Rational Numbers

(i) The given statement shows the associative property for addition.

(ii) The given statement shows the commutative property for multiplication.

(iii) The given statement shows the distributive property for multiplication over addition

Additive inverse property:

Case 1: If x is positive number, x+(-x)=0. So, (-x) is called as additive inverse of x.

Case 2: If x is negative number, -x+(-(-x))=0. So, (x) is called as additive inverse of x.

According to the above property,

The additive inverse of 6 is (-6).

The additive inverse of 9 is (-9).

The additive inverse of 123 is

(-123).

The additive inverse of -76 is 76.

The additive inverse of -85 is 85.

The additive inverse of 1000 is

(-1000).

(i) M + 6 = 8

⇒ m = 8 - 6

⇒m = 2

(ii) M + 25 = 15

⇒m = 15 - 25

⇒m = -10

(iii) M – 40 = -26

⇒m = 40 - 26

⇒m = 14

(iv) M + 28 = -49

⇒m = - 28 - 49

⇒m = -77

Increasing order of above numbers,

-333<-210<-26<-8<0<21<33<85<2011.

Decreasing order of above numbers,

2011>528>364>210>85>12>-58>-1024>-10000.

The equivalent rational number can be calculated by multiplying and dividing with a positive number. In the question, it is given that the denominator should not be greater than 80.

First equivalent rational number of when multiplying and dividing with 2 is

Second equivalent rational number of when multiplying and dividing with 3 is

Third equivalent rational number of when multiplying and dividing with 4 is

Similarly, other equivalent rational numbers of are

The equivalent rational number can be calculated by multiplying and dividing with a positive number and it is given that numerator should not be greater than 180.

First equivalent rational number of when multiplying and dividing with 2 is

Second equivalent rational number of when multiplying and dividing with 3 is

Third equivalent rational number of when multiplying and dividing with 4 is

Similarly, other equivalent rational numbers of are

The positive rational numbers having the sum of numerator and denominator as 11 are:

According to the question, increasing order of above rational numbers is

The positive rational numbers having numerator-denominator as -2 with their increasing order are:

Rational numbers are those numbers which can be written in the form of , where p and q are the integers and .

Now considering , which is written in the form of . Where . Here both are integers and .

Hence, the number is a rational number.

Yes the 0.9 and 0.8 can also be written in rational numbers.

0.9 can also be written as Which is a rational number.

0.8 can also be written as Which is a rational number.

(i) This is showing addition property.

(ii) This is showing multiplication property.

(iii) This is showing additive identity.

(iv) This is showing multiplicative identity.

(v) This is showing additive inverse.

(vi) This is showing multiplicative inverse.

Commutative property of addition is

(i) Let LHS(Left Hand Side) is

Let RHS(Right Hand Side) is

Hence, LHS=RHS i.e. Which is showing commutative property of addition.

(ii) Let LHS(Left Hand Side) is

Let RHS(Right Hand Side) is

Hence, LHS=RHS i.e. Which is showing commutative property of addition.

(iii) Let LHS(Left Hand Side) is

Let RHS(Right Hand Side) is

Hence, LHS=RHS i.e. Which is showing commutative property of addition.

Commutative property of multiplication is

(i) Let LHS(Left Hand Side) is

Let RHS(Right Hand Side) is

Hence, LHS=RHS i.e. Which is showing commutative property of multiplication.

(ii) Let LHS(Left Hand Side) is

Let RHS(Right Hand Side) is

Hence, LHS=RHS i.e. Which is showing commutative property of multiplication.

(iii) Let LHS(Left Hand Side) is

Let RHS(Right Hand Side) is

Hence, LHS=RHS i.e. Which is showing commutative property of multiplication.

theDistributive property of multiplication over addition is .

(i) Let LHS(Left Hand Side) is

Let RHS (Right Hand Side) is

Hence, LHS=RHS i.e. a*(b+c)=a*b+a*c. Which is showing distributive property of multiplication over addition.

(ii) Let LHS(Left Hand Side) is

Let RHS (Right Hand Side) is

Hence, LHS=RHS i.e. . Which is showing distributive property of multiplication over addition.

(iii) Let LHS(Left Hand Side) is

Let RHS (Right Hand Side) is

Hence, LHS=RHS i.e. . Which is showing distributive property of multiplication over addition.

