Question

Write the condition to be satisfied by q so that a rational number p/q has a terminating decimal expension

Answer 1

The condition is q should be in $2^{a} \times 5^{b}$ form.

Given:

p/q contains a terminating expansion in decimal.

To find:

The necessary condition to be satisfied by q.

Solution:

The necessary condition which is to be satisfied by q such that the rational number p/q contains a terminating expansion in decimal is:

The denominator, that is, q should be in the form of $\left(2^{2} \times 5^{b}\right),$

Where, a , b denotes some non-negative integer, that is, a, b are positive integers.

Answer 2

Answer:

$x=\frac{p}{q}\:be \:a\\rational\:number, \:such \:that\\the \: prime\: factorisation\\of\:q\:is\:of\:the\\form\:2^{n}\times 5^{m},\:where\:n\:and\:m\\are \: non-negative\: integers.\\Then \:x\:has\:a\: decimal\: expansion\\which\: terminates.$

Step-by-step explanation:

$Let,\:x=\frac{p}{q}\:be \:a\\rational\:number, \:such \:that\\the \: prime\: factorisation\\of\:q\:is\:of\:the\\form\:2^{n}\times 5^{m},\:where\:n\:and\:m\\are \: non-negative\: integers.\\Then \:x\:has\:a\: decimal\: expansion\\which\: terminates.$

$Example:\\\frac{3}{20}$

$Denominator (q)=20=2^{2}\times 5^{1}\:is \\of \:the\:form \:2^{n}\times 5^{m}$

Therefore,

$\frac{3}{20}=\frac{3\times 5}{2^{2}\times 5^{2}}\\=\frac{15}{100}\\=0.15\:(terminating\:decimal)$

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