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Whole numbers are not closed under ......... operation.\[\]
A. Addition \[\]
B. Subtraction \[\]
C. Multiplication \[\]
D. none of these\[\]

Answer
VerifiedVerified
562.5k+ views
Hint: We recall the definition of closure property and when the set is closed under an operation. We recall the elements of the whole number set $W$, take any two whole numbers $a, b \in W$ and then add, subtract, multiply them to check whether the result is also a whole number or not to check the closure property. \[\]

Complete step-by-step solution:
We know from arithmetic that a set is closed under an operation if the output of that operation on members of the set always produces a member of that set. Let us define a binary operation $o$ which operates between the elements of the set $A$, then we say $o$ satisfies the closure property or set $A$ is closed under the operation $o$ if $aob\in A$ for all$a,b\in A$.\[\]
The whole number set $W$ takes all the natural numbers and the number zero. So we have
\[W=\left\{ 0,1,2,3,... \right\}\]
If we take any two elements from the whole number set and add them we will get the sum also a whole number, for example, we take 0 and 1 then the sum will be$0+1=1$. So we can say for all $a, b \in W$ we will find $a+b\in W.$ It means the whole number set is closed under the operation addition. So option A is not correct. \[\]
If we take any two elements from the whole number set and subtract one from the other we may not get a whole number, for example, $0-1=-1$ where the result $-1$ is outside the whole number set in the set of integers. We can never get a whole number when we subtract greater numbers from the smaller number. It means if $ b > a $, $a-b\notin W$. So the whole number set is not closed under subtraction and option B is correct. \[\]
If we take any two elements from the whole number set and multiply them we will get the product also a whole number, for example, $0\times 1=0$ where 0 is the whole number. So we can say for all $a, b \in W$ we will find $a\times b\in W.$ It means the whole number set is closed under the operation multiplication. So option C is not correct. \[\]
So the only correct option is B.\[\]

Note: We note that we have only considered here binary operation. We can also define closure property for $n-$nary operation which takes $n$ elements simultaneously. The whole number set is also not a closed number under division because when the divisor is not a factor of dividend or the divisor is 0 we may not get the whole number as a result of the division
.\[\]