Case 1: If is positive number, . So, is called as additive inverse of

Case 2: If is negative number, . So, is called as additive inverse of

The additive inverse of

The additive inverse of

The additive inverse of

The additive inverse of

The additive inverse of

The multiplicative inverse of a digit is the reciprocal of given digit i.e. ,

The multiplicative inverse of

The multiplicative inverse of

The multiplicative inverse of

The multiplicative inverse

The multiplicative inverse of

(i) on a number line is shown below

(ii) on the number, line is shown below

(iii) on the number, line is shown below

(iv) on the number, line is shown below

(v) on the number, line is shown below

The ascending order of the above rational numbers is

The rational numbers can also be written as and

So, the five rational number lies between are

The positive rational numbers less than 1 whose sum of numerator and denominator does not exceed 10 are

There are 15 rational numbers.

I didn’t remember the concept used in this question.

There will be three cases for the question given above.

Case 1: when m/n >1 and p/q >1, this situation means that p > q and m > n

And this means that m + p > n + q and And this will mean > 1 and it will lie between m/p and n/q

Case 2: when m/n <1 and p/q <1, this situation means m < n and p < q

And thus m + p < n + q and this will mean < 1

And thus will lie less than m/p and n/q

Case 3: when any one is less than 1, then the case gets ambiguous.

Let m/n <1 and p/q >1

Now, m < n and p > q

Now we don’t know the nature of

And so this depends on the numbers taken.

The rational numbers between 0 and 1 are in the form of having denominator as 80.

Value of ranges from 1 to 79. The rational numbers are

So there are 79 rational numbers lies between 0 and 1 having denominator as 80.

The rational number is in the form of and ,

According to the question,

So the rational numbers are .

There are 34 rational numbers following the conditions and .

(a) The number 0 is not in the set of natural numbers.

0 is added to set of natural number (1,2,3,4,…..) and it becomes set of whole numbers (0,1,2,3,4,….)

(b) The least number in the set of all the whole numbers is 0.

0 is the smallest number which when added to the set of natural number make it a set of whole number.

(c) The least number in the set of all even natural numbers is 2.

The set of even natural number starts from (2,4,6,….) so the least number in the set of all even natural numbers is number 2 .

(d) The successor of 8 in the set of all natural numbers is 9.

The number succeeding number 8 is number 9 in the set of all natural numbers as the set of natural number consist of term (1,2,3,4,5,6,7,8,9,10,11,12,…..)

(e) The sum of two odd integers is always an even number

for e.g;- 3 + 5 = 8 , 5 + 7 = 12

Let m and n be odd integers. By definition of odd we have that m = 2a + 1 and

n = 2b + 1. Consider the sum m + n = (2a + 1) + (2b + 1) = 2ab + 2b + 2 = 2k, where

k = a + b + 1 is an integer. Therefore by definition of even we have shown that

m + n is even.

(f) The product of two odd integers is odd.

Let a and b be two odd integers.

By definition of odd we know that a = 2m + 1 and b = 2n + 1.

Now, the product ab = (2m + 1)(2n + 1)

= 4mn + 2m + 2n + 1

= 2(2mn + m + n) + 1 = 2k + 1, where k = (2mn + m + n ) is an integer. Therefore by

Definition of odd number, the product of two odd integers is also odd.

(a) The set of all even natural numbers is a smallest element – False

The set of all even natural number consist of elements (2, 4, 6, 8,10,……) but the smallest element is 0 which is not included in the set of natural numbers and the smallest natural number is 1 which is also not included in set even numbers.

(b) Every non-empty subset of ℤ has the smallest element. – true

Every non empty subset of Z have the smallest element. It depends on the conditions attached to the non-empty subset such as the set of even natural numbers has 2 as its least element. The set of prime numbers bigger than 5 has 7 as its least element

(c ) Every integer can be identified with a rational number- true

Every integer is a rational number as each integer n can be written in the form .

For example 7 = 7/1 and thus 7 is a rational number.

(d) For each rational number, one can find the next rational number.- False

One cannot always find the next rational number for each rational number as there also lies irrational numbers in-between.

(e) There is the largest rational number.- False

The largest rational number cannot be determined as there are infinites numbers and each number can be expressed as a rational number.

(f) Every integer is either even or odd – true

We know n is even if n = 2k for some integer k and n is odd if n = 2k + 1 for some integer k.

Every integer can be expressed as 2k or 2k + 1 hence every integer is either odd or even.

