how to prove it - ch3, sec3.4(Proofs Involving Conjunctions and Biconditionals) ex

Proof strategies for conjunctions and biconditionals introduced in this sections are pretty trivial. Here they are...

To prove a goal of the form $P\wedge Q$: simply prove P and Q separately

To prove a goal of the form $P\leftrightarrow Q$: simply prove $P\to Q$ and $Q\to P$ separately

To use a given of the form $P\wedge Q$ assume as if P and Q are both given. To use a given of the form $P\leftrightarrow Q$ assume as if $P\to Q$ and $Q\to P$ are both given.

Some times when proving $P\leftrightarrow Q$, steps to prove $P\to Q$ and $Q\to P$ are same except the order is reversed. In those cases, to keep the proof concise, we can go direct $P\leftrightarrow R\leftrightarrow Q$ where R is a common form that comes while proving $P\to Q$ and $Q\to P$

Other Notes:
This is Set specific. A = B can be proved by either proving $\forall (x\in A\leftrightarrow x\in B)$ or by proving $(A\subseteq B)\wedge (B\subseteq A)$ where A and B are sets.

Another thing when proofs involve even/odd numbers, we can use following definitions for even/odd.
An integer x is even if $\exists k\in Z(x=2k)$
An integer x is odd if $\exists k\in Z(x=2k+1)$


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Ex-1
($\to$)Suppose $\forall x\left(P\right(x)\wedge Q(x\left)\right)$ is true. Let y be arbitrary, so P(y) is true. Since y is arbitrary, we can conclude $\forall xP\left(x\right)$ is true. By using similar argument we can prove that $\forall xQ\left(x\right)$ is also true. Hence $\forall xP\left(x\right)\wedge \forall xQ\left(x\right)$ is true.
($\leftarrow$)Suppose $\forall xP\left(x\right)\wedge \forall xQ\left(x\right)$ is true. Let y be arbitrary, it follows that P(y) is true and Q(y) is true as well, so $P\left(y\right)\wedge Q\left(y\right)$ is true. Since y is arbitrary, we can conclude $\forall x\left(P\right(x)\wedge Q(x\left)\right)$ is true.


Ex-2
Suppose $A\subseteq B$, $A\subseteq C$ and let x be an arbitrary element of A. Since $A\subseteq B$, therefore $x\in B$. Also since $A\subseteq C$, therefore $x\in C$ also. So $x\in B\wedge x\in C$ and therefore $x\in B\cap C$. Since x is arbitrary, we can conclude $A\subseteq B\cap C$. Thus if $A\subseteq B$ and $A\subseteq C$ then $A\subseteq B\cap C$


Ex-3 Let C be an arbitrary set and x be an arbitrary element of $C\backslash B$. It follows that $x\in C$ and $x\notin B$. Since $A\subseteq B$, it means every element of A is an element of B also or in other words(the contrapositive) a element that is not in B is not in A also. Since $x\notin B$, therefore $x\notin A$. Since $x\in C$ and $x\notin A$, therefore $x\in C\backslash A$. Since x is arbitrary, we can conclude $C\backslash B\subseteq C\backslash A$


Ex-4 Suppose $A\subseteq B$ and $\neg (A\subseteq C)$. Since $\neg (A\subseteq C)$, there exists atleast one x s.t. x is in A but not in C. Since $A\subseteq B$, x should be in B also. So x is in B but not in C, Thus $\neg (B\subseteq C)$.


Ex-5 Suppose $A\subseteq B\backslash C$ and A is non-empty. Let $a$ be an element of A. It follows that $a\in B\backslash C$, that is $a\in B$ and $a\notin C$. We've found an element that is in B but not in C, so $\neg (B\subseteq C)$.


Ex-6
Let x be an arbitrary element of $A\backslash (B\cap C)$. It follows
$x\in A\backslash (B\cap C)$ iff
$x\in A\wedge \neg x\in (B\cap C)$ iff
$x\in A\wedge \neg (x\in B\wedge x\in C)$ iff
$x\in A\wedge (x\notin B\vee x\notin C)$ iff
$(x\in A\wedge x\notin B)\vee (x\in A\wedge x\notin C)$ iff
$(x\in A\backslash B)\vee (x\in A\backslash C)$ iff
$x\in (A\backslash B)\cup (A\backslash C)$
Since x is arbitrary, we can conclude $A\backslash (B\cap C)$ = $(A\backslash B)\cup (A\backslash C)$


