At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
i misspelled your name above, sorry
No problems. I'll be right back.
First part looks good for nonempty.
cool. the first step is the easy one
Yep, you can always just show that the zero vector is in the set as a backup. Your way is fine though.
Second part is a little convoluted, but I'll get back to that shortly, I think you are on the right track. :)
how would you show the first step. i thought x1 was just 3 times x2 and since its in the x y plane its non empty
First step I would just say (0,0) is in the set since 0 = 3*0. It's kind of "OK, duh..." but it works to prove the point.
oh, wow ok yeah. makes sense
For the second part, let's be clear first on what we will start with and what the goal is: START: x = (x1,x2) where x1 = 3x2. GOAL: alpha x = (alpha x1, alpha x2) where alpha x1 = 3(alpha x2). Notice the pattern.
multiplying by alpha doesnt really change anything. is that what you are asking?
so it is still 3x2 = x2 so it would pass, right?
Yes, I think it is just a matter of writing style at this point. :) Here's how I would phrase the actual proof for clarity.
Suppose we have x = (x1,x2). Then x1 = 3*x2. Now, alpha*x1 = alpha*3*x2, that is, alpha*x1 = 3*(alpha*x2). So, alpha*x = (alpha*x1, alpha*x2) is in the set. I would prefer to not convert x1 into 3*x2 in the proof. That turned me around a bit. But I see what you did. :)
ok. so we have the same idea for step 2, just worked it differently?
Wait... you say that "and that equals alpha*3*x2 = alpha *x2" but that isn't actually true.
ok. if i go with what i was doing thats where i get a little confused for step 3 x + y for my way of working would be x = (3x2,x2) y = (3y2, y2) (3x2+3y2) would be 3(x2 + y2)...... should i just do what you did and make a note taht x1 = 3x2 and just use x1 in the proof?
OK, let's hold off on part 2 for the time being. I want to try to do it your way for step 3. I think that might be more insightful.
With your way, you have x = (3x2, x2) and y = (3y2, y2). OK, that's fine. What happens now is that we have no equation to work with. x1 = 3x2 is totally pointless right now since you already used it to convert. So, for x + y we need to show that x+y = (x1+y1, x2+y2) satisfies (x1+y1) = 3*(x2+y2). Is that actually true? Well you would say that x + y = (3x2 + 3y2, x2 + y2). And, yes, the first component 3x2 + 3y2 = 3*(x2 + y2) the second component. Thus, x + y is in the set. I have to say, it's a little unorthodox, but it seems to work.
Your teacher is probably going to expect to see something with x1's in the proof, but I don't want to discount your way of doing it either. :)
ok. i think your way would be better because its shorter. and i guess easier. i thought i had to use the 3x2 because the original problem said x1 = 3x2. so i thought we had to use it throughout
Yeah, the equation is really just used as a "check" to make sure that the components work like they are supposed to.
ok. im going to redo it and just make a note that x1 = 3x2 and use x1 in the proof
i scrolled up and was reading and i think that is a lot less confusing than what i was trying
Alright, I'd be happy to check it. It's always good to experiment with your own methods first so that you can sort out the process in your head. :)
ok. ill be back in a few and post what i come up with
just a quick question. we are saying \[\alpha x=0 \] right?
for the second step
We need to show that \[\alpha x\] satisfies \[\alpha x_1 = 3*(\alpha x_2)\]
ok. so for step 2 (it applies to all problems depending on condition of x1 and x2) should i write \[\alpha x = 3*\alpha x2\]
i think i answered my own question. yes
before working it
ok so for the third step (not doing what i originally did) would i again say x1=3x2 then X = (x1,x2) Y=(y1,y2) then X + Y = (x1+y1), (x2+y2) but that doesnt make sense. since i am trying to show x1 = 3x2. or would that actually be the proof because i cant think of a example that would counter that. (unless there is one and i cant think of it)