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. Let R be a ring and let x € R. We inductively define¹ multiplication of elements of R by ositive integers: for x ER, we set 1x = x (here 1 = 1z is the multiplicative identity of Z), nd for ne N, we set (n+1)x= nx+x. (a) Prove that (m+n)x= mx +nx for all m, n € N and x € R. (b) Prove that n(x + y) = nx+ny for all n € N and x, y € R. (c) Prove that (mn)x= m(nx) for all m, n € N and x € R. ote that it is not the case that these properties are special cases of the distributive and ssociative properties in the ring R, because m and n are positive integers, and cannot be ssumed to be elements of R, an arbitrary ring. With the notation of the preceding problem, we have 2r= (1+1)r= lr+r=r+r

. Let R be a ring and let x € R. We inductively define¹ multiplication of elements of R by ositive integers: for x ER, we set 1x = x (here 1 = 1z is the multiplicative identity of Z), nd for ne N, we set (n+1)x= nx+x. (a) Prove that (m+n)x= mx +nx for all m, n € N and x € R. (b) Prove that n(x + y) = nx+ny for all n € N and x, y € R. (c) Prove that (mn)x= m(nx) for all m, n € N and x € R. ote that it is not the case that these properties are special cases of the distributive and ssociative properties in the ring R, because m and n are positive integers, and cannot be ssumed to be elements of R, an arbitrary ring. With the notation of the preceding problem, we have 2r= (1+1)r= lr+r=r+r

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter3: Groups
Section3.5: Isomorphisms
Problem 4E: Let G=1,i,1,i under multiplication, and let G=4=[ 0 ],[ 1 ],[ 2 ],[ 3 ] under addition. Find an...
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1. Let R be a ring and let x € R. We inductively define multiplication of elements of R by positive integers: for x ER, we set 1x = x (here 1 = 1z is the multiplicative identity of Z), and for n € N, we set (n+1)x= nx+x. = (a) Prove that (m + n)x = mx + nx for all m, n E N and x € R. (b) Prove that n(x + y) = nx + ny for all n N and x, y € R. (c) Prove that (mn)x= m(nx) for all m, n E N and x € R. Note that it is not the case that these properties are special cases of the distributive and associative properties in the ring R, because m and n are positive integers, and cannot be assumed to be elements of R, an arbitrary ring. With the notation of the preceding problem, we have Similarly, we have 2x = (1+1)x= 1x + x = x+x. 3x = = (2+1)x = 2x + x = (x+x) + x = x+x+x. And so on. This is why, for n € N, the element nx is sometimes described informally as the element of R obtained by adding n summands, all equal to x; but this description is just that: informal. Such a description is not sufficiently rigorous to establish the important properties of the multiplication operation of N on R.
Transcribed Image Text:1. Let R be a ring and let x € R. We inductively define multiplication of elements of R by positive integers: for x ER, we set 1x = x (here 1 = 1z is the multiplicative identity of Z), and for n € N, we set (n+1)x= nx+x. = (a) Prove that (m + n)x = mx + nx for all m, n E N and x € R. (b) Prove that n(x + y) = nx + ny for all n N and x, y € R. (c) Prove that (mn)x= m(nx) for all m, n E N and x € R. Note that it is not the case that these properties are special cases of the distributive and associative properties in the ring R, because m and n are positive integers, and cannot be assumed to be elements of R, an arbitrary ring. With the notation of the preceding problem, we have Similarly, we have 2x = (1+1)x= 1x + x = x+x. 3x = = (2+1)x = 2x + x = (x+x) + x = x+x+x. And so on. This is why, for n € N, the element nx is sometimes described informally as the element of R obtained by adding n summands, all equal to x; but this description is just that: informal. Such a description is not sufficiently rigorous to establish the important properties of the multiplication operation of N on R.
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