# Check if equation defines an operation on the set of integers.

#### beesee

Hi,

I have the following exercise, it is required to check if
the following equation defines an operation on the set of integers, and if so, check if it is associative, commutative and identity element.

$m * n = mn + 1$

I have taken $m = 2, n = 3$
$2*3 = (2 \cdot 3) + 1 = 6 + 1 = 7 \in \mathbb{Z}$
so that equation defines a valid operation.

now I check if it is associative, commutative, identity:

i) associative.
if $m=2, n=3, o=4$
I have to prove that
$(m*n)*o = m*(n*o)$

checking the left side:
\begin{align}(2*3)*4 &= ((2 \cdot 3) + 1)*4 \\ &= 7 * 4 \\ &= (7 \cdot 4) + 1 \\ &= 28 + 1 \\ &= 29\end{align}
checking the right side:
\begin{align}2 * (3*4) &= 2 * ((3 \cdot 4) + 1) \\ &= 2* (12+1) \\ &= 2 * 13 \\ &= (2 \cdot 13) + 1 \\ &= 26 + 1 \\ &= 27 \end{align}

therefore it is not associative because:
$(m*n)*o \ne m*(n*o)$

ii) commutative:
if $m = -2, n = 7$
we have to prove if:
$m*n = n * m$
checking the left side:
\begin{align} (-2)*7 &= (-2 \cdot 7) + 1 \\ &= -14 + 1 \\ &= - 13 \end{align}
checking the right side:
\begin{align} 7 * (-2) &= (7 \cdot -2) + 1 \\ &= -14 + 1 \\ &= -13\end{align}

therefore it is commutative because:
$m*n = n*m$

iii) identity element
I cite the following:
0 is an identity element for addition of integers, and 1 is an identity element for multiplication of integers.
so I have taken $e = 1$ because it is considered inside a product in $\mathbb{Z}$.

if $m = 2$,
we have to prove if:
$m*e = e * m = m$
checking the left side:
\begin{align} m * e &= (m \cdot e) + 1 \\ &= (m \cdot 1) + 1 \\ &= (2 \cdot 1) +1 \\ &= 2+1 \\ &= 3 \end{align}
checking the right side:
\begin{align} e * m &= (e \cdot m) + 1 \\ &= (1 \cdot m) + 1 \\ &= (1 \cdot 2) + 1 \\ &= 2+1 \\ &= 3 \end{align}

I have also tried with $m = -2, m = 0$, and I have obtained that the identity element exists,
but, on my text it is written that there is not identity element. Please, can you tell me where is the error? Many thanks!

#### romsek

Math Team
you can't just pick a set of values to prove the various properties. What holds true for one set of values may not hold true for another.

Is $*$ an operation on the integers?

a) Does it take integer(s) as operands? Yes.

b) Does it produce an integer?

Multiplication and addition are closed under the integers so yes this operation produces an integer and thus it is an operation over the integers.

Checking associativity

$(a*b)*c \overset{?}{=} a*(b*c)$

$(a*b)*c = (ab + 1)*c = (ab+1)c + 1 = abc + c + 1$

$a*(b*c) = a*(bc + 1) = a(bc+1)+1 = abc + a + 1$

$(a*b)*c \neq a*(b*c)$ so $*$ is not associative

Checking commutativity

$a*b \overset{?}{=} b*a$

$ab + 1 = ba + 1$ so yes $*$ is commutative

For the identity

$a*I=a$

$aI+1 = a$

$I = \dfrac{a-1}{a} = 1 - \dfrac 1 a \not \in \mathbb{Z},~a\neq 1$

so in general this operation has no identity element