Q: If \$p\$ is prime, then \$\sqrt{p}\$ is irrational. So \$\sqrt{7}\$ is:

A rational number can be represented in the form of \$\dfrac pq,\$ where \$p\$ and \$q\$ are integers and \$q\$ is a non-zero integer.Here, \$\sqrt 7\$ is not a perfect square and thus cannot be expressed in the form of \$\dfrac pq,\$ thus it is an irrational number.

Q: Which of the following are non-terminating numbers?

\$ We\quad know\quad the\quad factors\quad of\quad the\quad denominator\quad of\quad a\quad terminating\quad number\quad are\quad \\ only\quad 2\quad and\quad 5.\quad If\quad it\quad has\quad any\quad other\quad number\quad then\quad it\quad will\quad be\quad non\quad terminating.\\ Option\quad A\longrightarrow \quad denominator\quad is\quad 3,\quad therefore\quad it\quad is\quad non\quad terminating.\\ Option\quad B\longrightarrow \quad denominator\quad is\quad 3\times 5\quad therefore\quad it\quad is\quad non\quad terminating.\\ Option\quad C\longrightarrow \quad denominator\quad is\quad 5,\quad therefore\quad it\quad is\quad terminating.\\ Option\quad D\longrightarrow denominator\quad is\quad 13,\quad therefore\quad it\quad is\quad non\quad terminating.\\ Answer-\quad Option\quad A,B,D\\ \$

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

The form of q is \$2^n*5^m\$q can be \$1,2,5,10,20,40....\$Any integer divided by these numbers will always give a terminating decimal number.

Q: Which of the following will have a terminating decimal expansion?

We know that the divisor of the forms \$2^n5^m\$ always form a terminating decimal number. Let us simplify the expressions and find if the expansion have a terminating decimal expansion. Option A : \$\dfrac{77}{210} \$\$=\dfrac{7\times11}{2\times 3\times 5 \times 7 }\$\$=\dfrac{11}{2\times 3 \times 5}\$Since there is a factor of 3 in the denominator, the decimal expansion will not be terminating.Option B : \$\dfrac{23}{30} \$\$=\dfrac{23}{2\times 3\times 5}\$Since, the denominator contains a power of 3 , it is non-terminating. Option C : \$\dfrac{125}{441} \$\$=\dfrac{5\times5\times5}{3\times3\times7\times7}\$This is also non-terminating. Option D : \$\dfrac{23}{8}\$\$=\dfrac{23}{2\times2\times2}\$This contains power of 2 in the denominator. Hence, the decimal expansion is terminating.

Q: \$3+2\sqrt{5}\$ a rational number.

Let's assume that \$3+2\sqrt5\$ is rational..... then \$3+2\sqrt5 = p/q \$\$\sqrt5 =( p-3q)/(2q) \$ now take \$p-3q\$ to be P and \$2q\$ to be Q........where P and Q are integers which means, \$\sqrt5= P/Q\$...... But this contradicts the fact that \$\sqrt5\$ is rational So our assumption is wrong and \$3+2\sqrt5\$ is irrational.

Q: State true or false:\$\dfrac{1}{\sqrt{2}}\$ is an irrational number.

Let us assume that \$\sqrt 2\$ is irrational \$\dfrac{1}{\sqrt 2} = \dfrac pq\$ (where \$p\$ and \$q\$ are co prime)\$\dfrac qp = \sqrt 2\$\$q = \sqrt 2p\$squaring both sides\$q^2 = 2p^2\$ \$...(1)\$By theorem, \$q\$ is divisible by \$2\$\$\therefore\ q = 2c\$ (where \$c\$ is an integer)putting the value of \$q\$ in equitation \$(1)\$\$2p^2= q^2 =4c^2\$\$p^2=\dfrac{4c^2}{2} = 2c^2\$\$\dfrac{p^2}{2}= c^2\$ By theorem, \$p\$ is also divisible by \$2\$But \$p\$ and \$q\$ are coprimeThis is a contradiction which has arisen due to our wrong assumption.Hence \$\dfrac{ 1}{\sqrt2}\$ is irrational.

