Q: State the following statement is True or False\$\dfrac{987}{10500}\$ will have terminating decimal expansion.

Terminating decimal expansion, because \$ \dfrac{987}{10500}=\dfrac{329}{3500}=\dfrac{329}{2^2.5^3.7}=\dfrac{47}{2^2.5^3}\$

Q: H.C.F. of \$26\$ and \$91\$ is:

Prime factors of \$26=2\times 13\$Prime factors of \$91=7\times 13\$As \$13\$ is the only common factor.Therefore,\$HCF(26,91)=13\$

Q: State True or False:\$6+\sqrt{2}\$ is a rational number.

Let's assume that \$6+\sqrt2\$ is rational..... then \$6+\sqrt2 = p/q \$\$\sqrt2 =( p-6q)/(q) \$ now take \$p-6q\$ to be P and \$q\$ to be Q........where P and Q are integers which means, \$\sqrt2= P/Q\$...... But this contradicts the fact that \$\sqrt2\$ is rational So our assumption is wrong and \$6+\sqrt2\$ is irrational.

Q: Use Euclids division algorithm to find the HCF of \$441, 567, 693.\$

The Euclidean Algorithm for finding HCF \$(A,B)\$ is as follows:If \$A = 0\$ then HCF \$(A,B)=B,\$ since the HCF \$(0,B)=B,\$ and we can stop. If \$B = 0\$ then HCF \$(A,B)=A,\$ since the HCF \$(A,0)=A,\$ and we can stop. Write \$A\$ in quotient remainder form \$(A = BQ + R)\$Find HCF \$(B,R)\$ using the Euclidean Algorithm since HCF \$(A,B) = HCF(B,R)\$Here, HCF of \$441\$ and \$567\$ can be found as follows:-\$567=441\times 1+126\$\$\Rightarrow\$ \$441=126\times 3+63\$\$\Rightarrow\$ \$126=63\times 2 + 0\$Since remainder is \$0\$, therefore, H.C.F of \$\left( 441,567 \right) \$ is \$=63\$Now H.C.F of \$63\$ and \$693\$ is\$693=63\times11 +0\$Therefore, H.C.F of \$\left( 63,693 \right) =63\$Thus, H.C.F of \$\left( 441,567,693 \right) =63.\$

Q: The product of a non-zero rational and an irrational number is

Let \$x\$ be a rational number and \$y\$ be an irrational number. Let \$xy = a \$.Let us assume that \$a\$ is rational. Since, \$a\$ is rational it can be expressed as \$\dfrac{p}{q} \$, where \$p\$ and \$q\$ are integers. Let \$ x =\dfrac{m}{n}\$ , where \$m\$ and \$n\$ are integers. Now, \$xy = a \$\$\dfrac{my}{n} = \dfrac{p}{q}\$.On cross multipliying we get, \$\Rightarrow y =\dfrac{pn}{qm}\$.Now, \$pn\$ and \$qm\$ are integers. Hence, \$\dfrac{pn}{qm}\$ is a rational number. However, \$y\$ is irrational. Hence, our assumption is incorrect. Hence, the product of a non-zero rational and an irrational number is always an Irrational number.So, option \$A\$.

Q: Euclids division lemma states that if a and b are any two \$+\$ ve integers, then there exists unique integers q and r such that :

Division Algorithm: Euclids division lemme states that if a and b are any two positive integers then there exist unique integers q and r such that, \$ a = q b + r \, \, 0 \leq r < b \$\$ \therefore \$ option C is correct.Proof: We begin by proving that the set s = {a - xb | x is an integer: a-xb \$\geq\$ 0 }is non-empty. To do this, if suffices to exhibit a value of x making a-xb nonnegative a value of x making a-xb nonnegative. Because the integer b \$\geq\$1 , we have |a| b \$\geq\$ |a| and so - \$ a - (-|a|) b = a + |a| b \geq a + |a| \geq 0 \$For the choice x = -|a|, then a - xb lies in s. This paves the way for an application of well ordering principle, from which we infer that the set S contains a smallest integer : call it r. By the definition S, \$\exists\$ an integer q satisfying , r = a - qb , \$0\leq r\$we argue that r < b. If this were not this case, then r \$\geq\$ 0 and a - ( q + 1 ) b = ( a - q b ) - b = r - b \$\geq\$ 0This implication is that the integer a - ( q + 1 ) b has the proper form to belong to the set S. But a - ( q + 1 ) b = r - b < r , leading to a contradiction of the choice of r as the smallest member of S. Hence r < b .Next we turn to the task of showing the uniqueness of q and r. Suppose that a has two representation of the desired form say,\$ a = q b + r = q^{'}b + r^{'} \$Where \$ 0 \leq r < b, \, 0 \leq r^{'} < b . \$ Then \$r^{'}- r = b (q - q^{'}) \$ and, owing to the fact that\$ | r^{'} - r | = b | q - q^{'} | \$ upon adding two inequalities \$-b < -r \leq 0 \$ and \$ 0 \leq r^{'} < b \$ we obtain \$ -b < r^{'}-r < b \Rightarrow |r^{'} - r | < b \$Thus, \$b|q-q^{'}| < b \$ which yields \$ 0 \leq |q-q^{'} | < 1 \$As, \$ |q-q^{'} | \$ is positive, the only possibility is that \$ |q-q^{'} | = 0 \$ where \$q = q^{'} : \$ this is turns give \$ r = r^{'} \$ and ends the proof.