Ex-7
($\to$)Suppose A and B are arbitrary sets, let x be an arbitrary element of $P(A\cap B)$. It follows that $x\subseteq (A\cap B)$. Let y be an arbitrary element of x, then $y\in (A\cap B)$ and therefore $y\in A$ and $y\in B$. Since y is arbitrary so $x\subseteq A$ and also $x\subseteq B$, therefore $x\in P\left(A\right)$ and $x\in P\left(B\right)$. Thus, $x\in P\left(A\right)\cap P\left(B\right)$.
($\leftarrow$)Suppose A and B are arbitrary sets, let x be an arbitrary element of $P\left(A\right)\wedge P\left(B\right)$. It follows that $x\in P\left(A\right)\wedge x\in P\left(B\right)$, therefore $x\subseteq A$ and $x\subseteq B$. Let y be an arbitrary element of x, then $y\in A$ and $y\in B$. So, $y\in A\cap B$. Since y is arbitrary, we conclude $x\subseteq (A\cap B)$. Thus, $x\in P(A\cap B)$.


Ex-8
($\to$)Suppose $A\subseteq B$. Let x be an arbitrary element of $P\left(A\right)$, Since $x\in P\left(A\right)$ it follows $x\subseteq A$. Since $A\subseteq B$, so $x\subseteq B$ also. Therefore $x\in P\left(B\right)$. Since x is arbitrary, we can conclude $P\left(A\right)\subseteq P\left(B\right)$.
($\leftarrow$)Suppose $P\left(A\right)\subseteq P\left(B\right)$. Let x be an arbitrary element of A, so there should exist atleast one set y in $P\left(A\right)$ s.t. $x\in y$. Since $P\left(A\right)\subseteq P\left(B\right)$, so $y\in P\left(B\right)$. It means $y\subseteq B$ and $x\in y$, therefore $x\in B$ also. Since x is arbitrary, we can conclude $A\subseteq B$.


Ex-9 Suppose x and y are odd integers. So there exists a $k_1\in Z$ s.t. $x=2k_1+1$ and a $k_2\in Z$ s.t. $y=2k_2+1$. On multiplying both we get $xy=2(2k_1{k}_2+k_1+k_2)+1$. Since $(2k_1{k}_2+k_1+k_2)\in Z$, therefore xy is odd.


Ex-10 Let n be an arbitrary integer.
($\to$)We want to prove that If $n^3$ is even then so is n. We'll prove the contrapositive that is, If n is odd then $n^3$ is also odd. Suppose n is odd, then there exists a $k\in Z$ s.t. $n=2k+1$. So $n^3=(2k+1{)}^3$ and then $n^3=8k^3+1+6k(2k+1)$, then $n^3=8k^3+12k^2+6k+1$, then $n^3=2(4k^3+6k^2+3k)+1$. Since $(4k^3+6k^2+3k)\in Z$, so $n^3$ is odd.
($\leftarrow$)Suppose n is even, then there exists a $k\in Z$ s.t. $n=2k$, so $n^3=2\left(4k^3\right)$. Since $4k^3\in Z$, so $n^3$ is even.


Ex-11
(a) The flaw is in chosing the same k for both m and n.
(b) It is incorrect. A counterexample is n = 5, m = 2.


Ex-12 Suppose x is an arbitrary real number.
($\to$) Suppose, for a particular real number y, (x + y = xy)
=> y = $\frac{x}{x-1}$
it is defined only when $(x-1)\ne 0$, that is $x\ne 1$. Since x is arbitrary, we can conclude that $\forall x\in R\exists y\in R\left(\right(x+y=xy)\to (x\ne 1\left)\right)$

($\leftarrow$) Suppose, $x\ne 1$, Let us take a real number y = $\frac{x}{x-1}$. It is defined because $x\ne 1$ and hence $(x-1)\ne 0$.
Now check (x + y)
= $x+\frac{x}{x-1}$
= $\frac{x(x-1)+x}{x-1}$
= $\frac{x^2}{x-1}$
= xy
Since x is arbitrary, we can conclude that $\forall x\in R\exists y\in R\left(\right(x\ne 1)\to (x+y=xy\left)\right)$


Ex-13 Let z = 1, and now prove
$\forall x\in R^{+}[\exists y\in R(y-x=\frac{y}{x})\leftrightarrow x\ne 1]$

Suppose x be arbitrary positive real number and then we need to prove following
$\exists y\in R(y-x=\frac{y}{x})\leftrightarrow x\ne 1$

($\to$) $\exists y\in R(y-x=\frac{y}{x})\to x\ne 1$
we prove the contrapositive
which is , $x=1\to \forall y\in R(y-x\ne \frac{y}{x})$
Suppose, x = 1 and y be arbitrary real number. clearly $y-1\ne y$. So, that establishes first half of the biconditional statement.

