bupbebeo wrote:IF P is a set of integer and # is in P, is every positive multiple of 3 in P.
(1) For any integer in P, the sum of 3 and that integer is also in P
( 2) For any integer in , that integer minus 3 is also in P
Why would you do 1+2 ?dxgamez wrote:Is 3 considered in P?
3 could be in P too right? Then st 2 would be zero for int=3.
How does it satisfy 1+2? I'm quite confused.
You need to add or sub 3 from an integer .(1) For any integer in P, the sum of 3 and that integer is also in P
( 2) For any integer in , that integer minus 3 is also in P
How can you conclude that a number picked from set is multiple of 3.Now if you take any multiple of 3 and add or subtract 3 from it you will always get a multiple of 3 .
That is = > x (mul of 3)+/- 3 = y (mul of 3) . Y = integer .
rockeyb wrote:Why would you do 1+2 ?dxgamez wrote:Is 3 considered in P?
3 could be in P too right? Then st 2 would be zero for int=3.
How does it satisfy 1+2? I'm quite confused.You need to add or sub 3 from an integer .(1) For any integer in P, the sum of 3 and that integer is also in P
( 2) For any integer in , that integer minus 3 is also in P
. Lets say the question was changed to :eaakbari wrote:Exactly I do feel the same. What do you say rockeyb?
I would say E is correct answer.Lets say the question was changed to :
IF P is a set of integer , is every multiple of 3 in P.
(1) For any integer in P, the sum of 3 and that integer is also in P
( 2) For any integer in , that integer minus 3 is also in P
Then in that case my answer would be C and not E .
Why? see my explanation above .