Q: Write whether every positive integer can be of the form \$4q + 2\$, where \$q\$ is an integer.

No, all positive integers cannot be written in the form of \$ 4q+2\$Because, \$4q+2=2\left( 2q+1 \right) \$Therefore, \$4q+2\$ is an even number.So, we can't write odd numbers in the form \$4q+2\$ (where \$q\$ is an integer)Also \$2q+1\$ is an odd number, hence the maximum power of \$2\$ that divides \$4q+1\$ is \$1\$. Therefore, we can't represent the numbers divisible by \$4\$ this form.

Q: Say true or false:A positive integer is of the form 3q + 1, q being a natural number, then you write its square in any form other than 3m + 1, i.e.,3m or 3m + 2 for some integer m.

Let the positive integer \$n\$ is of the form \$3q, 3q+1,\$ and \$ 3q+2\$If \$n=3q\$Squaring both sides, we get, \$=>{ n }^{ 2 }=9{ q }^{ 2 }\$ \$=>{ n }^{ 2 }=3\left( { 3q }^{ 2 } \right) \$ \$=>{ n }^{ 2 }=3m\$, where \$m=3{ q }^{ 2 }\$Now, if \$n=3q+1\$ \$=>{ n }^{ 2 }={ \left( 3q+1 \right) }^{ 2 }\$ \$=>{ n }^{ 2 }=9{ q }^{ 2 }+6q+1\$ \$=>{ n }^{ 2 }={ 3q\left( 3q+2 \right) }+1\$ \$=>{ n }^{ 2 }=3m+1 ,\$ where \$ m=q\left( 3q+2 \right) \$Now, if \$n=3q+2\$ \$=>{ n }^{ 2 }={ \left( 3q+2 \right) }^{ 2 }\$ \$=>{ n }^{ 2 }=9{ q }^{ 2 }+12q+4\$ \$=>{ n }^{ 2 }={ 3q\left( 3q+4 \right) }+4\$ \$=>{ n }^{ 2 }={ 3q\left( 3q+4 \right) }+3+1\$ \$=>{ n }^{ 2 }=3m+1\$ where \$m=\left( 3{ q }^{ 2 }+4q+1 \right) \$Hence, \${ n }^{ 2 }\$ integer is of the form \$3m\$ and \$3m+1\$ not \$3m+2\$

Q: H.C.F. of \$6, 72\$ and \$120\$ is:

Factors of 6 are \$1,2,3,6\$Factors of 72 are \$1,2,3,4,6,8,9,12,18,24,36,72\$Factors of 120 are \$1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120\$The highest factor in all the three is \$6\$ .

Question 1

State the following statement is True or False$\dfrac{987}{10500}$ will have terminating decimal expansion.

Accepted Answer

Terminating decimal expansion, because $ \dfrac{987}{10500}=\dfrac{329}{3500}=\dfrac{329}{2^2.5^3.7}=\dfrac{47}{2^2.5^3}$

Question 2

H.C.F. of $26$ and $91$ is:

Accepted Answer

Prime factors of $26=2\times 13$Prime factors of $91=7\times 13$As $13$ is the only common factor.Therefore,$HCF(26,91)=13$

Question 3

State True or False:$6+\sqrt{2}$ is a rational number.

Accepted Answer

Let's assume that $6+\sqrt2$ is rational..... then $6+\sqrt2 = p/q $$\sqrt2 =( p-6q)/(q) $ now take $p-6q$ to be P and $q$ to be Q........where P and Q are integers which means, $\sqrt2= P/Q$...... But this contradicts the fact that $\sqrt2$ is rational So our assumption is wrong and $6+\sqrt2$ is irrational.