($\leftarrow$) $x\ne 1\to \exists y\in R(y-x=\frac{y}{x})$
clearly, given that $x\ne 1$, then you can choose real number $y=\frac{x^2}{x-1}$ for which $y-x=\frac{y}{x}$. So, that establishes second half of the biconditional statement.

Ex-14 Let x be an arbitrary element of $\cup \{A\backslash B\mid A\in F\}$. It follows that there exists a set $a\in F$ s.t. $x\in a\backslash B$. Moreover, $\neg (a\subseteq B)$ is true otherwise $a\backslash B$ will be empty. Since $\neg (a\subseteq B)$, so $a\notin P\left(B\right)$. Since $a\in F$ and $a\notin P\left(B\right)$, so $a\in F\backslash P\left(B\right)$ and hence $x\in \cup (F\backslash P(B\left)\right)$. Since x is arbitrary, we can conclude $\cup \{A\backslash B\mid A\in F\}\subseteq \cup (F\backslash P(B\left)\right)$


Ex-15
given: every element of $F$ is disjoint from some element of $G$
goal: $\cup F$ and $\cap G$ are disjoint $\equiv$ $\forall x\in \cup F(\forall y\in \cap Gx\ne y)$

Formal Proof:
Let x be an arbitrary element of $\cup F$, so there exists a set $f\in F$ s.t. $x\in f$. Let y be an arbitrary element of $\cap G$, so y is in every set in $G$. Since every element of $F$ is disjoint from some element of $G$, so there exists a $g\in G$ s.t. $f$ and $g$ are disjoint. But $y\in g$ also as y is in every set in $G$ and since $x\in f$ and $f$ and $g$ are disjoint, therefore $x\ne y$. Since x and y are arbitrary, we can conclude that $\cup F$ and $\cap G$ are disjoint.


Ex-16 We will prove that $A\subseteq \cup P\left(A\right)$ and also $\cup P\left(A\right)\subseteq A$.
Let us prove $A\subseteq \cup P\left(A\right)$. Let x be an arbitrary element of A. Since $A\in P\left(A\right)$, so by definition of $\cup P\left(A\right)$, x should be an element of $\cup P\left(A\right)$ also. Since x is arbitrary, $A\subseteq P\left(A\right)$.
Let us prove $\cup P\left(A\right)\subseteq A$. Let x be an arbitrary element of $\cup P\left(A\right)$, so there exists an $a\in P\left(A\right)$ s.t. $x\in a$. Since $a\in P\left(A\right)$, therefore $a\subseteq A$ and hence $x\in A$ also. Since x is arbitrary we can conclude that $\cup P\left(A\right)\subseteq A$.
Thus $A=P\left(A\right)$.


Ex-17
(a) Let x be an arbitrary element of $\cup (F\cap G)$. It follows, there exists a set $A\in (F\cap G)$ s.t. $x\in A$. Since $A\in (F\cap G)$, so $A\in F$ and hence $x\in \cup F$ also. By similar argument $x\in \cup G$ also. Since $x\in \cup F$ and $x\in \cup G$, therefore $x\in (\cup F)\cap (\cup G)$. Since x is arbitrary, we can conclude $\cup (F\cap G)\subseteq (\cup F)\cap (\cup G)$

(b) The flaw is in the fact that we are chosing the same A in both families.

(c) $F$ = { {1},{2} } and $G$ = { {2}, {1,2} }
$F\cap G$ = { {2} }
$\cup (F\cap G)$ = { 2 }... (i)

$\cup F$ = { 1,2 }
$\cup G$ = { 1,2 }
$(\cup F)\cap (\cup G)$ = { 1,2 }... (ii)

clearly (i) and (ii) are not same.

