Explanation
Suppose $${ A }_{ 1 },{ A }_{ 2 },...,{ A }_{ 30 }$$ are thirty sets, each with five elements and $${ B }_{ 1 },{ B }_{ 2 },...,{ B }_{ 30 }$$ are $$n$$ sets ecah with three elements. Let $$\displaystyle \bigcup _{ i=1 }^{ 30 }{ { A }_{ i }= } \bigcup _{ j=1 }^{ n }{ { B }_{ j } } =S$$
If each element of $$S$$ belongs to exactly ten of the $${ A }_{ i }'s$$ and exactly none of the $${ B }_{ j }'s$$ then $$n=$$
Let $$n$$ be a fixed positive integer. Let a relation $$R$$ defined on $$I$$ (the set of all integers) as follows: $$aRb$$ iff $$n/(a-b)$$, that is, iff $$a-b$$ is divisible by $$n$$, then, the relation $$R$$ is
Rational numbers are those numbers which can be expressed in the form $$ \frac {p}{q} $$, where p and q are integers and $$ q \neq 0 $$
Numbers which are not rational numbers are called irrational numbers. From the given set of numbers, $$-6, -5\frac{3}{4}, -\frac{3}{5},-\frac{3}{8}, 0, \frac {4}{5}, 1, 1, \frac{2}{3}, 3.01, 8.47$$ are clearly rational numbers are they can be written in $$ \frac {p}{q} $$ form. Now, $$ - \sqrt {4} = -2 $$ which is also a rational number. And, $$ \sqrt {8} = 2\sqrt {2} $$ is not a rational number. Also, $$ \pi $$ is not a rational number. Hence, the irrational numbers in the given set are {$$ \sqrt {8}, \pi $$}
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