Q: The number of possible pairs of number, whose product is 5400 and the HCF is 30 is

\$ Given\quad that\quad product\quad of\quad the\quad number\quad is\quad 5400=30\times 3\times 2\times 30.\\ \therefore \quad Possible\quad pairs\quad as\quad per\quad the\quad requirment\quad are-\\ (1)\quad 30\times (3\times 2\times 30)=30\times 180\\ (2)\quad (30\times 3)\times (2\times 30)=90\times 60\\ \therefore \quad Total\quad number\quad of\quad pairs=2\quad \quad (Ans) \$

Q: The denominator of \$34.12345\$ is of the form \$2^n \times 5^m\$, where \$m, n\$ are non-negative integers.

\$34.12345\$ is a terminating decimal fraction.

Question 1

If $p$ is prime, then $\sqrt{p}$ is irrational. So $\sqrt{7}$ is:

Accepted Answer

A rational number can be represented in the form of $\dfrac pq,$ where $p$ and $q$ are integers and $q$ is a non-zero integer.Here, $\sqrt 7$ is not a perfect square and thus cannot be expressed in the form of $\dfrac pq,$ thus it is an irrational number.

Question 2

Which of the following are non-terminating numbers?

Accepted Answer

$ We\quad know\quad the\quad factors\quad of\quad the\quad denominator\quad of\quad a\quad terminating\quad number\quad are\quad \ only\quad 2\quad and\quad 5.\quad If\quad it\quad has\quad any\quad other\quad number\quad then\quad it\quad will\quad be\quad non\quad terminating.\ Option\quad A\longrightarrow \quad denominator\quad is\quad 3,\quad therefore\quad it\quad is\quad non\quad terminating.\ Option\quad B\longrightarrow \quad denominator\quad is\quad 3\times 5\quad therefore\quad it\quad is\quad non\quad terminating.\ Option\quad C\longrightarrow \quad denominator\quad is\quad 5,\quad therefore\quad it\quad is\quad terminating.\ Option\quad D\longrightarrow denominator\quad is\quad 13,\quad therefore\quad it\quad is\quad non\quad terminating.\ Answer-\quad Option\quad A,B,D\  $

Question 3

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

Accepted Answer

The form of q is $2^n*5^m$q can be $1,2,5,10,20,40....$Any integer divided by these numbers will always give a terminating decimal number.

Question 4

Which of the following will have a terminating decimal expansion?

Accepted Answer

We know that the divisor of the forms $2^n5^m$ always form a terminating decimal number. Let us simplify the expressions and find if the expansion have a terminating decimal expansion. Option A : $\dfrac{77}{210} $$=\dfrac{7\times11}{2\times 3\times 5 \times 7 }$$=\dfrac{11}{2\times 3 \times 5}$Since there is a factor of 3 in the denominator, the decimal expansion will not be terminating.Option B : $\dfrac{23}{30} $$=\dfrac{23}{2\times 3\times 5}$Since, the denominator contains a power of 3 , it is non-terminating. Option C : $\dfrac{125}{441} $$=\dfrac{5\times5\times5}{3\times3\times7\times7}$This is also non-terminating. Option D : $\dfrac{23}{8}$$=\dfrac{23}{2\times2\times2}$This contains power of 2 in the denominator. Hence, the decimal expansion is terminating.

Question 5

$3+2\sqrt{5}$ a rational number.

Accepted Answer

Let's assume that $3+2\sqrt5$ is rational..... then $3+2\sqrt5 = p/q $$\sqrt5 =( p-3q)/(2q) $ now take $p-3q$ to be P and $2q$ to be Q........where P and Q are integers which means, $\sqrt5= P/Q$...... But this contradicts the fact that $\sqrt5$ is rational So our assumption is wrong and $3+2\sqrt5$ is irrational.

Question 6

State true or false:$\dfrac{1}{\sqrt{2}}$ is an irrational number.

Accepted Answer

Let us assume that $\sqrt 2$ is irrational $\dfrac{1}{\sqrt 2} = \dfrac pq$ (where $p$ and $q$ are co prime)$\dfrac qp = \sqrt 2$$q     = \sqrt 2p$squaring both sides$q^2   = 2p^2$             $...(1)$By theorem, $q$ is divisible by $2$$\therefore\  q = 2c$ (where $c$ is an integer)putting the value of $q$ in equitation $(1)$$2p^2= q^2  =4c^2$$p^2=\dfrac{4c^2}{2} = 2c^2$$\dfrac{p^2}{2}= c^2$ By theorem, $p$ is also divisible by $2$But $p$ and $q$ are coprimeThis is a contradiction which has arisen due to our wrong assumption.Hence $\dfrac{ 1}{\sqrt2}$ is irrational.

