Mathematics Asked on February 7, 2021
Would this proof be sufficient for just proving B ⊆ A alone?
$A = {x ∈ Bbb Z : ∃y∈Bbb Z,x = 5y + 1}, $$B = {x ∈ Bbb Z : ∃y∈Bbb Z, x = 10y − 9}$
Suppose some arbitrary element x is in B. if x∈B, then x∈A by definition of B ⊆ A.
Also, by the definition of set B,
x = 10y – 9 for some y
= 10y – 9 + 10 -10
= 10y – 10 + 1
= 5(2y-5) + 1
Since, 2y-5 is an integer, x is in A by definition.
Therefore B ⊆ A.
Alternative proof. aZ + b = { an + b : n in Z }
A = 5Z + 1
B = 10Z - 9 = 10Z - 10 + 1 = 10Z + 1
10Z subset 5Z
10Z + 1 subset 5Z + 1
The tricky part is to show 10Z - 10 = 10Z.
Answered by William Elliot on February 7, 2021
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