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When can the fundamental arithmetical operation on \[2\] recurring decimals be performed directly without converting them to vulgar fraction?
$A)$ Only in addition and subtraction
$B)$ Only in addition and multiplication
$C)$ Only in addition, subtraction and multiplication
$D)$ In all the four-arithmetic operation:

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Last updated date: 25th Jul 2024
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Answer
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Hint: First of all, we need to know something about the arithmetic operators and which are the four major types of it; the addition operation will sum of the two or more than two numbers, or values, or the variables.
A subtraction operator will do minus the two numbers with respect to the major big signs with positive or negative; and multiplication is the operator of the times by the number multiplied and hence the division is inverse of the multiplication operator.

Complete step-by-step solution:
Since by these four operators only we are going to find the answer for the given question which is that fundamentals of the (four) arithmetical variables or operation on two recurring decimals can be directly performed without converting them to vulgar fraction. Since in rational numbers solving of the numerator divides the denominator terms, we get the decimals or else in irrational numbers too,
Since vulgar refers to the rational number’s numerator and denominator but with one condition that both need to be integers not anywhere of fractions. Hence in the given question we need to operate two recurse decimals but without converting them to a vulgar fraction.
This can be obtained in two operators; since in addition, even the sum of two terms yields no decimals or fraction only with integers division and decimals also at the subtraction operations we get the same results. But in multiplication it is the times of the number multiplied so that a decimal fact will occur due to the division is the inverse of multiplication on the without vulgarity. Hence option $A)$only in addition and subtraction is correct.

Note: Since arithmetic has four operations; if we eliminate the multiplicative operator, division is the inverse of multiplication so we are able to give the same reason to it too. And since options$B)$only in addition and multiplication $C)$only in addition, subtraction and multiplication$D)$in all the four arithmetic operation are all wrong, Because of multiplications in B and C and D (all arithmetic contains multiplication).