Let sets R and T be defined as
R = {x ∈ Z | x is divisible by 2}
T = {x ∈ Z | x is divisible by 6}. Then T ⊂ R
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YES, WE CAN SAY THAT T ⊂ R BY THE FOLLOWING PROOF:
GIVEN: T = {x ∈ Z | x is divisible by 6}.
LET x be 6k (as is it the multiple of 6.)
6k = 2(3k) = 2p ----1
given that R = {x ∈ Z | x is divisible by 2}
that means x is multiple of 2 and that x has even numbers
therefore x=2q ----2
we can observe that set R consists of all even numbers including 6 and all other its multiples too.
therefore we can say that T ⊂ R
GIVEN: T = {x ∈ Z | x is divisible by 6}.
LET x be 6k (as is it the multiple of 6.)
6k = 2(3k) = 2p ----1
given that R = {x ∈ Z | x is divisible by 2}
that means x is multiple of 2 and that x has even numbers
therefore x=2q ----2
we can observe that set R consists of all even numbers including 6 and all other its multiples too.
therefore we can say that T ⊂ R
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