
cause u+v=v+u does hold in cases
so the 1st for all is wrong
Like... just pick 2 points that arent the same
$(0,0)+(0,1)=(0,0)\neq (0,1)=(0,1)+(0,0)$
true coralBOT
Omegabet_
red sinew
there's a clear counterexample
ashen quarry
So u+v does hold?
ashen quarry
What for
red sinew
the thing we've been talking about for the past 20 minutes
why a's + isnt commutative.
red sinew
I explained that
red sinew
cause u+v=v+u does hold in cases
.
ashen quarry
But not this case
red sinew
there are specific choices of u and v that makes u+v=v+u hold
you said it fails for every choice of u and v
again, use a concrete counterexample
ashen quarry
**Omegabet\_**
I should write the counterexample as my proof?
ashen quarry
My linear algebra teacher never did that is all
red sinew
(a,b)+(c,d)=(a,b)
(c,d)+(a,b)=(c,d)
this was just to see what makes the axiom fail
I mean, it's common sense this is how you disprove something.
"Every horse is brown"
If you find a horse that isn't brown, then that statement is false
ashen quarry
Ok
red sinew
yeah
red sinew
good idea
ashen quarry
Do i have to do the scalar part? Since i (you) already proved it doesnt belong to the subset ig it counts for all of it right?
red sinew
what
ashen quarry
a) k(a,b)=ka,kb
red sinew
what about it?
ashen quarry
I dont have to prove that part right
red sinew
prove what
ashen quarry
Nvm
ashen quarry
I couldnt do it
red sinew
I couldnt do it
which?
For b, it should be clear that if something will go wrong, itll be with the weird scalar multiplication
and c likely some distributive rule will fail (since you get stuff like k(a+c+1) after you scale an addition)
Specific axioms that should fail:
b) ||(b+c)u=bu+cu||
c) ||k(u+v)=ku+kv||
young sedgeBOT
@ashen quarry
young sedgeBOT
@red sinew The thread creator still needs help with this help request.