Note: How I found it?
I realized for $x\in (\cup F)\cap (\cup G)$, $x\in (\cup F)\wedge x\in (\cup G)$ and hence there exist particular $f\in F$ and $g\in G$ s.t. $x\in f$ and $x\in g$. Note that $f$ and $g$ could be *different* sets.
However for $x\in \cup (F\cap G)$, there exist a particular $a\in (F\cap G)$ s.t. $x\in a$. Note that for this case the *same* set $a$ has to be in both the families.


Ex-18
($\to$)
given:
goal: If $(\cup F)\cap (\cup G)\subseteq \cup (F\cap G)$ then $\forall A\in F\forall B\in G(A\cap B\subseteq \cup (F\cap G\left)\right)$

Assume the hypotheses

given: $(\cup F)\cap (\cup G)\subseteq \cup (F\cap G)$
goal: $\forall A\in F\forall B\in G(A\cap B\subseteq \cup (F\cap G\left)\right)$

Let A be an arbitrary element of $F$ and B be an arbitrary element of $G$.

given: $(\cup F)\cap (\cup G)\subseteq \cup (F\cap G)$
goal: $A\cap B\subseteq \cup (F\cap G)$ $\equiv$ $\forall x(x\in (A\cap B)\to x\in \cup (F\cap G\left)\right)$

Let x be arbitrary

given: $(\cup F)\cap (\cup G)\subseteq \cup (F\cap G)$
goal: $x\in (A\cap B)\to x\in \cup (F\cap G)$

Assume the hypotheses

given: $(\cup F)\cap (\cup G)\subseteq \cup (F\cap G)$ , $x\in (A\cap B)$
goal: $x\in \cup (F\cap G)$

Formal Proof:
Suppose $(\cup F)\cap (\cup G)\subseteq \cup (F\cap G)$. Let A be an arbitrary element of $F$ and B be an arbitrary element of $G$ and x be an arbitrary element of $A\cap B$. It follows $x\in A$ and $x\in B$. Since $x\in A$ and $A\in F$, so $x\in \cup F$. Similarly, $x\in \cup G$. So $x\in (\cup F)\cap (\cup G)$. Since $(\cup F)\cap (\cup G)\subseteq \cup (F\cap G)$, hence $x\in \cup (F\cap G)$. Since x is arbitrary, we can conclude that $A\cap B\subseteq \cup (F\cap G)$. And since A and B are arbitrary, therefore $\forall A\in F\forall B\in G(A\cap B\subseteq \cup (F\cap G\left)\right)$

($\leftarrow$)
Suppose $\forall A\in F\forall B\in G(A\cap B\subseteq \cup (F\cap G\left)\right)$. Let x be an arbitrary element of $(\cup F)\cap (\cup G)$. Then $x\in \cup F$, it follows that there exists a set $f\in F$ s.t. $x\in f$. Similarly there exists a set $g\in G$ s.t. $x\in g$. So $x\in f\cap g$. Since $\forall A\in F\forall B\in G(A\cap B\subseteq \cup (F\cap G\left)\right)$, so $(f\cap g)\subseteq \cup (F\cap G)$ and hence $x\in \cup (F\cap G)$. Since x is arbitrary, we conclude that $(\cup F)\cap (\cup G)\subseteq \cup (F\cap G)$


Ex-19
($\to$)We will prove it by contradiction. Suppose $\cup F$ and $\cup G$ are disjoint. Let A be an arbitrary element of $F$ and B be an arbitrary element of $G$ and A and B are not disjoint. Then, there exists a x s.t. $x\in A$ and $x\in B$. Since $x\in A$ and $A\in F$, so $x\in \cup F$ and similarly $x\in \cup G$ also. This means $\cup F$ and $\cup G$ have a common element x and hence not disjoint and that contradicts our assumption. Hence A and B have to be disjoint. Since A and B are arbitrary, we conclude that for all $A\in F$ and $B\in G$, A and B are disjoint.
($\leftarrow$)We will prove it by contradiction. Suppose for all $A\in F$ and $B\in G$, A and B are disjoint. Assume $\cup F$ and $\cup G$ are not disjoint, then there exists a x s.t. $x\in \cup F$ and $x\in \cup G$. It follows there exist $A\in F$ and $B\in G$ s.t. $x\in A$ as well as $x\in B$ but this contradicts our earlier assumption. Hence $\cup F$ and $\cup G$ are disjoint.


