NCERT Solutions · Class 11 Mathematics Sets

Roster form first, then compare element by element. \(\displaystyle \mathrm{X}\subset \mathrm{Y}\) means every element of \(\displaystyle \mathrm{X}\) is an element of \(\displaystyle \mathrm{Y}\), so the sets must be listed before they can be compared.Solve the equation defining \(\displaystyle \mathrm{A}\): \[x^{2}-8x+12=0\ \Rightarrow\ (x-2)(x-6)=0\ \Rightarrow\ x=2\ \text{or}\ x=6,\] so \(\displaystyle \mathrm{A}=\{2,6\}\). The others are already in roster form: \[\mathrm{B}=\{2,4,6\},\qquad \mathrm{C}=\{2,4,6,8,\ldots\}\ (\text{all even natural numbers}),\qquad \mathrm{D}=\{6\}.\]Now test each pair: No other inclusion holds: \(\displaystyle 4\in \mathrm{B}\) but \(\displaystyle 4\notin \mathrm{A}\); \(\displaystyle 8\in \mathrm{C}\) but \(\displaystyle 8\notin \mathrm{B}\); \(\displaystyle 2\in \mathrm{A}\) but \(\displaystyle 2\notin \mathrm{D}\).D ⊂ A ⊂ B ⊂ C — that is, the subset relations among these four sets are \(\displaystyle \mathrm{D}\subset \mathrm{A}\), \(\displaystyle \mathrm{D}\subset \mathrm{B}\), \(\displaystyle \mathrm{D}\subset \mathrm{C}\), \(\displaystyle \mathrm{A}\subset \mathrm{B}\), \(\displaystyle \mathrm{A}\subset \mathrm{C}\) and \(\displaystyle \mathrm{B}\subset \mathrm{C}\).

Prove it or break it with one counterexample. Keep the two relations apart while reading each part: \(\displaystyle \in\) is membership (being one of the listed elements) and \(\displaystyle \subset\) is inclusion (every element of the first set being in the second). Most of these parts turn on exactly that difference.(i) If \(\displaystyle x\in \mathrm{A}\) and \(\displaystyle \mathrm{A}\in \mathrm{B}\), then \(\displaystyle x\in \mathrm{B}\) — False. Take \(\displaystyle \mathrm{A}=\{1\}\) and \(\displaystyle \mathrm{B}=\big\{\{1\},\,2\big\}\). Then \(\displaystyle 1\in \mathrm{A}\), and \(\displaystyle \mathrm{A}=\{1\}\) is itself one of the elements of \(\displaystyle \mathrm{B}\), so \(\displaystyle \mathrm{A}\in \mathrm{B}\). But the elements of \(\displaystyle \mathrm{B}\) are \(\displaystyle \{1\}\) and \(\displaystyle 2\), and \(\displaystyle 1\) is neither of them, so \(\displaystyle 1\notin \mathrm{B}\).(ii) If \(\displaystyle \mathrm{A}\subset \mathrm{B}\) and \(\displaystyle \mathrm{B}\in \mathrm{C}\), then \(\displaystyle \mathrm{A}\in \mathrm{C}\) — False. Take \(\displaystyle \mathrm{A}=\{1\}\), \(\displaystyle \mathrm{B}=\{1,2\}\), \(\displaystyle \mathrm{C}=\big\{\{1,2\},\,3\big\}\). Then \(\displaystyle \mathrm{A}\subset \mathrm{B}\) and \(\displaystyle \mathrm{B}\in \mathrm{C}\), but the only elements of \(\displaystyle \mathrm{C}\) are \(\displaystyle \{1,2\}\) and \(\displaystyle 3\), so \(\displaystyle \mathrm{A}=\{1\}\notin \mathrm{C}\).(iii) If \(\displaystyle \mathrm{A}\subset \mathrm{B}\) and \(\displaystyle \mathrm{B}\subset \mathrm{C}\), then \(\displaystyle \mathrm{A}\subset \mathrm{C}\) — True. Let \(\displaystyle x\in \mathrm{A}\). Since \(\displaystyle \mathrm{A}\subset \mathrm{B}\), \(\displaystyle x\in \mathrm{B}\); since \(\displaystyle \mathrm{B}\subset \mathrm{C}\), \(\displaystyle x\in \mathrm{C}\). As \(\displaystyle x\) was arbitrary, every element of \(\displaystyle \mathrm{A}\) lies in \(\displaystyle \mathrm{C}\), i.e. \(\displaystyle \mathrm{A}\subset \mathrm{C}\).(iv) If \(\displaystyle \mathrm{A}\not\subset \mathrm{B}\) and \(\displaystyle \mathrm{B}\not\subset \mathrm{C}\), then \(\displaystyle \mathrm{A}\not\subset \mathrm{C}\) — False. Take \(\displaystyle \mathrm{A}=\{1\}\), \(\displaystyle \mathrm{B}=\{2\}\), \(\displaystyle \mathrm{C}=\{1,3\}\). Here \(\displaystyle 1\notin \mathrm{B}\), so \(\displaystyle \mathrm{A}\not\subset \mathrm{B}\); and \(\displaystyle 2\notin \mathrm{C}\), so \(\displaystyle \mathrm{B}\not\subset \mathrm{C}\). Yet \(\displaystyle 1\in \mathrm{C}\), so \(\displaystyle \mathrm{A}\subset \mathrm{C}\) — the conclusion fails.(v) If \(\displaystyle x\in \mathrm{A}\) and \(\displaystyle \mathrm{A}\not\subset \mathrm{B}\), then \(\displaystyle x\in \mathrm{B}\) — False. Take \(\displaystyle \mathrm{A}=\{3,5,7\}\) and \(\displaystyle \mathrm{B}=\{3,4,6\}\). Since \(\displaystyle 5\notin \mathrm{B}\), \(\displaystyle \mathrm{A}\not\subset \mathrm{B}\). Now choose \(\displaystyle x=5\): it satisfies \(\displaystyle x\in \mathrm{A}\), but \(\displaystyle 5\notin \mathrm{B}\). (\(\displaystyle \mathrm{A}\not\subset \mathrm{B}\) says some element of \(\displaystyle \mathrm{A}\) escapes \(\displaystyle \mathrm{B}\); it says nothing forcing a given \(\displaystyle x\) into \(\displaystyle \mathrm{B}\).)(vi) If \(\displaystyle \mathrm{A}\subset \mathrm{B}\) and \(\displaystyle x\notin \mathrm{B}\), then \(\displaystyle x\notin \mathrm{A}\) — True. This is just the contrapositive of \(\displaystyle \mathrm{A}\subset \mathrm{B}\). Suppose, on the contrary, that \(\displaystyle x\in \mathrm{A}\). Then \(\displaystyle \mathrm{A}\subset \mathrm{B}\) would give \(\displaystyle x\in \mathrm{B}\), contradicting \(\displaystyle x\notin \mathrm{B}\). Hence \(\displaystyle x\notin \mathrm{A}\).(i) False (ii) False (iii) True (iv) False (v) False (vi) True

