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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:
Note:
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 associative if a*(b*c) = (a*b)*c
[4] Addition forms a binary operation over the set of Natural numbers.
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