Sets: Is A ⊆ B? A={x ∈ ℤ | x ≡ 7 (mod 8)} B={x ∈ ℤ | x ≡ 3 (mod 4)}

Dustinsfl

A={x ∈ ℤ | x ≡ 7 (mod 8)}
B={x ∈ ℤ | x ≡ 3 (mod 4)}

Is A ⊆ B? Yes

Since x ∈ A, then x_a = 7 + 8a = 8a + 7 = 2(4a + 3) +1. And since the ∈ B are of the form x_b = 3 + 4b = 4b + 3 = 2(2b + 1) + 1, both ∈ A,B are odd. A ⊆ B since the ∈ of both sets are of 2p + 1. Q.E.D.

Is this correct?

Mathnerdmo

No. You want to show that every element of A is an element of B. All you showed is that neither contains an even integer.

Try writing 7 + 8a in a form that shows it is in B.

Dustinsfl

I don't know how to go about doing that.

Mark44

What is this character - ℤ ?

7 + 8a = 3 + 4 + 4(2a) = 3 + 4(1 + 2a)

Dustinsfl

Integers

VeeEight

So do you see how the conclusion follows? It is pretty straight forward now.

Dustinsfl

Since x ∈ A, then x = 7 + 8a = 8a + 7 = 4(2a) + 4 + 3 = 4(2a + 1) + 3. And since the ∈ B are of the form x = 3 + 4b = 4b + 3, all x = 4p + 3 and A ⊆ B. Q.E.D.

Correct?

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Mathnerdmo	No. You want to show that every element of A is an element of B. All you showed is that neither contains an even integer. Try writing 7 + 8a in a form that shows it is in B.

Mark44 Mentor	Sets: Is A ⊆ B? A={x ∈ ℤ \| x ≡ 7 (mod 8)} B={x ∈ ℤ \| x ≡ 3 (mod 4)} What is this character - ℤ ? 7 + 8a = 3 + 4 + 4(2a) = 3 + 4(1 + 2a)

VeeEight	So do you see how the conclusion follows? It is pretty straight forward now.

Dustinsfl	Since x ∈ A, then x = 7 + 8a = 8a + 7 = 4(2a) + 4 + 3 = 4(2a + 1) + 3. And since the ∈ B are of the form x = 3 + 4b = 4b + 3, all x = 4p + 3 and A ⊆ B. Q.E.D. Correct?