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In the event that $\ast$ is not a binary operation, give justification for this: On $Z^+,$ defined $\;\ast\;by\;a\;\ast\;b=|a-b|.$

Note: This is part 4 of a 5 part question, split as 5 separate questions here.

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  • A binary operation $∗$ on a set $A$ is a function $∗$ from $A \times A$ to $ A$. Therefore, if $a,b \in A \Rightarrow a*b \in A\; \forall\; a,b, \in A$
Given $Z^+$ define $*$ by $a*b=|a-b|$
For each pair of element $a,b \in R$, then exists an unique element $|a-b| \in Z^+$
Therefore $*$ on defined (a,b) defined by $a*b=|a-b| $ carries a unique value in $z+$ and is a binary operation.
answered Mar 13, 2013 by sreemathi.v
edited Mar 19, 2013 by balaji.thirumalai

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