Consider a binary operation - on set 1-2-3-4-5 given by the following multiplication table-iCompute 2 - 3 - 4 and 2 -3 - 4 -begin array - c-c-c-c-c-c- Whinelast - 1 - 2 - 3 - 4 - 5 illhline - 1 - 1 - 1 - 1 - 111 hline2 - 1 - 2 - 1 - 2 - 111 hline3 - 1 - 1 - 3 - 1 - 111 hline4 - 1 - 2 - 1 - 4 - 1 hline5 - 1 - 1 - 1 - 1 - 511 hline e n d array ii Is - commutative- -begin array - c-c-c-c-c-c- Whinelast - 1 - 2 - 3 - 4 - 5 lllhline1 - 1 - 1 - 1 - 1 - 1 Whline2 - 1 - 2 - 1 - 2 - 1 hline3 - 1 - 1 - 3 - 1 - 111 hline4 - 1 - 2 - 1 - 4 - 1 Whline 5 - 1 - 1 - 1 - 1 - 511 hline e n d a r r a y iiiCompute 2 - 3 -4ast5 -beginarray -c-c-c-c-c-c-hline1 - 1 - 1 - 1 - 1 - 1 illhline2 - 1 - 2 - 1 - 2 - 111 hline

We have (2 ∗ 3) ∗ 4 =(1) ∗ 4 =1 and 2 ∗ (3 ∗ 4) =2 ∗ 1 =1 For every a,b ∈ 1,2,3,4,5,we have a ∗ b =b ∗ a. Therefore, the operation ∗ is commutative We have (2 ∗ ...

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Byju's Answer

Standard XII

Mathematics

Permutation: n Different Things Taken All at a Time When All Are Not Different.

Consider a bi...

Question

Consider a binary operation $$ on set{1,2,3,4,5} given by the following multiplication table:
(i)Compute $(2 3) * 4$ and $2 * (3 * 4)$

$\begin{matrix} * & 1 & 2 & 3 & 4 & 5 1 & 1 & 1 & 1 & 1 & 1 2 & 1 & 2 & 1 & 2 & 1 3 & 1 & 1 & 3 & 1 & 1 4 & 1 & 2 & 1 & 4 & 1 5 & 1 & 1 & 1 & 1 & 5 \end{matrix}$

(ii) Is $$ commutative?

$\begin{matrix} & 1 & 2 & 3 & 4 & 5 1 & 1 & 1 & 1 & 1 & 1 2 & 1 & 2 & 1 & 2 & 1 3 & 1 & 1 & 3 & 1 & 1 4 & 1 & 2 & 1 & 4 & 1 5 & 1 & 1 & 1 & 1 & 5 \end{matrix}$

(iii)Compute $(2 * 3) \times (4 a s t 5)$
$\begin{matrix} * & 1 & 2 & 3 & 4 & 5 1 & 1 & 1 & 1 & 1 & 1 2 & 1 & 2 & 1 & 2 & 1 3 & 1 & 1 & 3 & 1 & 1 4 & 1 & 2 & 1 & 4 & 1 5 & 1 & 1 & 1 & 1 & 5 \end{matrix}$

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Solution

We have $(2 * 3) * 4 = (1) * 4 = 1$
and $2 * (3 * 4) = 2 * 1 = 1$

For every $a, b \in$ {1,2,3,4,5},we have $a * b = b * a$ .
Therefore, the operation $$ is commutative

We have $(2 3) = 1$ and $(4 * 5) = 1$
Therefore, $(2 * 3) * (4 * 5) = 1 * 1 = 1$

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Similar questions

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Consider a binary operation $*$ on set{1,2,3,4,5} given by the following multiplication table:
(iii)Compute $(2 * 3) \times (4 * 5)$
$\begin{matrix} * & 1 & 2 & 3 & 4 & 5 1 & 1 & 1 & 1 & 1 & 1 2 & 1 & 2 & 1 & 2 & 1 3 & 1 & 1 & 3 & 1 & 1 4 & 1 & 2 & 1 & 4 & 1 5 & 1 & 1 & 1 & 1 & 5 \end{matrix}$

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We have (2∗3)∗4=(1)∗4=1 and 2∗(3∗4)=2∗1=1 For every a,b∈ {1,2,3,4,5},we have a∗b=b∗a. Therefore, the operation ∗ is commutative We have (2∗3)=1 and (4∗5)=1 Therefore, (2∗3)∗(4∗5)=1∗1=1

We have $(2 * 3) * 4 = (1) * 4 = 1$
and $2 * (3 * 4) = 2 * 1 = 1$

For every $a, b \in$ {1,2,3,4,5},we have $a * b = b * a$ .
Therefore, the operation $$ is commutative

We have $(2 3) = 1$ and $(4 * 5) = 1$
Therefore, $(2 * 3) * (4 * 5) = 1 * 1 = 1$