Ex-20
(a) Suppose x be an arbitrary element of $(\cup F)\backslash (\cup G)$. It follows that $x\in \cup F$ and $x\notin \cup G$. Since $x\in \cup F$, therefore there exists $A\in F$ s.t. $x\in A$. Since $x\notin \cup G$, x can not be an element of any set in $G$. Since $x\in A$, so $A\notin G$. Since $A\in F$ and $A\notin G$, so $A\in F\backslash G$. Since $x\in A$ and $A\in F\backslash G$, therefore $x\in \cup (F\backslash G)$. Since x is arbitrary, we can conclude $(\cup F)\backslash (\cup G)\subseteq \cup (F\backslash G)$

(b) The flaw is in the statement, "Since $x\in A$ and $A\notin G$, $x\notin \cup G$". On the contrary, Since A is a particular set in $G$ and there *can* be another set $B\in G$ s.t. $x\in B$ and in that case $x\in \cup G$.

(c)
($\to$)
given:
goal: If $\cup (F\backslash G)\subseteq (\cup F)\backslash (\cup G)$ then $\forall A\in (F\backslash G)\forall B\in G(A\cap B=\varnothing )$

Assume the hypotheses

given: $\cup (F\backslash G)\subseteq (\cup F)\backslash (\cup G)$
goal: $\forall A\in (F\backslash G)\forall B\in G(A\cap B=\varnothing )$

Let A be an arbitrary element of $F\backslash G$ and B be an arbitrary element of $G$.

given: $\cup (F\backslash G)\subseteq (\cup F)\backslash (\cup G)$
goal: $A\cap B=\varnothing$ $\equiv$ $\forall x(x\notin A)\vee (x\notin B)$

Let x be arbitrary

given: $\cup (F\backslash G)\subseteq (\cup F)\backslash (\cup G)$
gaol: $(x\notin A)\vee (x\notin B)$ $\equiv$ $x\in A\to x\notin B$

Assume the hypotheses

given: $\cup (F\backslash G)\subseteq (\cup F)\backslash (\cup G)$ , $x\in A$
goal: $x\notin B$

Formal Proof:
Suppose $\cup (F\backslash G)\subseteq (\cup F)\backslash (\cup G)$. Let A be an arbitrary element of $F\backslash G$ and B be an arbitrary element of $G$ and x be an arbitrary element of A. Since $x\in A$ and $A\in (F\backslash G)$, therefore $x\in \cup (F\backslash G)$. Since $\cup (F\backslash G)\subseteq (\cup F)\backslash (\cup G)$, therefore $x\in (\cup F)\backslash (\cup G)$, therefore $x\notin (\cup G)$. It means x is not an element of any element in $G$. Hence x can not be element of B or $x\notin B$. Thus $x\in A\to x\notin B$, therefore $x\notin A\vee x\notin B$, therefore $\neg (x\in A\wedge x\in B)$, therefore $x\notin (A\cap B)$. Since x is arbitrary, we can conclude $A\cap B=\varnothing$

($\leftarrow$)
given:
goal: If $\forall A\in (F\backslash G)\forall B\in G(A\cap B=\varnothing )$ then $\cup (F\backslash G)\subseteq (\cup F)\backslash (\cup G)$

Assume the hypotheses

given: $\forall A\in (F\backslash G)\forall B\in G(A\cap B=\varnothing )$
goal:
$\cup (F\backslash G)\subseteq (\cup F)\backslash (\cup G)$
$\equiv$ $\forall x(x\in \cup (F\backslash G)\to x\in (\cup F)\backslash (\cup G\left)\right)$

Let x be arbitrary

given: $\forall A\in (F\backslash G)\forall B\in G(A\cap B=\varnothing )$
goal: $x\in \cup (F\backslash G)\to x\in (\cup F)\backslash (\cup G)$

Assume the hypotheses

given: $\forall A\in (F\backslash G)\forall B\in G(A\cap B=\varnothing )$ , $x\in \cup (F\backslash G)$
goal: $x\in (\cup F)\backslash (\cup G)$

Formal Proof:
Suppose $\forall A\in (F\backslash G)\forall B\in G(A\cap B=\varnothing )$. Let x be an arbitrary element of $\cup (F\backslash G)$. It follows that there exists a set $y\in F\backslash G$ s.t. $x\in y$.
Since $y\in F\backslash G$, so $y\in F$. Also $x\in y$, hence $x\in \cup F$.
Since $y\in F\backslash G$, so y is disjoint from every set in $G$ because of our assumpion made in the starting of the proof and hence x can not belong to any set in $G$. Thus, $x\notin \cup G$.
As we've proven that $x\in \cup F$ and $x\notin \cup G$, so $x\in (\cup F)\backslash (\cup G)$. Since x is arbitrary, we can conclude $\cup (F\backslash G)\subseteq (\cup F)\backslash (\cup G)$

