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EXERCISE 1.5 1–7 (part 5 of 6)

  1. Exercise 1

    Let U={1,2,3,4,5,6,7,8,9},A={1,2,3,4},B={2,4,6,8}\displaystyle \mathrm{U}=\{1,2,3,4,5,6,7,8,9\}, \mathrm{A}=\{1,2,3,4\}, \mathrm{B}=\{2,4,6,8\} and C={3,4,5,6}\displaystyle \mathrm{C}=\{3,4,5,6\}. Find
    (i)
    A′\displaystyle \mathrm{A}^{\prime}
    (ii)
    B′\displaystyle \mathrm{B}^{\prime}
    (iii)
    (A∪C)′\displaystyle (\mathrm{A} \cup \mathrm{C})^{\prime}
    (iv)
    (A∪B)′\displaystyle (\mathrm{A} \cup \mathrm{B})^{\prime}
    (v)
    (A′)′\displaystyle \left(\mathrm{A}^{\prime}\right)^{\prime}
    (vi)
    (B - C)'

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    Complement. \(\displaystyle \mathrm{A}' = \mathrm{U} - \mathrm{A} \) is the set of elements of the universal set that are not in \(\displaystyle \mathrm{A} \). Here
    \[\mathrm{U} = \{1,2,3,4,5,6,7,8,9\},\ \mathrm{A} = \{1,2,3,4\},\ \mathrm{B} = \{2,4,6,8\},\ \mathrm{C} = \{3,4,5,6\}. \]
    (i)
    Delete \(\displaystyle 1,2,3,4 \) from U:
    \[\mathrm{A}' = \{5,6,7,8,9\} \]
    (ii)
    Delete \(\displaystyle 2,4,6,8 \) from U:
    \[\mathrm{B}' = \{1,3,5,7,9\} \]
    (iii)
    First \(\displaystyle \mathrm{A} \cup \mathrm{C} = \{1,2,3,4,5,6\} \), so
    \[(\mathrm{A} \cup \mathrm{C})' = \{7,8,9\} \]
    (iv)
    First \(\displaystyle \mathrm{A} \cup \mathrm{B} = \{1,2,3,4,6,8\} \), so
    \[(\mathrm{A} \cup \mathrm{B})' = \{5,7,9\} \]
    (v)
    \(\displaystyle \mathrm{A}' = \{5,6,7,8,9\} \), and taking its complement returns the original set (the law \(\displaystyle (\mathrm{A}')' = \mathrm{A} \)):
    \[(\mathrm{A}')' = \{1,2,3,4\} = \mathrm{A} \]
    (vi)
    \(\displaystyle \mathrm{B} - \mathrm{C} \) keeps the elements of B not in C: \(\displaystyle \mathrm{B} - \mathrm{C} = \{2,8\} \). Hence
    \[(\mathrm{B} - \mathrm{C})' = \{1,3,4,5,6,7,9\} \]
    (i) {$\displaystyle 5,6,7,8,9$} (ii) {$\displaystyle 1,3,5,7,9$} (iii) {$\displaystyle 7,8,9$} (iv) {$\displaystyle 5,7,9$} (v) {$\displaystyle 1,2,3,4$} (vi) {$\displaystyle 1,3,4,5,6,7,9$}

  2. Exercise 2

    If U={a,b,c,d,e,f,g,h}\displaystyle \mathrm{U}=\{a, b, c, d, e, f, g, h\}, find the complements of the following sets :
    (i)
    A={a,b,c}\displaystyle \mathrm{A}=\{a, b, c\}
    (ii)
    B={d,e,f,g}\displaystyle \mathrm{B}=\{d, e, f, g\}
    (iii)
    C={a,c,e,g}\displaystyle \mathrm{C}=\{a, c, e, g\}
    (iv)
    D={f,g,h,a}\displaystyle \mathrm{D}=\{f, g, h, a\}

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    Complement. With \(\displaystyle \mathrm{U} = \{a,b,c,d,e,f,g,h\} \), the complement of a set is what is left of U after deleting that set's elements.
    (i)
    \(\displaystyle \mathrm{A} = \{a,b,c\} \): delete \(\displaystyle a,b,c \) from U
    \[\mathrm{A}' = \{d,e,f,g,h\} \]
    (ii)
    \(\displaystyle \mathrm{B} = \{d,e,f,g\} \): delete \(\displaystyle d,e,f,g \)
    \[\mathrm{B}' = \{a,b,c,h\} \]
    (iii)
    \(\displaystyle \mathrm{C} = \{a,c,e,g\} \): delete \(\displaystyle a,c,e,g \)
    \[\mathrm{C}' = \{b,d,f,h\} \]
    (iv)
    \(\displaystyle \mathrm{D} = \{f,g,h,a\} \): delete \(\displaystyle a,f,g,h \)
    \[\mathrm{D}' = \{b,c,d,e\} \]
    (i) {d, e, f, g, h} (ii) {a, b, c, h} (iii) {b, d, f, h} (iv) {b, c, d, e}

  3. Exercise 3

    Taking the set of natural numbers as the universal set, write down the complements of the following sets:
    (i)
    {x:x\displaystyle \{x: x is an even natural number }\displaystyle \}
    (ii)
    {x:x\displaystyle \{x: x is an odd natural number }\displaystyle \}
    (iii)
    {x:x\displaystyle \{x: x is a positive multiple of 3\displaystyle 3}\displaystyle \}
    (iv)
    {x:x\displaystyle \{x: x is a prime number }\displaystyle \}
    (v)
    {x:x\displaystyle \{x: x is a natural number divisible by 3\displaystyle 3 and 5\displaystyle 5}\displaystyle \}
    (vi)
    {x:x\displaystyle \{x: x is a perfect square }\displaystyle \}
    (vii)
    {x:x\displaystyle \{x: x is a perfect cube }\displaystyle \}
    (viii)
    {x:x+5=8}\displaystyle \{x: x+5=8\}
    (ix)
    {x:2x+5=9}\displaystyle \{x: 2 x+5=9\}
    (x)
    {x:x≥7}\displaystyle \{x: x \geq 7\}
    (xi)
    {x:x∈N\displaystyle \{x: x \in N and 2x+1>10}\displaystyle 2 x+1>10\}

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    NCERT’s answer
    (i)
    $\displaystyle \{x: x$ is an odd natural number $\displaystyle \}$ (ii) $\displaystyle \{x: x$ is an even natural number $\displaystyle \}$ (iii) $\displaystyle \{x: x \in \mathbf{N}$ and $\displaystyle x$ is not a multiple of $\displaystyle 3 \}$ (iv) $\displaystyle \{x: x$ is a positive composite number or $\displaystyle x=1\}$ (v) $\displaystyle \{x: x$ is a positive integer which is not divisible by $\displaystyle 3$ or not divisible by $\displaystyle 5 \}$ (vi) $\displaystyle \{x: x \in \mathbf{N}$ and $\displaystyle x$ is not a perfect square $\displaystyle \}$ (vii) $\displaystyle \{x: x \in \mathbf{N}$ and $\displaystyle x$ is not a perfect cube $\displaystyle \}$ (viii) $\displaystyle \{x: x \in \mathbf{N}$ and $\displaystyle x \neq 3\}$ (ix) $\displaystyle \{x: x \in \mathbf{N}$ and $\displaystyle x \neq 2\}$ (x) $\displaystyle \{x: x \in \mathbf{N}$ and $\displaystyle x<7\}$ (xi) $\displaystyle \left\{x: x \in \mathbf{N}\right.$ and $\displaystyle \left.x \leq \frac{9}{2}\right\}$
    Complement in N. With \(\displaystyle \mathrm{U} = \mathbf{N} \), the complement of \(\displaystyle \mathrm{A} \) is \(\displaystyle \mathrm{A}' = \{x \in \mathbf{N} : x \notin \mathrm{A}\} \) — negate the defining property, keeping \(\displaystyle x \) a natural number.
    (i)
    A natural number that is not even is odd:
    \[\{x : x \text{ is an odd natural number}\} \]
    (ii)
    Likewise, not odd means even:
    \[\{x : x \text{ is an even natural number}\} \]
    (iii)
    In \(\displaystyle \mathbf{N} \) every multiple of \(\displaystyle 3 \) is positive, so the complement is
    \[\{x : x \in \mathbf{N} \text{ and } x \text{ is not a multiple of } 3\} \]
    (iv)
    The naturals that are not prime are \(\displaystyle 1 \) together with the composite numbers:
    \[\{x : x \in \mathbf{N},\ x \text{ is not a prime}\} = \{1\} \cup \{x : x \text{ is a composite number}\} \]
    (v)
    A number divisible by both \(\displaystyle 3 \) and \(\displaystyle 5 \) is divisible by \(\displaystyle 15 \), so the set is \(\displaystyle \{15,30,45,\ldots\} \) and
    \[\{x : x \in \mathbf{N} \text{ and } x \text{ is not divisible by } 15\} \]
    (vi)
    \(\displaystyle \{x : x \in \mathbf{N} \text{ and } x \text{ is not a perfect square}\} \)
    (vii)
    \(\displaystyle \{x : x \in \mathbf{N} \text{ and } x \text{ is not a perfect cube}\} \)
    (viii)
    \(\displaystyle x + 5 = 8 \Rightarrow x = 3 \), so the set is \(\displaystyle \{3\} \) and
    \[\{x : x \in \mathbf{N} \text{ and } x \neq 3\} \]
    (ix)
    \(\displaystyle 2x + 5 = 9 \Rightarrow 2x = 4 \Rightarrow x = 2 \), so the set is \(\displaystyle \{2\} \) and
    \[\{x : x \in \mathbf{N} \text{ and } x \neq 2\} \]
    (x)
    The set is \(\displaystyle \{7,8,9,\ldots\} \), so its complement is
    \[\{x : x \in \mathbf{N} \text{ and } x < 7\} = \{1,2,3,4,5,6\} \]
    (xi)
    \(\displaystyle 2x + 1 > 10 \Rightarrow 2x > 9 \Rightarrow x > 4.5 \), so the set is \(\displaystyle \{5,6,7,\ldots\} \) and its complement is
    \[\{x : x \in \mathbf{N} \text{ and } x \leq 4\} = \{1,2,3,4\} \]
    (i) odd naturals (ii) even naturals (iii) naturals not multiples of $\displaystyle 3$ (iv) $\displaystyle 1$ and the composite numbers (v) naturals not divisible by $\displaystyle 15$ (vi) naturals that are not perfect squares (vii) naturals that are not perfect cubes (viii) {x ∈ N : x ≠ $\displaystyle 3$} (ix) {x ∈ N : x ≠ $\displaystyle 2$} (x) {$\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$, $\displaystyle 5$, $\displaystyle 6$} (xi) {$\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$}

  4. Exercise 4

    If U={1,2,3,4,5,6,7,8,9},A={2,4,6,8}\displaystyle \mathrm{U}=\{1,2,3,4,5,6,7,8,9\}, \mathrm{A}=\{2,4,6,8\} and B={2,3,5,7}\displaystyle \mathrm{B}=\{2,3,5,7\}. Verify that
    (i)
    (A∪B)′=A′∩B′\displaystyle (\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}
    (ii)
    (A∩B)′=A′∪B′\displaystyle (\mathrm{A} \cap \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}

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    De Morgan's laws, verified by listing. With \(\displaystyle \mathrm{U} = \{1,2,3,4,5,6,7,8,9\} \), \(\displaystyle \mathrm{A} = \{2,4,6,8\} \) and \(\displaystyle \mathrm{B} = \{2,3,5,7\} \), first record the two complements:
    \[\mathrm{A}' = \{1,3,5,7,9\},\qquad \mathrm{B}' = \{1,4,6,8,9\} \]
    (i)
    Left side: \(\displaystyle \mathrm{A} \cup \mathrm{B} = \{2,3,4,5,6,7,8\} \), so
    \[(\mathrm{A} \cup \mathrm{B})' = \{1,9\} \]
    Right side: the elements common to \(\displaystyle \mathrm{A}' \) and \(\displaystyle \mathrm{B}' \) are \(\displaystyle 1 \) and \(\displaystyle 9 \), so
    \[\mathrm{A}' \cap \mathrm{B}' = \{1,9\} \]
    Both sides are \(\displaystyle \{1,9\} \), so \(\displaystyle (\mathrm{A} \cup \mathrm{B})' = \mathrm{A}' \cap \mathrm{B}' \). Verified.
    (ii)
    Left side: the only element in both A and B is \(\displaystyle 2 \), so \(\displaystyle \mathrm{A} \cap \mathrm{B} = \{2\} \) and
    \[(\mathrm{A} \cap \mathrm{B})' = \{1,3,4,5,6,7,8,9\} \]
    Right side:
    \[\mathrm{A}' \cup \mathrm{B}' = \{1,3,5,7,9\} \cup \{1,4,6,8,9\} = \{1,3,4,5,6,7,8,9\} \]
    Both sides agree, so \(\displaystyle (\mathrm{A} \cap \mathrm{B})' = \mathrm{A}' \cup \mathrm{B}' \). Verified.
    (i) (A ∪ B)′ = A′ ∩ B′ = {$\displaystyle 1$, $\displaystyle 9$} (ii) (A ∩ B)′ = A′ ∪ B′ = {$\displaystyle 1$, $\displaystyle 3$, $\displaystyle 4$, $\displaystyle 5$, $\displaystyle 6$, $\displaystyle 7$, $\displaystyle 8$, $\displaystyle 9$}

  5. Exercise 5

    Draw appropriate Venn diagram for each of the following :
    (i)
    (A∪B)′\displaystyle (\mathrm{A} \cup \mathrm{B})^{\prime},
    (ii)
    A′∩B′\displaystyle \mathrm{A}^{\prime} \cap \mathrm{B}^{\prime},
    (iii)
    (A∩B)′\displaystyle (\mathrm{A} \cap \mathrm{B})^{\prime},
    (iv)
    A′∪B′\displaystyle \mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}

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    Shading a Venn diagram. For each part draw the same picture: a rectangle for the universal set \(\displaystyle \mathrm{U}\) with two overlapping circles \(\displaystyle \mathrm{A}\) and \(\displaystyle \mathrm{B}\) inside it. Those circles cut the rectangle into exactly four regions — only \(\displaystyle \mathrm{A}\), only \(\displaystyle \mathrm{B}\), the lens-shaped overlap \(\displaystyle \mathrm{A}\cap\mathrm{B}\), and the part of the rectangle lying outside both circles. Deciding which of these four regions to shade is the whole question.(i) \(\displaystyle (\mathrm{A}\cup\mathrm{B})'\). \(\displaystyle \mathrm{A}\cup\mathrm{B}\) is the two circles taken together, so its complement is everything in \(\displaystyle \mathrm{U}\) that is left over: shade only the part of the rectangle outside both circles.(ii) \(\displaystyle \mathrm{A}'\cap\mathrm{B}'\). A point lies here when it is outside \(\displaystyle \mathrm{A}\) and outside \(\displaystyle \mathrm{B}\) — that is again the part of the rectangle outside both circles. So the shading is identical to (i); the picture is exactly De Morgan's law \(\displaystyle (\mathrm{A}\cup\mathrm{B})'=\mathrm{A}'\cap\mathrm{B}'\).(iii) \(\displaystyle (\mathrm{A}\cap\mathrm{B})'\). \(\displaystyle \mathrm{A}\cap\mathrm{B}\) is only the lens-shaped overlap, so its complement is everything else: shade the whole rectangle except the overlap — that is, only-\(\displaystyle \mathrm{A}\), only-\(\displaystyle \mathrm{B}\) and the region outside both circles.(iv) \(\displaystyle \mathrm{A}'\cup\mathrm{B}'\). A point lies here when it is outside \(\displaystyle \mathrm{A}\) or outside \(\displaystyle \mathrm{B}\); the only points excluded are those lying in both circles at once, i.e. the overlap. So the shading is identical to (iii), which is De Morgan's other law \(\displaystyle (\mathrm{A}\cap\mathrm{B})'=\mathrm{A}'\cup\mathrm{B}'\).Sketch the rectangle-with-two-circles once and shade a fresh copy of it four times — the four regions named above are what the answer refers to.(i) and (ii) both shade the region of \(\displaystyle \mathrm{U}\) outside both circles; (iii) and (iv) both shade all of \(\displaystyle \mathrm{U}\) except the overlap \(\displaystyle \mathrm{A}\cap\mathrm{B}\).

