
For any two sets prove that:
1) $A\cup (A\cap B)=A$
2) $A\cap (A\cup B)=A$
Answer
609.3k+ views
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.
$A\cup (A\cap B=(A\cup A)\cap (A\cup B)$ 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.
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