TEST BANK FOR Mathematics For Machine Learning By C. S. (Cheng Soon) Ong, M. P. (Marc Peter)

2.1 We consider (Rnf􀀀1g; ?), where
a ? b := ab + a + b; a; b 2 Rnf􀀀1g (2.1)
a. Show that (Rnf􀀀1g; ?) is an Abelian group.
b. Solve
3 ? x ? x = 15
in the Abelian group (Rnf􀀀1g; ?), where ? is defined in (2.1).
a. First, we show that Rnf􀀀1g is closed under ?: For all a; b 2 Rnf􀀀1g:
a ? b = ab + a + b + 1 􀀀 1 = (a + 1) | {z }
6=0
(b + 1) | {z }
6=0
􀀀1 6= 􀀀1
) a ? b 2 Rnf􀀀1g
Next, we show the group axioms
Associativity: For all a; b; c 2 Rnf􀀀1g:
(a ? b) ? c = (ab + a + b) ? c
= (ab + a + b)c + (ab + a + b) + c
= abc + ac + bc + ab + a + b + c
= a(bc + b + c) + a + (bc + b + c)
= a ? (bc + b + c)
= a ? (b ? c)
Commutativity:
8a; b 2 Rnf􀀀1g : a ? b = ab + a + b = ba + b + a = b ? a
Neutral Element: n = 0 is the neutral element since
8a 2 Rnf􀀀1g : a ? 0 = a = 0 ? a
Inverse Element: We need to find a, such that a ? a = 0 = a ? a.
a ? a = 0 () aa + a + a = 0
() a(a + 1) = 􀀀a
a6=􀀀1
() a = 􀀀
a
a + 1
= 􀀀1 +
1
a + 1
6= 􀀀1 2 Rnf􀀀1g
468
This material will be published by Cambridge University Press as Mathematics for Machine Learning
by Marc Peter Deisenroth, A. Aldo Faisal, and Cheng Soon Ong. This pre-publication version is
free to view and download for personal use only. Not for re-distribution, re-sale or use in derivative
works. c by M. P. Deisenroth, A. A. Faisal, and C. S. Ong, 2020. https://mml-book.com.
Exercises 469
b.
3 ? x ? x = 15 () 3 ? (x2 + x + x) = 15
() 3x2 + 6x + 3 + x2 + 2x = 15
() 4x2 + 8x 􀀀 12 = 0
() (x 􀀀 1)(x + 3) = 0
() x 2 f􀀀3; 1g
2.2 Let n be in Nnf0g. Let k; x be in Z. We define the congruence class k of the
integer k as the set
k = fx 2 Z j x 􀀀 k = 0 (modn)g
= fx 2 Z j 9a 2 Z: (x 􀀀 k = n a)g :
We now define Z=nZ (sometimes written Zn) as the set of all congruence
classes modulo n. Euclidean division implies that this set is a finite set containing
n elements:
Zn = f0; 1; : : : ; n 􀀀 1g
For all a; b 2 Zn, we define
a b := a + b
a. Show that (Zn;) is a group. Is it Abelian?
b. We now define another operation 
 for all a and b in Zn as
a 
 b = a b ; (2.2)
where a b represents the usual multiplication in Z.
Let n = 5. Draw the times table of the elements of Z5nf0g under 
, i.e.,
calculate the products a 
 b for all a and b in Z5nf0g.
Hence, show that Z5nf0g is closed under 
 and possesses a neutral
element for 
. Display the inverse of all elements in Z5nf0g under 
.
Conclude that (Z5nf0g;
) is an Abelian group.
c. Show that (Z8nf0g;
) is not a group.
d. We recall that the B´ezout theorem states that two integers a and b are
relatively prime (i.e., gcd(a; b) = 1) if and only if there exist two integers
u and v such that au + bv = 1. Show that (Znnf0g;
) is a group if and
only if n 2 Nnf0g is prime.
a. We show that the group axioms are satisfied:
Closure: Let a; b be in Zn. We have:
a b = a + b
= (a + b) mod n
by definition of the congruence class, and since [(a + b) mod n] 2
f0; : : : ; n 􀀀1g, it follows that ab 2 Zn. Thus, Zn is closed under .