Absorption, then the distributive law. The trick is to start from \(\displaystyle \mathrm{B}\), open it up with the absorption law \(\displaystyle \mathrm{B}=\mathrm{B}\cap(\mathrm{A}\cup \mathrm{B})\), and then feed in the two hypotheses one at a time until \(\displaystyle \mathrm{C}\) appears.\[\begin{aligned} \mathrm{B} &= \mathrm{B}\cap(\mathrm{A}\cup \mathrm{B}) && \text{(absorption: } \mathrm{B}\subset \mathrm{A}\cup \mathrm{B}\text{)}\\ &= \mathrm{B}\cap(\mathrm{A}\cup \mathrm{C}) && \text{(given } \mathrm{A}\cup \mathrm{B}=\mathrm{A}\cup \mathrm{C}\text{)}\\ &= (\mathrm{B}\cap \mathrm{A})\cup(\mathrm{B}\cap \mathrm{C}) && \text{(distributive law)}\\ &= (\mathrm{A}\cap \mathrm{C})\cup(\mathrm{B}\cap \mathrm{C}) && \text{(given } \mathrm{A}\cap \mathrm{B}=\mathrm{A}\cap \mathrm{C}\text{)}\\ &= (\mathrm{A}\cup \mathrm{B})\cap \mathrm{C} && \text{(distributive law, taking } \mathrm{C}\text{ common)}\\ &= (\mathrm{A}\cup \mathrm{C})\cap \mathrm{C} && \text{(given } \mathrm{A}\cup \mathrm{B}=\mathrm{A}\cup \mathrm{C}\text{)}\\ &= \mathrm{C} && \text{(absorption: } \mathrm{C}\subset \mathrm{A}\cup \mathrm{C}\text{)} \end{aligned} \]Note that both hypotheses were needed — the union condition alone, or the intersection condition alone, is not enough.B = C