Question 4

According to Euclid's division algorithm, using Euclid's division lemma for any two positive integers $a$ and $b$ with $a > b$ enables us to find the

Accepted Answer

According to Euclid's division lemma, For each pair of positive integers $a$ and $b$, we can find unique integers $q$ and $r$ satisfying the relation $a = bq + r , 0 &#8804; r < b$If $r = 0$ then q will be HCF of $a$ and $b$The basis of the Euclidean division algorithm is Euclid&#8217;s division lemma. To calculate the Highest Common Factor (HCF) of two positive integers $a$ and $b$ we use Euclid&#8217;s division algorithm. HCF is the largest number which exactly divides two or more positive integers. By exactly we mean that on dividing both the integers $a$ and $b$ by HCF, the remainder is zero.So correct answer is option A

Question 5

Use Euclids division algorithm to find the HCF of $441, 567, 693.$

Accepted Answer

The Euclidean Algorithm for finding HCF $(A,B)$ is as follows:If $A = 0$ then HCF $(A,B)=B,$ since the HCF $(0,B)=B,$ and we can stop.  If $B = 0$ then HCF $(A,B)=A,$ since the HCF $(A,0)=A,$ and we can stop.  Write $A$ in quotient remainder form $(A = BQ + R)$Find HCF $(B,R)$ using the Euclidean Algorithm since HCF $(A,B) = HCF(B,R)$Here, HCF of $441$ and $567$ can be found as follows:-$567=441\times 1+126$$\Rightarrow$ $441=126\times 3+63$$\Rightarrow$ $126=63\times 2 + 0$Since remainder is $0$, therefore, H.C.F of $\left( 441,567 \right) $ is $=63$Now H.C.F of $63$ and $693$ is$693=63\times11 +0$Therefore, H.C.F of $\left( 63,693 \right) =63$Thus, H.C.F of $\left( 441,567,693 \right) =63.$

Question 6

The product of a non-zero rational and an irrational number is

Accepted Answer

Let $x$ be a rational number and $y$ be an irrational number. Let $xy = a $.Let us assume that $a$ is rational. Since, $a$ is rational it can be expressed as $\dfrac{p}{q} $, where $p$ and $q$ are integers. Let $ x =\dfrac{m}{n}$ , where $m$ and $n$ are integers. Now, $xy = a $$\dfrac{my}{n} = \dfrac{p}{q}$.On cross multipliying we get, $\Rightarrow y =\dfrac{pn}{qm}$.Now, $pn$ and $qm$ are integers. Hence, $\dfrac{pn}{qm}$ is a rational number. However, $y$ is irrational. Hence, our assumption is incorrect. Hence, the product of a non-zero rational and an irrational number is always an Irrational number.So, option $A$.

Question 7

Euclids division lemma states that if a and b are any two $+$ ve integers, then there exists unique integers q and r such that :

Accepted Answer

Division Algorithm: Euclids division lemme states that if a and b are any two positive integers then there exist unique integers q and r such that, $ a = q b + r \, \, 0 \leq r < b $$ \therefore $ option C is correct.Proof: We begin by proving that the set s = {a - xb | x is an integer: a-xb $\geq$ 0 }is non-empty. To do this, if suffices to exhibit a value of x making a-xb nonnegative a value of x making a-xb nonnegative. Because the integer b $\geq$1 , we have |a| b $\geq$ |a| and so - $ a - (-|a|) b = a + |a| b \geq a + |a| \geq 0 $For the choice x = -|a|, then a - xb lies in s. This paves the way for an application of well ordering principle, from which we infer that the set S contains a smallest integer : call it r. By the definition S, $\exists$ an integer q satisfying , r = a - qb , $0\leq r$we argue that r < b. If this were not this case, then r $\geq$ 0 and a - ( q + 1 ) b = ( a - q b ) - b = r - b $\geq$ 0This implication is that the integer a - ( q + 1 ) b has the proper form to belong to the set S. But a - ( q + 1 ) b = r - b < r , leading to a contradiction of the choice of r as the smallest member of S. Hence r < b .Next we turn to the task of showing the uniqueness of q and r. Suppose that a has two representation of the desired form say,$ a = q b + r = q^{'}b + r^{'} $Where $ 0 \leq r < b, \, 0 \leq r^{'} < b . $ Then $r^{'}- r = b (q - q^{'}) $ and, owing to the fact that$ | r^{'} - r | = b | q - q^{'} | $ upon adding two inequalities $-b < -r \leq 0 $ and $ 0 \leq r^{'} < b $ we obtain $ -b < r^{'}-r < b \Rightarrow |r^{'} - r | < b $Thus, $b|q-q^{'}| < b $ which yields $ 0 \leq |q-q^{'} | < 1 $As, $ |q-q^{'} |  $ is positive, the only possibility is that $  |q-q^{'} | = 0 $ where $q = q^{'} : $ this is turns give $ r = r^{'} $ and ends the proof.

Question 8

Write whether every positive integer can be of the form $4q + 2$, where $q$ is an integer.

