Question

proof of function of multiplication of integers!

Answer 1

problem 15

Answer 2

This isn't a hard proof. There are only 4 cases. \[x\in A\]\[x\notin A\]\[x\in B\]\[x\notin B\]

Answer 3

Oops, I mean \[x\in A \wedge x\in B\]\[x\in A \wedge x\notin B\]\[x\notin A \wedge x\in B\]\[x\notin A \wedge x\notin B\]

Answer 4

@math_proof Do you understand?

Answer 5

not ready

Answer 6

which part are you talking about

Answer 7

not really*

Answer 8

Okay, there are only those four cases

Answer 9

but which part are you talking about?

Answer 10

part (i)

Answer 11

ok

Answer 12

So for each case, show what the result will be.

Answer 13

For \[x\in A \wedge x\in B\]You get\[1\cdot 1 = 1\]We know that \[x \in A\cap B\]By definition. So for this case we confirm it is true. Just do the same with the other 3 cases.

Answer 14

then for \[x \in A ∧ x notinB\]

Answer 15

is\[1\times0 =0\]

Answer 16

Yeah

Answer 17

so its \[x inA cupB?\]

Answer 18

Which is also correct, because \(x \notin A\cap B\)

Answer 19

SO ALL of them will be true?

Answer 20

Yeah, they won't make you prove it if it isn't true.

Answer 21

This is an easy proof because there are only 4 cases to consider. Thus you can exhaustively confirm it for every case.

Answer 22

what about part b?

Answer 23

Okay for part b, what you need to do is see what happens for each case.

Answer 24

Do you get 1 or 0?

Answer 25

Then find a set operation on A and B in which \(x\in A \text{ some_operation } B\)

Answer 26

still lost

Answer 27

those are the same cases as in part a?

Answer 28

These are the four cases:
\[x\in A \wedge x\in B\]
\[x\in A \wedge x\notin B\]
\[x\notin A \wedge x\in B\]
\[x\notin A \wedge x\notin B\]
What does \(X_C(x)\) equal for each case?

Answer 29

1, 0, 0, 0?

Answer 30

For case \(x\in A \wedge x\notin B\) we have 1+0-1*0... does that equal 0? Try again.

Answer 31

how are you getting the numbers?

Answer 32

how is that 1+0-1*0

Answer 33

\[x\in A \wedge x\notin B\Rightarrow X_A(x)=1\wedge X_B(x) = 0\]\[X_C(x) = X_A(x)+X_B(x) - X_A(x)X_B(x) = 1+0-1*0 = 1\]

Answer 34

So we have 1, 1, 1, 0. So what set operation is this?

Answer 35

ooooo

Answer 36

is it a cartesian product (A × B) ∪ (A × C)

Answer 37

No, it's just \[A\cup B\]