Prove a cycle of implications. To show four statements are equivalent it is enough to prove the single loop \[(\text{i})\Rightarrow(\text{ii})\Rightarrow(\text{iii})\Rightarrow(\text{iv})\Rightarrow(\text{i}),\] because then any one of them reaches any other by travelling round the loop. That is four proofs instead of the twelve a pairwise approach would need.(i) \(\displaystyle \Rightarrow\) (ii). Assume \(\displaystyle \mathrm{A}\subset \mathrm{B}\). If \(\displaystyle \mathrm{A}-\mathrm{B}\) were non-empty there would be an \(\displaystyle x\) with \(\displaystyle x\in \mathrm{A}\) and \(\displaystyle x\notin \mathrm{B}\), contradicting \(\displaystyle \mathrm{A}\subset \mathrm{B}\). Hence \(\displaystyle \mathrm{A}-\mathrm{B}=\varphi\).(ii) \(\displaystyle \Rightarrow\) (iii). Assume \(\displaystyle \mathrm{A}-\mathrm{B}=\varphi\). The inclusion \(\displaystyle \mathrm{B}\subset \mathrm{A}\cup \mathrm{B}\) always holds. For the other direction let \(\displaystyle x\in \mathrm{A}\cup \mathrm{B}\). If \(\displaystyle x\in \mathrm{B}\) there is nothing to do; if \(\displaystyle x\in \mathrm{A}\) but \(\displaystyle x\notin \mathrm{B}\), then \(\displaystyle x\in \mathrm{A}-\mathrm{B}=\varphi\), which is impossible. So \(\displaystyle \mathrm{A}\cup \mathrm{B}\subset \mathrm{B}\), and therefore \(\displaystyle \mathrm{A}\cup \mathrm{B}=\mathrm{B}\).(iii) \(\displaystyle \Rightarrow\) (iv). Assume \(\displaystyle \mathrm{A}\cup \mathrm{B}=\mathrm{B}\). Then \[\mathrm{A}\cap \mathrm{B}=\mathrm{A}\cap(\mathrm{A}\cup \mathrm{B})=\mathrm{A},\] the last step being the absorption law \(\displaystyle \mathrm{A}\cap(\mathrm{A}\cup \mathrm{B})=\mathrm{A}\).(iv) \(\displaystyle \Rightarrow\) (i). Assume \(\displaystyle \mathrm{A}\cap \mathrm{B}=\mathrm{A}\). Let \(\displaystyle x\in \mathrm{A}\). Then \(\displaystyle x\in \mathrm{A}\cap \mathrm{B}\), so in particular \(\displaystyle x\in \mathrm{B}\). Hence \(\displaystyle \mathrm{A}\subset \mathrm{B}\).The loop closes, so each condition implies all the others.(i) \(\displaystyle \mathrm{A}\subset \mathrm{B}\), (ii) \(\displaystyle \mathrm{A}-\mathrm{B}=\varphi\), (iii) \(\displaystyle \mathrm{A}\cup \mathrm{B}=\mathrm{B}\) and (iv) \(\displaystyle \mathrm{A}\cap \mathrm{B}=\mathrm{A}\) are equivalent.

Element chase. To prove one set is contained in another, take an arbitrary element of the first and show it must lie in the second.Let \(\displaystyle x\in \mathrm{C}-\mathrm{B}\). By the definition of difference this means \[x\in \mathrm{C}\quad\text{and}\quad x\notin \mathrm{B}.\]Now use the hypothesis \(\displaystyle \mathrm{A}\subset \mathrm{B}\). If \(\displaystyle x\) belonged to \(\displaystyle \mathrm{A}\), then \(\displaystyle \mathrm{A}\subset \mathrm{B}\) would force \(\displaystyle x\in \mathrm{B}\), contradicting \(\displaystyle x\notin \mathrm{B}\). Hence \(\displaystyle x\notin \mathrm{A}\).So \(\displaystyle x\in \mathrm{C}\) and \(\displaystyle x\notin \mathrm{A}\), which is precisely \(\displaystyle x\in \mathrm{C}-\mathrm{A}\). Since \(\displaystyle x\) was an arbitrary element of \(\displaystyle \mathrm{C}-\mathrm{B}\), every element of \(\displaystyle \mathrm{C}-\mathrm{B}\) lies in \(\displaystyle \mathrm{C}-\mathrm{A}\).If \(\displaystyle \mathrm{A}\subset \mathrm{B}\), then \(\displaystyle \mathrm{C}-\mathrm{B}\subset \mathrm{C}-\mathrm{A}\).