Accepted Answer

No, all positive integers cannot be written in the form of $ 4q+2$Because, $4q+2=2\left( 2q+1 \right) $Therefore, $4q+2$ is an even number.So, we can't write odd numbers in the form $4q+2$ (where $q$ is an integer)Also $2q+1$ is an odd number, hence the maximum power of $2$ that divides $4q+1$ is $1$. Therefore, we can't represent the numbers divisible by $4$  this form.

Question 9

Say true or false:A positive integer is of the form 3q + 1, q being a natural number, then you write its square in any form other than  3m + 1, i.e.,3m or 3m + 2  for some integer m.

Accepted Answer

Let the positive integer $n$ is of the form $3q, 3q+1,$ and $ 3q+2$If $n=3q$Squaring both sides, we get,    $=>{ n }^{ 2 }=9{ q }^{ 2 }$    $=>{ n }^{ 2 }=3\left( { 3q }^{ 2 } \right) $    $=>{ n }^{ 2 }=3m$, where $m=3{ q }^{ 2 }$Now, if $n=3q+1$    $=>{ n }^{ 2 }={ \left( 3q+1 \right)  }^{ 2 }$    $=>{ n }^{ 2 }=9{ q }^{ 2 }+6q+1$    $=>{ n }^{ 2 }={ 3q\left( 3q+2 \right)  }+1$    $=>{ n }^{ 2 }=3m+1 ,$ where $  m=q\left( 3q+2 \right) $Now, if $n=3q+2$    $=>{ n }^{ 2 }={ \left( 3q+2 \right)  }^{ 2 }$    $=>{ n }^{ 2 }=9{ q }^{ 2 }+12q+4$    $=>{ n }^{ 2 }={ 3q\left( 3q+4 \right)  }+4$    $=>{ n }^{ 2 }={ 3q\left( 3q+4 \right)  }+3+1$    $=>{ n }^{ 2 }=3m+1$ where $m=\left( 3{ q }^{ 2 }+4q+1 \right) $Hence, ${ n }^{ 2 }$ integer is of the form $3m$ and $3m+1$ not $3m+2$

Question 10

H.C.F. of $6, 72$ and $120$ is:

Accepted Answer

Factors of 6 are $1,2,3,6$Factors of 72 are $1,2,3,4,6,8,9,12,18,24,36,72$Factors of 120 are $1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120$The highest factor in all the three is $6$ .

Real Numbers

Maths
8038 Views

State the following statement is True or False
$\frac{9 8 7}{1 0 5 0 0}$ will have terminating decimal expansion.

H.C.F. of $26$ and $91$ is:

State True or False:
$6 + 2$ is a rational number.

According to Euclid's division algorithm, using Euclid's division lemma for any two positive integers $a$ and $b$ with $a > b$ enables us to find the

Use Euclids division algorithm to find the HCF of $441, 567, 693 .$

The product of a non-zero rational and an irrational number is

Euclids division lemma states that if a and b are any two $+$ ve integers, then there exists unique integers q and r such that :

Write whether every positive integer can be of the form $4 q + 2$ , where $q$ is an integer.

Say true or false:
A positive integer is of the form 3q + 1, q being a natural number, then you write its square in any form other than 3m + 1, i.e.,3m or 3m + 2 for some integer m.

H.C.F. of $6, 72$ and $120$ is:

Real Numbers

Maths 8038 Views

State the following statement is True or False10500987​ will have terminating decimal expansion.

H.C.F. of 26 and 91 is:

State True or False:6+2​ is a rational number.

According to Euclid's division algorithm, using Euclid's division lemma for any two positive integers a and b with a>b enables us to find the

Use Euclids division algorithm to find the HCF of 441,567,693.

The product of a non-zero rational and an irrational number is

Euclids division lemma states that if a and b are any two + ve integers, then there exists unique integers q and r such that :

Write whether every positive integer can be of the form 4q+2, where q is an integer.

Say true or false:A positive integer is of the form 3q + 1, q being a natural number, then you write its square in any form other than 3m + 1, i.e.,3m or 3m + 2 for some integer m.

H.C.F. of 6,72 and 120 is:

Maths
8038 Views

State the following statement is True or False
$\frac{9 8 7}{1 0 5 0 0}$ will have terminating decimal expansion.

H.C.F. of $26$ and $91$ is:

State True or False:
$6 + 2$ is a rational number.

According to Euclid's division algorithm, using Euclid's division lemma for any two positive integers $a$ and $b$ with $a > b$ enables us to find the

Use Euclids division algorithm to find the HCF of $441, 567, 693 .$

Euclids division lemma states that if a and b are any two $+$ ve integers, then there exists unique integers q and r such that :

Write whether every positive integer can be of the form $4 q + 2$ , where $q$ is an integer.

Say true or false:
A positive integer is of the form 3q + 1, q being a natural number, then you write its square in any form other than 3m + 1, i.e.,3m or 3m + 2 for some integer m.

H.C.F. of $6, 72$ and $120$ is: