Let * be the binary operation on N given by a * b = LCM of a and b. Find |
(1) 5 * 7, 20 * 16. |
(ii) Is * commutative? |
(iii) Is * associative? |
(iv) Find the identity of * in N. |
(v) Which elements of N are invertible for the operation *? |
OR |
Let N denote the set of all natural numbers and R be the relation on \[N\times N\] defined by d(a, b) R (c, d), if \[ad(b+c)\text{ }=\text{ }bc(a+b).\] Show that R is an equivalence relation. |
Answer:
The binary operation * on N is defined as a * b = LCM of a and b. (i) Consider, 5 * 7= LCM of 5 and 7 = 35 and 20 * 16= LCM of 20 and 16 = 80 (ii) It is known that, LCM of a and b = LCM of b and \[a,\,\,\forall \,\,a,\,\,b\in N.\] Thus, the operation '*? is commutative. (iii) For \[a,\,\,b,\,\,c\in N,\] we have (a * b)* c = (LCM of a and b)* c = LCM of a b and c a * (b * c) = a * (LCM of b and c) = LCM of a b and c Therefore, \[(a*b)*c=a\,\,*(b\,\,*c).\] \[\therefore \] \[a*b=b\,\,*a.\] Thus the operation \[*\]is associative. (iv) We know that, LCM of a and 1 = a = LCM of 1 and \[a,\,\,\forall \,\,a\in N,\] i.e. \[a*1=a=1*a,\] \[\forall a\in N\] Thus, 1 is the identity of * in N. (v) An element a in N is invertible with respect to the operation *, if there exists an element b in N such that a * b = e = b * a. Here, e = 1. This means that, LCM of a and b = 1 = LCM of b and a This case is possible only when a and b are equal to 1. Thus, 1 is the only invertible element of N with respect to the operation*. OR We observe the following properties of relation R. Reflexivity Let (a, b) be an arbitrary element of \[N\times N\] Then,\[(a,b)\in N\times N\Rightarrow a,\,\,b\in N\Rightarrow ab(b+a)=ba(a+b)\] [By commutativity of addition and multiplication on N] \[\Rightarrow \] \[(a,\,\,b)R(a,\,\,b)\] Thus, (a, b)R(a, b)for all \[(a,\,\,b)\in N\,\times N.\] So, R is reflexive on \[N\times N.\] Symmetry Let \[(a,\,\,b),(c,\,\,d)\in N\,\,\times N\] be such that \[(a,\,\,b)R(c,\,\,d).\] Then, (a, b)R(c, d) \[\Rightarrow \] \[ad(b+c)=bc(a+b)\] \[\Rightarrow \] \[cb(d+a)=da(c+b)\] [By commutativity of addition and multiplication on N] \[\Rightarrow \] (c, d)R(a, b) Thus, (a, b)R(c, d) \[\Rightarrow \] (c, d)R(a, b) for all(a, b) \[(c,\,\,d)\in N\times N.\] So, R is Symmetry on \[N\,\times N.\] Transitivity Let(a, b) (c, d)\[(e,\,\,f)\in N\times N\] such that (a, b)R(c, d)and (c, d)R(e, f). Then, \[(a,\,\,b)R(c,\,\,d)\Rightarrow ad(b+c)=bc(a+d)\] \[\Rightarrow \frac{b+c}{bc}=\frac{a+d}{ad}\] \[\frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}\] ?(i) and \[(c,\,\,d)R(e,\,\,f)\Rightarrow cf(d+e)=de(c+f)\] \[\Rightarrow \] \[\frac{d+e}{de}=\frac{c+f}{cf}\] \[\Rightarrow \] \[\frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f}\] ?(ii) On addition Eqs. (i) and (ii), we get \[\left( \frac{1}{b}+\frac{1}{c} \right)+\left( \frac{1}{d}+\frac{1}{e} \right)=\left( \frac{1}{a}+\frac{1}{d} \right)+\left( \frac{1}{c}+\frac{1}{f} \right)\] \[\Rightarrow \] \[\frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}\] \[\Rightarrow \] \[\frac{b+e}{be}=\frac{a+f}{af}\] \[\Rightarrow \] \[af(b+e)=be(a+f)\] \[\Rightarrow \] \[(a,\,\,b)R(e,\,\,f)\] Thus, (a, b)R(c, d) and (c, d)R(e, f) \[\Rightarrow \] \[(a,\,\,b)R(e,\,\,f)\in N\,\,\times N\] So, R is transitive on \[N\,\,\times N.\] Hence, R is an equivalence relation \[N\,\,\times N.\] The binary operation * on N is defined as a * b = LCM of a and b. (i) Consider, 5 * 7= LCM of 5 and 7 = 35 and 20 * 16= LCM of 20 and 16 = 80 (ii) It is known that, LCM of a and b = LCM of b and \[a,\,\,\forall \,\,a,\,\,b\in N.\] Thus, the operation '*? is commutative. (iii) For \[a,\,\,b,\,\,c\in N,\] we have (a * b)* c = (LCM of a and b)* c = LCM of a b and c a * (b * c) = a * (LCM of b and c) = LCM of a b and c Therefore, \[(a*b)*c=a\,\,*(b\,\,*c).\] \[\therefore \] \[a*b=b\,\,*a.\] Thus the operation \[*\]is associative. (iv) We know that, LCM of a and 1 = a = LCM of 1 and \[a,\,\,\forall \,\,a\in N,\] i.e. \[a*1=a=1*a,\] \[\forall a\in N\] Thus, 1 is the identity of * in N. (v) An element a in N is invertible with respect to the operation *, if there exists an element b in N such that a * b = e = b * a. Here, e = 1. This means that, LCM of a and b = 1 = LCM of b and a This case is possible only when a and b are equal to 1. Thus, 1 is the only invertible element of N with respect to the operation*. OR We observe the following properties of relation R. Reflexivity Let (a, b) be an arbitrary element of \[N\times N\] Then,\[(a,b)\in N\times N\Rightarrow a,\,\,b\in N\Rightarrow ab(b+a)=ba(a+b)\] [By commutativity of addition and multiplication on N] \[\Rightarrow \] \[(a,\,\,b)R(a,\,\,b)\] Thus, (a, b)R(a, b)for all \[(a,\,\,b)\in N\,\times N.\] So, R is reflexive on \[N\times N.\] Symmetry Let \[(a,\,\,b),(c,\,\,d)\in N\,\,\times N\] be such that \[(a,\,\,b)R(c,\,\,d).\] Then, (a, b)R(c, d) \[\Rightarrow \] \[ad(b+c)=bc(a+b)\] \[\Rightarrow \] \[cb(d+a)=da(c+b)\] [By commutativity of addition and multiplication on N] \[\Rightarrow \] (c, d)R(a, b) Thus, (a, b)R(c, d) \[\Rightarrow \] (c, d)R(a, b) for all(a, b) \[(c,\,\,d)\in N\times N.\] So, R is Symmetry on \[N\,\times N.\] Transitivity Let(a, b) (c, d)\[(e,\,\,f)\in N\times N\] such that (a, b)R(c, d)and (c, d)R(e, f). Then, \[(a,\,\,b)R(c,\,\,d)\Rightarrow ad(b+c)=bc(a+d)\] \[\Rightarrow \frac{b+c}{bc}=\frac{a+d}{ad}\] x\[\frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}\] ?(i) and \[(c,\,\,d)R(e,\,\,f)\Rightarrow cf(d+e)=de(c+f)\] \[\Rightarrow \] \[\frac{d+e}{de}=\frac{c+f}{cf}\] \[\Rightarrow \] \[\frac{1}{d}+\frac{1}{e}=\frac{1}{c}+\frac{1}{f}\] ?(ii) On addition Eqs. (i) and (ii), we get \[\left( \frac{1}{b}+\frac{1}{c} \right)+\left( \frac{1}{d}+\frac{1}{e} \right)=\left( \frac{1}{a}+\frac{1}{d} \right)+\left( \frac{1}{c}+\frac{1}{f} \right)\] \[\Rightarrow \] \[\frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}\] \[\Rightarrow \] \[\frac{b+e}{be}=\frac{a+f}{af}\] \[\Rightarrow \] \[af(b+e)=be(a+f)\] \[\Rightarrow \] \[(a,\,\,b)R(e,\,\,f)\] Thus, (a, b)R(c, d) and (c, d)R(e, f) \[\Rightarrow \] \[(a,\,\,b)R(e,\,\,f)\in N\,\,\times N\] So, R is transitive on \[N\,\,\times N.\] Hence, R is an equivalence relation \[N\,\,\times N.\]
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