Question

Prove that \({A \cup B} = {B \cup A}\).
Hmm... is this accepted as a proof?

Answer 1

Are you trying to prove this proposition?

Answer 2

Yes, with a rigorous proof; not a visual one (if you'd talk about the Venn Diagram).

Answer 3

http://www.proofwiki.org/wiki/Union_is_Commutative

Answer 4

Let \({\rm A} = \{a_1,a_2,a_3\cdots a_n\}\) and \({\rm B}= \{b_1,b_2,b_3\cdots b_n\}\).\[{\rm A \cup B} = \{a_1 , a_2, a_3\cdots a_n, b_1,b_2\cdots b_n\}\]and\[{\rm B \cup A} = \{b_1,b_2\cdots b_n,a_1,a_2,a_3\cdots a_n\}\]It is seen that both sets have the same elements, therefore \(\rm B \cup A = A \cup B\).

Answer 5

Is the above a nice proof?

Answer 6

I always look for proofs at proofwiki

Answer 7

Well, I try to practice doing proofs so I become good at them. :)

Answer 8

My prof removed marks when we don't state any laws or axioms.

Answer 9

Seem like a direct proof by example.

Answer 10

Hmm, so how do I remove the example part?

Answer 11

Wait—let me look at the proof at proofwiki.

Answer 12

You fall back on the axioms for math

Answer 13

Well, I can explain the lemmas used :)

Answer 14

That will work. You can use lemmas

Answer 15

\[{\rm A = B} \ \ \ \ {\rm iff} \ \ \ \ \text{all the elements in A are contained within B, and B contains no more.}\]

Answer 16

Using this will work

Answer 17

I don't see a reason to remove marks since you are using what is proven already.

Answer 18

I see the proof there; though it's pretty tricky since it utilizes the commutative rule for logical operators :P

Answer 19

Yeah, logical operators are the foundation.

Answer 20

I like it though because it is simple to understand

Answer 21

I do agree with your statement. Thanks =)

Answer 22

\[{\rm A \cup B} = \{a_1 , a_2, a_3,\dots a_n, b_1,b_2,\dots b_n\}\]\[\qquad\quad=\{a_1,b_1,a_2,b_2,a_3,b_3,\dots,a_n,b_n\}\]\[\qquad\quad= \{b_1,b_2,\dots b_n,a_1,a_2,a_3,\dots a_n\}\]\[\qquad\quad={\rm B \cup A} \]