Turn each difference into an intersection with a complement, then distribute. The one rewriting that makes both identities routine is \[\mathrm{X}-\mathrm{Y}=\mathrm{X}\cap \mathrm{Y}'.\]First identity. \[\begin{aligned} (\mathrm{A}\cap \mathrm{B})\cup(\mathrm{A}-\mathrm{B}) &=(\mathrm{A}\cap \mathrm{B})\cup(\mathrm{A}\cap \mathrm{B}')\\ &=\mathrm{A}\cap(\mathrm{B}\cup \mathrm{B}') && \text{(distributive law)}\\ &=\mathrm{A}\cap \mathrm{U} && \text{(complement law }\mathrm{B}\cup \mathrm{B}'=\mathrm{U}\text{)}\\ &=\mathrm{A}. \end{aligned} \] In words: each element of \(\displaystyle \mathrm{A}\) either lies in \(\displaystyle \mathrm{B}\) — and is then counted by \(\displaystyle \mathrm{A}\cap \mathrm{B}\) — or does not, and is then counted by \(\displaystyle \mathrm{A}-\mathrm{B}\). Nothing outside \(\displaystyle \mathrm{A}\) is in either piece.Second identity. \[\begin{aligned} \mathrm{A}\cup(\mathrm{B}-\mathrm{A}) &=\mathrm{A}\cup(\mathrm{B}\cap \mathrm{A}')\\ &=(\mathrm{A}\cup \mathrm{B})\cap(\mathrm{A}\cup \mathrm{A}') && \text{(distributive law)}\\ &=(\mathrm{A}\cup \mathrm{B})\cap \mathrm{U}\\ &=\mathrm{A}\cup \mathrm{B}. \end{aligned} \]\(\displaystyle \mathrm{A}=(\mathrm{A}\cap \mathrm{B})\cup(\mathrm{A}-\mathrm{B})\) and \(\displaystyle \mathrm{A}\cup(\mathrm{B}-\mathrm{A})=\mathrm{A}\cup \mathrm{B}\).

Create the pattern the distributive law needs, using the identity elements. The two facts \(\displaystyle \mathrm{A}=\mathrm{A}\cap \mathrm{U}\) and \(\displaystyle \mathrm{A}=\mathrm{A}\cup \varphi\) let us rewrite the lone \(\displaystyle \mathrm{A}\) so that the distributive law can be applied. These two results are the absorption laws.(i) \[\begin{aligned} \mathrm{A}\cup(\mathrm{A}\cap \mathrm{B}) &=(\mathrm{A}\cap \mathrm{U})\cup(\mathrm{A}\cap \mathrm{B}) && (\mathrm{A}=\mathrm{A}\cap \mathrm{U})\\ &=\mathrm{A}\cap(\mathrm{U}\cup \mathrm{B}) && \text{(distributive law)}\\ &=\mathrm{A}\cap \mathrm{U} && (\mathrm{U}\cup \mathrm{B}=\mathrm{U}\text{, since }\mathrm{B}\subset \mathrm{U})\\ &=\mathrm{A}. \end{aligned} \](ii) \[\begin{aligned} \mathrm{A}\cap(\mathrm{A}\cup \mathrm{B}) &=(\mathrm{A}\cup \varphi)\cap(\mathrm{A}\cup \mathrm{B}) && (\mathrm{A}=\mathrm{A}\cup \varphi)\\ &=\mathrm{A}\cup(\varphi\cap \mathrm{B}) && \text{(distributive law)}\\ &=\mathrm{A}\cup \varphi && (\varphi\cap \mathrm{B}=\varphi)\\ &=\mathrm{A}. \end{aligned} \]\(\displaystyle \mathrm{A}\cup(\mathrm{A}\cap \mathrm{B})=\mathrm{A}\) and \(\displaystyle \mathrm{A}\cap(\mathrm{A}\cup \mathrm{B})=\mathrm{A}\).