(d)
$F$ = { {1}, {1,2} }
$G$ = { {1} }
$F\backslash G$ = { {1,2} }
$\cup (F\backslash G)$ = {1,2) ...(i)

$\cup F$ = {1,2}
$\cup G$ = {1}
$(\cup F)\backslash (\cup G)$ = {2} ...(ii)

clearly (i) and (ii) are not same.


Ex-21 We will prove the contrapositive, which says If for all $A\in F$ there exists some $B\in G$ s.t. $A\subseteq B$ then $\cup F\subseteq \cup G$. Suppose for all $A\in F$ there exists some $B\in G$ s.t. $A\subseteq B$. Let x be an arbitrary element of $\cup F$. It follows that there exists some $A\in F$ s.t. $x\in A$. But, there has to exist a $B\in G$ s.t. $A\subseteq B$. Then $x\in B$. So we have $x\in B$ and $B\in G$, therefore $x\in \cup B$. Since x is arbitrary, we can conclude $\cup F\subseteq \cup G$


Ex-22
(a) Following strategies have been used...
To prove a goal of the form $\forall xP\left(x\right)$
To prove a goal of the form $P\to Q$
To use a given of the form $P\wedge Q$

(b) Let x be an arbitrary element of $B\backslash \left({\cup }_{i\in I}{A}_i\right)$. Then
$x\in B\backslash \left({\cup }_{i\in I}{A}_i\right)$ iff
$x\in B\wedge \neg (x\in ({\cup }_{i\in I}{A}_i\left)\right)$ iff
$x\in B\wedge \neg (\exists i\in I(x\in A_i\left)\right)$ iff
$x\in B\wedge (\forall i\in I(x\notin A_i\left)\right)$ iff
$\forall i\in I(x\in B)\wedge (x\notin A_i)$ iff
$\forall i\in I(x\in B\backslash A_i)$ iff
$x\in {\cap }_{i\in I}(B\backslash A_i)$

Since x is arbitrary, we can conclude $B\backslash \left({\cup }_{i\in I}{A}_i\right)$ = ${\cap }_{i\in I}(B\backslash A_i)$

(c)
The theorem is, $B\backslash \left({\cap }_{i\in I}{A}_i\right)$ = ${\cup }_{i\in I}(B\backslash A_i)$

Let x be an arbitrary element of $B\backslash \left({\cap }_{i\in I}{A}_i\right)$. Then
$x\in B\backslash \left({\cap }_{i\in I}{A}_i\right)$ iff
$x\in B\wedge \neg (x\in ({\cap }_{i\in I}{A}_i\left)\right)$ iff
$x\in B\wedge \neg (\forall i\in I(x\in A_i\left)\right)$ iff
$x\in B\wedge (\exists i\in I(x\notin A_i\left)\right)$ iff
$\exists i\in I(x\in B\wedge x\notin A_i)$ iff
$\exists i\in I(x\in B\backslash A_i)$ iff
$x\in {\cup }_{i\in I}(B\backslash A_i)$

Since x is arbitrary, we can conclude $B\backslash \left({\cap }_{i\in I}{A}_i\right)$ = ${\cup }_{i\in I}(B\backslash A_i)$


Ex-23
(a) Let x be an arbitrary element of ${\cup }_{i\in I}(A_i\backslash B_i)$. Then there exists $i\in I$ s.t. $x\in A_i\backslash B_i$, that is $x\in A_i$ and $x\notin B_i$. Since $i\in I$ and $x\in A_i$, so $x\in {\cup }_{i\in I}{A}_i$. Since $i\in I$ and $x\notin B_i$, so $x\notin {\cap }_{i\in I}{B}_i$. Thus $x\in \left({\cup }_{i\in I}{A}_i\right)\backslash \left({\cap }_{i\in I}{B}_i\right)$. Since x is arbitrary, we can conclude that ${\cup }_{i\in I}(A_i\backslash B_i)\subseteq \left({\cup }_{i\in I}{A}_i\right)\backslash \left({\cap }_{i\in I}{B}_i\right)$

