Hello
Iv got an exam in a week and am trying to get through my whole course in that week. And am off to a horrible start =/
Can someone please tell how i should go about proving questions like the follwoing.
Q1)
A = {x <element_of> R | cos x = 1}
A = {x <element_of> R | sin x = 0}
Show that A <subsest of> B
Q2)
X = {24k + 7 | k <element_of> Z}
Y = {4n + 3 | n <element_of> Z}
Z= {6m + 1 | m <element_of> Z}
prove: X <subset_of> Y and X <subset_of> Z but Y <not_subset_of> Z
Thanks and im sure alot more questions would be comming soon =(
Hey thanks for the quick reply =)
hmm...yess i tried using the basic idea but couldnt get a proper proof. Would this be a clear enough proof:
Let x <element_of> A, that is cos x = 1
ie. x = 2kPi where k <element_of> N
Now, sin(x) = sin(2kPi) = 0 and therefor x <element_of> B
That is, A <subset_of> B
?