Discrete Math: Writing Proofs with All Steps

Question

What would the full proof look like for $\subset $? If $A = \left\{6a + 4b\in\mathbb{Z} : a, b \in\mathbb{Z}\right\}$ and $B = \left\{2a \in\mathbb{Z} : a \in\mathbb{Z}\right\}$. Show that $A = B$

So far I have: $A\subseteq $

Suppose $x\in A$. $x = 6a + 4b$ for some $a, b \in\mathbb{Z}$. $x = 2(3a + 2b)$. Let $k1 = (3a + 2b)$ where $k1\in\mathbb{Z}$. x = 2k1. 2x, x ∈ ℤ. Then $x\in B$.

$B\subseteq $ Suppose $y\in B$. $y = 2a$ for some $a\in\mathbb{Z}$. $y = 6a + 4(−a)$. Then $y\in A$.

I need help with showing steps for $B\subseteq A$.

Thank you all for the help!

Answer 1

You have noticed that $6a+4b = 2(3a+2b)$.

If $x \in B$, take $a = x/2$ and $b=-x/2$.

This can easily be seen as the $gcd(3,2) = 1$. In particular, $3 \cdot 1 + 2 \cdot -1 = 1$ allowing one to write any number as a linear combination of $3$ and $2$ by multiplying both sides by the desired integer. Namely, any even number when the expression has the coeffcient $2$ as mentioned above.

The aforementioned idea leads to the insight that the problem is equivalent to the following:

$2(3a+2b) = 2c \implies 3a+2b = c$ where $c$ is any integer This is always satisfied by Bezout's lemma for some $a,b$ $\in \mathbb{Z}$.