Another method(WRONG):
Let x be an arbitrary element of ${\cup }_{i\in I}(A_i\backslash B_i)$. Then
$x\in {\cup }_{i\in I}(A_i\backslash B_i)$
=> $\exists i\in I(x\in A_i\backslash B_i)$
=> $\exists i\in I(x\in A_i\wedge x\notin B_i)$
=> $(\exists i\in I(x\in A_i\left)\right)\wedge (\exists i\in I(x\notin B_i\left)\right)$ (justify this step)
=> $(x\in {\cup }_{i\in I}{A}_i)\wedge \neg (\forall i\in I(x\in B_i\left)\right)$
=> $(x\in {\cup }_{i\in I}{A}_i)\wedge \neg (x\in {\cap }_{i\in I}{B}_i)$
=> $x\in \left({\cup }_{i\in I}{A}_i\right)\backslash \left({\cap }_{i\in I}{B}_i\right)$

The reason this is wrong is that, what we've proven with this method is that
${\cup }_{i\in I}(A_i\backslash B_i)$ = $\left({\cup }_{i\in I}{A}_i\right)\backslash \left({\cap }_{i\in I}{B}_i\right)$
which is not true and that is why one of the steps in above analysis is not justifiable.

(b)
$A_1$ = {1,2} $A_2$ = {3}
$B_1$ = {2} $B_2$ = {5}

$A_1\backslash B_1$ = {1}
$A_2\backslash B_2$ = {3}
${\cup }_{i\in I}(A_i\backslash B_i)$ = {1,3} ...(i)

${\cup }_{i\in I}{A}_i$ = {1,2,3}
${\cap }_{i\in I}{B}_i$ = $\varnothing$
$\left({\cup }_{i\in I}{A}_i\right)\backslash \left({\cap }_{i\in I}{B}_i\right)$ = {1,2,3} ...(ii)

clearly (i) and (ii) are not same.


Ex-24
(a) Let x be an arbitrary element of ${\cup }_{i\in I}(A_i\cap B_i)$. It follows that there exists a $i\in I$ s.t. $x\in A_i$ and $x\in B_i$. Since we have a particular i s.t. $x\in A_i$ and $i\in I$, therefore $x\in {\cup }_{i\in I}{A}_i$. By similar argument $x\in {\cup }_{i\in I}{B}_i$ as well. Thus $x\in \left({\cup }_{i\in I}{A}_i\right)\cap (x\in {\cup }_{i\in I}{B}_i)$. Since x is arbitrary, we can conclude that ${\cup }_{i\in I}(A_i\cap B_i)\subseteq \left({\cup }_{i\in I}{A}_i\right)\cap \left({\cup }_{i\in I}{B}_i\right)$

(b)
$A_1$ = {1,3} $A_2$ = {4}
$B_1$ = {1} $B_2$ = {3,4}

$A_1\cap B_1$ = {1}
$A_2\cap B_2$ = {4}
${\cup }_{i\in I}(A_i\cap B_i)$ = {1,4} ...(i)

${\cup }_{i\in I}{A}_i$ = {1,3,4}
${\cup }_{i\in I}{B}_i$ = {1,3,4}
$\left({\cup }_{i\in I}{A}_i\right)\cap \left({\cup }_{i\in I}{B}_i\right)$ = {1,3,4} ...(ii)

Clearly, (i) and (ii) are not same.


Ex-25
Let us try c = ab. Since a and b are integers, so clearly a|c and b|c as well.

Ex-26
(a)
($\to$) Suppose 15|n, so there exists $k\in Z$ s.t. n = 15k. So n = 3(5k). Since 5k is an integer so 3|n. Also n = 5(3k) and 3k is an integer, so 5|n. So if 15|n then 3|n and 5|n.
($\leftarrow$) Suppose 3|n and 5|n. So there exist $k\in Z$ s.t.
n = 3k ...(i)
and $l\in Z$ s.t. n = 5l ...(ii)
(sidenote: we want to find a and b s.t. 3a + 5b = 15(or some multiplie of it), on a few tries I came up with a = -5 and b = 6)

multiply (i) with -5, multiply (ii) with 6 and add them both
-5n + 6n = (-15k) + (30l)
=> n = 15(2l - k)
Since (2l - k) is also an integer, so 15|n

(b) It can be proven by finding just one counterexample. n = 30 is one and n = 90, 150 etc are also the counterexamples.