For any two sets prove that:

1) $A\cup (A\cap B)=A$
2) $A\cap (A\cup B)=A$

Hint: Here, we have to apply the distributive which states that:

$A\cup (A\cap B=(A\cup A)\cap (A\cup B)$ and $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$.
For the proof (i) we have to make use of the fact that $A\subset A\cup B$ and a set is a subset of itself. Next, for the proof (ii) we can apply the proof (i) along with the distributive law.

Complete step-by-step answer:
Here, we have to prove that $A\cup (A\cap B)=A$ and $A\cap (A\cup B)=A$.


1) $A\cup (A\cap B)=A$

We know that a set is a subset of itself. Hence, we can say that $A\subset A$.

Also we have $A\subset A\cup B$.

Here, first let us consider LHS, $A\cup (A\cap B)=A$.

Now, we can apply the distributive law which states that:

$A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$

Hence, by applying the distributive law we will get:

$A\cup (A\cap B=(A\cup A)\cap (A\cup B)$

We know that $A\cup A=A$.

Hence, we get it as,

$\Rightarrow A\cup (A\cap B=A\cap (A\cup B)$

Since, $A\subset A\cup B$, we can say that the set common for A and $A\cup B$ will be A itself. Hence we, obtain:

$A\cup (A\cap B)=A$


2) $A\cap (A\cup B)=A$

Here, also we have to apply the distributive law which states that:

$A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$

Now, let us consider the LHS, $A\cap (A\cup B)=A$. Here, have to apply the above given distributive formula:

$\Rightarrow A\cap (A\cup B)=(A\cap A)\cup (A\cap B)$

We know that $A\cap A=A$. Therefore, we can write:

$\Rightarrow A\cap (A\cup B)=A\cup (A\cap B)$

We have already proved by (i) that $A\cup (A\cap B)=A$. Hence, we will obtain:
$A\cap (A\cup B)=A$

Note: Here, we can also prove (i) this by showing that $A\cup (A\cap B)\subset A$ and $A\subset A\cup (A\cap B)$. Then, we will get $A\cup (A\cap B)=A$. Similarly, we can prove $A\cap (A\cup B)\subset A$ and $A\subset A\cap (A\cup B)$ to prove that $A\cap (A\cup B)=A$. Otherwise we can use the help of Venn diagram also for the proofs which is simpler and time consuming as compared to other methods.