  6. Exercise 6

    Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60\displaystyle 60°, what is A'?

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    NCERT’s answer
    A' is the set of all equilateral triangles.
    Complement inside the stated universal set. By definition \(\displaystyle \mathrm{A}'=\{x\in \mathrm{U}: x\notin \mathrm{A}\}\), so we must describe the triangles that are not in \(\displaystyle \mathrm{A}\).A triangle belongs to \(\displaystyle \mathrm{A}\) when at least one of its angles is different from \(\displaystyle 60^{\circ}\). Negating that: a triangle fails to belong to \(\displaystyle \mathrm{A}\) exactly when no angle of it is different from \(\displaystyle 60^{\circ}\), i.e. when all three of its angles equal \(\displaystyle 60^{\circ}\).This is consistent, since \(\displaystyle 60^{\circ}+60^{\circ}+60^{\circ}=180^{\circ}\), and a triangle is equiangular if and only if it is equilateral.A′ = the set of all equilateral triangles in the plane (equivalently, all triangles each of whose angles is \(\displaystyle 60^{\circ}\)).

  7. Exercise 7

    Fill in the blanks to make each of the following a true statement :
    (i)
    A∪A′=…\displaystyle \mathrm{A} \cup \mathrm{A}^{\prime}=\ldots.
    (ii)
    ϕ′∩A=…\displaystyle \phi^{\prime} \cap \mathrm{A}=\ldots
    (iii)
    A∩A′=…\displaystyle \mathrm{A} \cap \mathrm{A}^{\prime}=\ldots.
    (iv)
    U′∩A=…\displaystyle \mathrm{U}^{\prime} \cap \mathrm{A}=\ldots.

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    NCERT’s answer
    (i)
    U (ii) A (iii) $\displaystyle \phi$ (iv) $\displaystyle \phi$
    Complement laws. Recall the four facts being tested: \(\displaystyle \mathrm{A}\cup \mathrm{A}'=\mathrm{U}\), \(\displaystyle \mathrm{A}\cap \mathrm{A}'=\varphi\), \(\displaystyle \varphi'=\mathrm{U}\) and \(\displaystyle \mathrm{U}'=\varphi\).(i) Every element of \(\displaystyle \mathrm{U}\) either belongs to \(\displaystyle \mathrm{A}\) or does not, so \(\displaystyle \mathrm{A}\cup \mathrm{A}'=\mathrm{U}\).(ii) The complement of the empty set is the whole universal set, \(\displaystyle \varphi'=\mathrm{U}\); and \(\displaystyle \mathrm{U}\cap \mathrm{A}=\mathrm{A}\) because \(\displaystyle \mathrm{A}\subset \mathrm{U}\). So \(\displaystyle \varphi'\cap \mathrm{A}=\mathrm{A}\).(iii) No element can both belong to \(\displaystyle \mathrm{A}\) and not belong to \(\displaystyle \mathrm{A}\), so \(\displaystyle \mathrm{A}\cap \mathrm{A}'=\varphi\).(iv) The complement of the universal set is empty, \(\displaystyle \mathrm{U}'=\varphi\); and \(\displaystyle \varphi\cap \mathrm{A}=\varphi\). So \(\displaystyle \mathrm{U}'\cap \mathrm{A}=\varphi\).(i) \(\displaystyle \mathrm{U}\) (ii) \(\displaystyle \mathrm{A}\) (iii) \(\displaystyle \varphi\) (iv) \(\displaystyle \varphi\)