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Hint: A binary operation * on a set A is a function from $A\times A\to A$. Hence * should be defined on every element of $A\times A$ and the set A should be closed under * for * to be a binary operation on A.

Complete step-by-step answer:

The set ${{Z}^{+}}$ is defined as ${{Z}^{+}}=\left\{ n:n\in Z,n\ge 0 \right\}$

The operation * defined as a*b = a-b over ${{Z}^{+}}$ does not form a binary operation because ${{Z}^{+}}$ is not closed under ${{Z}^{+}}$.

Consider the case when a = 2 and b = 3.

We have $a\in {{Z}^{+}}$ and $b\in {{Z}^{+}}$.

So $\left( a,b \right)\in {{Z}^{+}}\times {{Z}^{+}}$

But a*b = a - b = 2 - 3 = -1

Since $-1\notin {{Z}^{+}}$, ${{Z}^{+}}$ is not closed under *.

In other words there exists an ordered pair $\left( a,b \right)\in {{Z}^{+}}\times {{Z}^{+}}$ such that $\left( a,b \right)$ is unmapped by *.

So, * is not a function from ${{Z}^{+}}\times {{Z}^{+}}\to {{Z}^{+}}$ and hence * is not a binary operation.

Note:

[1] In order to prove that a statement is incorrect it is sufficient to come up with a counter example. In the question we proved that * is not a binary operation by proving that the claim of * being a binary operation will be incorrect as we have come up with a counter example of a = 2 and b = 3.

[2] A binary operation * is said to be commutative if a*b = b*a

[3] A binary operation * is said to be commutative if a*(b*c) = (a*b)*c

[4] Addition forms a binary operation over the set of Natural numbers.

Complete step-by-step answer:

The set ${{Z}^{+}}$ is defined as ${{Z}^{+}}=\left\{ n:n\in Z,n\ge 0 \right\}$

The operation * defined as a*b = a-b over ${{Z}^{+}}$ does not form a binary operation because ${{Z}^{+}}$ is not closed under ${{Z}^{+}}$.

Consider the case when a = 2 and b = 3.

We have $a\in {{Z}^{+}}$ and $b\in {{Z}^{+}}$.

So $\left( a,b \right)\in {{Z}^{+}}\times {{Z}^{+}}$

But a*b = a - b = 2 - 3 = -1

Since $-1\notin {{Z}^{+}}$, ${{Z}^{+}}$ is not closed under *.

In other words there exists an ordered pair $\left( a,b \right)\in {{Z}^{+}}\times {{Z}^{+}}$ such that $\left( a,b \right)$ is unmapped by *.

So, * is not a function from ${{Z}^{+}}\times {{Z}^{+}}\to {{Z}^{+}}$ and hence * is not a binary operation.

Note:

[1] In order to prove that a statement is incorrect it is sufficient to come up with a counter example. In the question we proved that * is not a binary operation by proving that the claim of * being a binary operation will be incorrect as we have come up with a counter example of a = 2 and b = 3.

[2] A binary operation * is said to be commutative if a*b = b*a

[3] A binary operation * is said to be commutative if a*(b*c) = (a*b)*c

[4] Addition forms a binary operation over the set of Natural numbers.

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