- #1
a_skier
- 17
- 0
A-(B[itex]\bigcap[/itex]C)=(A-B)[itex]\bigcup[/itex](A-C)
If A-B={xlx[itex]\in[/itex]A and x[itex]\notin[/itex]B}
A-C={xlx[itex]\in[/itex]A and x[itex]\notin[/itex]C}
then (A-B)[itex]\bigcup[/itex](A-C)={xlx[itex]\in[/itex]A, x[itex]\notin[/itex](B and C)
Let X=A and Y=(B[itex]\bigcap[/itex]C)
X-Y={xlx[itex]\in[/itex]X and x[itex]\notin[/itex]Y}
x[itex]\notin[/itex]Y
x[itex]\notin[/itex](B[itex]\bigcap[/itex]C)
x[itex]\notin[/itex](B and C)
Therefore, A-(B[itex]\bigcap[/itex]C)=(A-B)[itex]\bigcup[/itex](A-C).
Is this anywhere close to being right? I want to self study some rigorous math this summer but the hard part is that I don't have anything to check against to see if my answers are correct or not.
If A-B={xlx[itex]\in[/itex]A and x[itex]\notin[/itex]B}
A-C={xlx[itex]\in[/itex]A and x[itex]\notin[/itex]C}
then (A-B)[itex]\bigcup[/itex](A-C)={xlx[itex]\in[/itex]A, x[itex]\notin[/itex](B and C)
Let X=A and Y=(B[itex]\bigcap[/itex]C)
X-Y={xlx[itex]\in[/itex]X and x[itex]\notin[/itex]Y}
x[itex]\notin[/itex]Y
x[itex]\notin[/itex](B[itex]\bigcap[/itex]C)
x[itex]\notin[/itex](B and C)
Therefore, A-(B[itex]\bigcap[/itex]C)=(A-B)[itex]\bigcup[/itex](A-C).
Is this anywhere close to being right? I want to self study some rigorous math this summer but the hard part is that I don't have anything to check against to see if my answers are correct or not.