A is an element of B, but not a subset (assuming the b's aren't coincidentally just equal to the a's).
$ but $\{A\} \in P(B)?)
Alright this last bit was what I needed to see. I need to have my foundations but just making sure:
That's correct (by definition of subset essentially)Alright this last bit was what I needed to see. I need to have my foundations but just making sure:
.
A is a proper subset of B, since there are elements in B that are not in A. You are claiming that A is the set B itself.Can I please have my proof checked?
The video solution was clever in how it used a Pythagorean identity here to match up A and B, however I did it by solving. Just want to check on its validity
Lol. Yeah the question first asked to prove it was just a subset before claiming it was a proper subset. So I put x=π on the end to contradict they're the same set.
just fyi on the second line of the A it reads "Let A be the set of x in the real numbers such that k is in the integers such that x = 2k*pi" which makes zero senseCan I please have my proof checked?
The video solution was clever in how it used a Pythagorean identity here to match up A and B, however I did it by solving. Just want to check on its validity
That was a part of why I put that question up. What's the difference between using | and ,just fyi on the second line of the A it reads "Let A be the set of x in the real numbers such that k is in the integers such that x = 2k*pi" which makes zero sense
try
To get the final answer your friend got (which looks correct), use an absorption law at the second last line of your proof. (Unfortunately your simplification in your last line isn't valid. But if we just apply the absorption law there we'll get the answer.
They don't have an answer so I am suspecting that my answer is wrong lol. I started from the outside in my working.
Edit: After line 2 one of my friends used associativity like this
But ends up with a final answer of justinstead. Is this justified?
)
Ahh I absorbed wrongTo get the final answer your friend got (which looks correct), use an absorption law at the second last line of your proof. (Unfortunately your simplification in your last line isn't valid. But if we just apply the absorption law there we'll get the answer.
(See: https://proofwiki.org/wiki/Absorption_Laws_(Set_Theory)/Union_with_Intersection.)
Yes what your friend did is valid. Since intersection is both associative and commutative, we can do intersections in any order (like, (X cap Y) cap Z = X cap (Y cap Z) = X cap (Z cap Y) = (X cap Z) cap Y, using associativity and commutativity. I used 'cap' to mean intersection symbol.). Then using absorption law finishes it.
They don't have an answer so I am suspecting that my answer is wrong lol. I started from the outside in my working.
Edit: After line 2 one of my friends used associativity like this
But ends up with a final answer of just
Don't view "," and "|" in the same way (btw ":" is a common alternative for "|" that is my personal preference). The former is basically informal formatting here, whilst the latter is part of the formal set builder notation syntax.
no, but then you're saying the set consists of x and k, when you really want to just have the x's that satisfy the condition
if i remember its easier to prove by Venn DiagramsThis bugger.
