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प्रश्न
Every integer is a rational number but every rational number need not be an integer.
पर्याय
True
False
MCQ
चूक किंवा बरोबर
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उत्तर
This statement is True.
Explanation:
Integers.... –3, –2, –1, 0, 1, 2, 3,...
Rational numbers:
`1, (-1)/2, 0, 1/2, 1, 3/2,`......
Hence, every integer is rational number, but every rational number is not an integer.
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संबंधित प्रश्न
Add and express the sum as a mixed fraction:
\[\frac{- 31}{6} \text{and} \frac{- 27}{8}\]
Re-arrange suitably and find the sum in each of the following:
\[\frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5}\]
Re-arrange suitably and find the sum in each of the following:
\[\frac{1}{8} + \frac{5}{12} + \frac{2}{7} + \frac{7}{12} + \frac{9}{7} + \frac{- 5}{16}\]
What should be subtracted from \[\left( \frac{3}{4} - \frac{2}{3} \right)\] to get\[\frac{- 1}{6}?\]
Simplify:
\[\frac{5}{6} + \frac{- 2}{5} - \frac{- 2}{15}\]
Simplify each of the following and express the result as a rational number in standard form:
\[\frac{7}{6} \times \frac{- 3}{28}\]
Simplify:
\[\left( - 5 \times \frac{2}{15} \right) - \left( - 6 \times \frac{2}{9} \right)\]
Fill in the blanks:
\[- 4 \times \frac{7}{9} = \frac{7}{9} \times . . . . . .\]
Fill in the blanks:
\[\frac{5}{11} \times \frac{- 3}{8} = \frac{- 3}{8} \times . . . . . .\]
Find (x + y) ÷ (x − y), if
\[x = \frac{2}{5}, y = \frac{1}{2}\]