Question 7

The number of possible pairs of number, whose product is 5400 and the HCF is 30 is

Accepted Answer

$ Given\quad that\quad product\quad of\quad the\quad number\quad is\quad 5400=30\times 3\times 2\times 30.\ \therefore \quad Possible\quad pairs\quad as\quad per\quad the\quad requirment\quad are-\ (1)\quad 30\times (3\times 2\times 30)=30\times 180\ (2)\quad (30\times 3)\times (2\times 30)=90\times 60\ \therefore \quad Total\quad number\quad of\quad pairs=2\quad \quad (Ans) $

Question 8

$\dfrac{35}{50}$ has a non-terminating decimal expansion.

Accepted Answer

Let us write the denominator of the given fraction as a product of its prime factors,$50=2\times5\times5$      $=2\times5^2$So, we can observe that the denominator of the given fraction is of the for $2^m \times 5^n$ so, fraction $\dfrac{35}{50}$ will have terminating decimal expansion.Hence, the given statement is false.

Question 9

According to Euclid's division algorithm, HCF of any two positive integers $a$ and $b$ with $a > b$ is obtained by applying Euclid's division lemma to $a$ and $b$ to find $q$ and $r$ such that $a = b q + r$ , where $r$ must satisfy

Accepted Answer

According to Euclid's division algorithm, HCF of any two positive integers $a$ and $b$ with $a > b$ is obtained by applying Euclid's division lemma to $a$ and $b$ to find $q$ and $r$ such that $a = b q + r$ , where $r$ must satisfy

A

1 < r < b

B

0 < r < b

C

0 \leq r < b

D

0 < r \leq b

Easy

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Question 10

The denominator of $34.12345$ is of the form $2^n \times 5^m$, where $m, n$ are non-negative integers.

Accepted Answer

$34.12345$ is a terminating decimal fraction.

Real Numbers

Maths
11272 Views

If $p$ is prime, then $p$ is irrational. So
$7$ is:

Which of the following are non-terminating numbers?

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

Which of the following will have a terminating decimal expansion?

$3 + 25$ a rational number.

State true or false:

$\frac{1}{2}$ is an irrational number.

The number of possible pairs of number, whose product is 5400 and the HCF is 30 is

$\frac{3 5}{5 0}$ has a non-terminating decimal expansion.

According to Euclid's division algorithm, HCF of any two positive integers $a$ and $b$ with $a > b$ is obtained by applying Euclid's division lemma to $a$ and $b$ to find $q$ and $r$ such that $a = b q + r$ , where $r$ must satisfy

The denominator of $34.12345$ is of the form $2^{n} \times 5^{m}$ , where $m, n$ are non-negative integers.

$34.12345$ is a terminating decimal fraction.

Real Numbers

Maths 11272 Views

If p is prime, then p​ is irrational. So 7​ is:

Which of the following are non-terminating numbers?

Let x=qp​ be a rational number, such that the prime factorization of q is of the form 2n5m, where n,m are non-negative integers. Then x has a decimal expansion which terminates.

Which of the following will have a terminating decimal expansion?

3+25​ a rational number.

State true or false: 2​1​ is an irrational number.

The number of possible pairs of number, whose product is 5400 and the HCF is 30 is

5035​ has a non-terminating decimal expansion.

According to Euclid's division algorithm, HCF of any two positive integers a and b with a>b is obtained by applying Euclid's division lemma to a and b to find q and r such that a=bq+r, where r must satisfy

The denominator of 34.12345 is of the form 2n×5m, where m,n are non-negative integers.

34.12345 is a terminating decimal fraction.

Maths
11272 Views

If $p$ is prime, then $p$ is irrational. So
$7$ is:

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

$3 + 25$ a rational number.

State true or false:

$\frac{1}{2}$ is an irrational number.

$\frac{3 5}{5 0}$ has a non-terminating decimal expansion.

According to Euclid's division algorithm, HCF of any two positive integers $a$ and $b$ with $a > b$ is obtained by applying Euclid's division lemma to $a$ and $b$ to find $q$ and $r$ such that $a = b q + r$ , where $r$ must satisfy

The denominator of $34.12345$ is of the form $2^{n} \times 5^{m}$ , where $m, n$ are non-negative integers.

$34.12345$ is a terminating decimal fraction.