A single counterexample settles a "need not imply". We are not asked to prove anything is always false — only to exhibit one triple of sets for which \(\displaystyle \mathrm{A}\cap \mathrm{B}=\mathrm{A}\cap \mathrm{C}\) holds while \(\displaystyle \mathrm{B}=\mathrm{C}\) fails.Take \[\mathrm{A}=\{1,2\},\qquad \mathrm{B}=\{2,3\},\qquad \mathrm{C}=\{2,4\}.\]Then \[\mathrm{A}\cap \mathrm{B}=\{2\},\qquad \mathrm{A}\cap \mathrm{C}=\{2\},\] so \(\displaystyle \mathrm{A}\cap \mathrm{B}=\mathrm{A}\cap \mathrm{C}\). But \(\displaystyle 3\in \mathrm{B}\) while \(\displaystyle 3\notin \mathrm{C}\), so \(\displaystyle \mathrm{B}\neq \mathrm{C}\).The reason is worth naming: intersecting with \(\displaystyle \mathrm{A}\) only records what \(\displaystyle \mathrm{B}\) and \(\displaystyle \mathrm{C}\) do inside \(\displaystyle \mathrm{A}\). Anything they do outside \(\displaystyle \mathrm{A}\) — here the elements \(\displaystyle 3\) and \(\displaystyle 4\) — is invisible to the test, so it can differ freely.\(\displaystyle \mathrm{A}\cap \mathrm{B}=\mathrm{A}\cap \mathrm{C}\) need not imply \(\displaystyle \mathrm{B}=\mathrm{C}\); e.g. \(\displaystyle \mathrm{A}=\{1,2\}\), \(\displaystyle \mathrm{B}=\{2,3\}\), \(\displaystyle \mathrm{C}=\{2,4\}\).

Absorption, then the distributive law (the route the hint points to). Show that each of \(\displaystyle \mathrm{A}\) and \(\displaystyle \mathrm{B}\) equals the same set \(\displaystyle \mathrm{A}\cap \mathrm{B}\).Start with \(\displaystyle \mathrm{A}\): \[\begin{aligned} \mathrm{A} &= \mathrm{A}\cap(\mathrm{A}\cup \mathrm{X}) && \text{(absorption: } \mathrm{A}\subset \mathrm{A}\cup \mathrm{X}\text{)}\\ &= \mathrm{A}\cap(\mathrm{B}\cup \mathrm{X}) && \text{(given } \mathrm{A}\cup \mathrm{X}=\mathrm{B}\cup \mathrm{X}\text{)}\\ &= (\mathrm{A}\cap \mathrm{B})\cup(\mathrm{A}\cap \mathrm{X}) && \text{(distributive law)}\\ &= (\mathrm{A}\cap \mathrm{B})\cup \varphi && \text{(given } \mathrm{A}\cap \mathrm{X}=\varphi\text{)}\\ &= \mathrm{A}\cap \mathrm{B}. \end{aligned} \]Exactly the same computation with the roles of \(\displaystyle \mathrm{A}\) and \(\displaystyle \mathrm{B}\) exchanged: \[\begin{aligned} \mathrm{B} &= \mathrm{B}\cap(\mathrm{B}\cup \mathrm{X}) = \mathrm{B}\cap(\mathrm{A}\cup \mathrm{X})\\ &= (\mathrm{B}\cap \mathrm{A})\cup(\mathrm{B}\cap \mathrm{X}) = (\mathrm{A}\cap \mathrm{B})\cup \varphi = \mathrm{A}\cap \mathrm{B}. \end{aligned} \]Both \(\displaystyle \mathrm{A}\) and \(\displaystyle \mathrm{B}\) have been shown equal to \(\displaystyle \mathrm{A}\cap \mathrm{B}\), hence equal to each other.A = B

Give each pair its own shared element. The three pairwise intersections must be non-empty, yet no element may sit in all three sets at once. So use three different elements and let each one belong to exactly two of the sets — never to the third.Let element \(\displaystyle 1\) be shared by \(\displaystyle \mathrm{A}\) and \(\displaystyle \mathrm{C}\), element \(\displaystyle 2\) by \(\displaystyle \mathrm{A}\) and \(\displaystyle \mathrm{B}\), element \(\displaystyle 3\) by \(\displaystyle \mathrm{B}\) and \(\displaystyle \mathrm{C}\): \[\mathrm{A}=\{1,2\},\qquad \mathrm{B}=\{2,3\},\qquad \mathrm{C}=\{1,3\}.\]Check the three pairwise intersections: \[\mathrm{A}\cap \mathrm{B}=\{2\},\qquad \mathrm{B}\cap \mathrm{C}=\{3\},\qquad \mathrm{A}\cap \mathrm{C}=\{1\},\] all non-empty as required.Check the triple intersection: \(\displaystyle 1\notin \mathrm{B}\), \(\displaystyle 2\notin \mathrm{C}\) and \(\displaystyle 3\notin \mathrm{A}\), so no element lies in all three sets, and \[\mathrm{A}\cap \mathrm{B}\cap \mathrm{C}=\varphi.\]\(\displaystyle \mathrm{A}=\{1,2\},\ \mathrm{B}=\{2,3\},\ \mathrm{C}=\{1,3\}\)