(g) Between any two rational numbers, there is an integer. - false

Rational number can be any number 5/1, 2/1, which can be expressed in form of fraction and integers are the whole numbers which are not a fraction so there cannot lie a whole number between two rational number which are fractions.

(i) 100(100 – 3) – (100 × 100 – 3)

= 100 × 100 - 100 × 3 – 100 × 100 + 3

= - 300 + 3 (cancellation property)

= - 297

(ii) (20 – (2011 – 201) + (2011– (201 – 20))

= 20 - 2011 + 201 + 2011 – 201 + 20

= 40 (cancellation property)

Given m is an integer and m ≠ -1 and -2

Let m = 1

Then

Comparing

⇒ (taking the LCM)

Hence

If m < -2

Then let m = -3

And

Here also

If m > -1

Let m = 4

And

Comparing

Taking LCM

⇒

For every integer this will hold true that

For the operation r*s = r + s –(r × s)
(i) Let r = 1/2 and s = 1/4
Now ,




Which is also a rational number.
Hence Q is closed under the operation * .

(ii) Let three rational number be o,p,q
For the operation r*s = r + s –(r × s)
(o*p)*q = [ o + p -( o × p)]*q
= [o + p - op]*q
= [(o + p - op) + q - (( o + p - op) × q)]
= [ o + p - op + q - oq - pq +opq ] ..... (1)
Now,
o*(p*q) = o* [ p + q - (p × q ) ]
= o* [ p + q - pq ]
= [( o + (p + q - pq) - (o ×(p + q - pq))]
= [ o + p + q - pq - op - oq + opq] ..... (2)
From 1 and 2
(o*p)*q = o*(p*q)
Hence * is an associative operation.

(iii) Let two rational number be p,q.
For the operation r*s = r + s –(r × s)
p*q = p + q - ( p × q )
= p + q - pq
Now,
q*p = q + p - ( q × p )
= q + p - qp
Hence * is a commutative operation.

(iv) For the operation r*s = r + s –(r × s)
a * 1 = a + 1 - ( a × 1)
= a + 1 - a
= 1
For any a in Q a * 1 is 1.

(v) The value of two integers would be a = 2 and b = 2

For the operation r*s = r + s –(r × s)

The multiplicative inverse of number is given by the reciprocal of the integer, hence the multiplicative inverse of the given numbers are :-

For arranging rational numbers in increasing or decreasing order, always make sure that the denominator is same.

For given numbers, make the denominator same by taking L.C.M of denominators.

Denominators are – 13, 26, 6, 43, 28, 11

Let us write down the prime factors of numbers

13 = 13 × 1

26 = 13 × 2

6 = 2 × 3

43 = 43 × 1

28 = 2 × 2 × 7

11 = 11 × 1

L.C.M of the given numbers will be = 13 × 2 × 3 × 2 × 11 × 43 × 7

= 516516

Now lets make the denominators equal by multiplying and dividing by terms.

And Similarly all our numbers will become as

Now arranging in increasing order we get,

Hence we have the order as

Now for arranging rational numbers in increasing or decreasing order make the denominator of numbers equal. But as we can see from the numbers that would make the whole process very long.

We can divide the numbers and can calculate the approximate values.

21/17 = 1.235

31/27 = 1.148

13/11 = 1.18

41/37 = 1.10

51/47 = 1.08

9/8 = 1.125

Now we can easily arrange the numbers as follows

21/17 > 13/11 > 31/27 > 9/8 > 41/37 > 51/47

(a) The additive inverse of 0 is 0 itself because 0 is neither positive nor negative number.

(b) The multiplicative inverse of 1 is 1 as 1 can be written as 1/1 which itself the reciprocal of the 1.

(c ) (1, -1) is a pair of integers having multiplicative inverse.

(i) In the set of all rational numbers associativity in addition holds true

e.g.; A) let

Now a + (b + c) = (a + b) + c

LHS

RHS

LHS = RHS hence associativity holds true over addition in rational numbers

(ii) let us check for subtraction a –( b - c) = (a- b) - c

RHS

LHS

Hence associativity doesn’t hold true with subtraction in rational numbers

(ii) For two rational number addition and multiplication are commutative and subtraction and division are not commutative

e.g

And

But are not same hence commutate property is not true with division and subtraction

(iii) E.g

LHS =

RHS =

LHS = RHS

e.g

RHS

LHS

LHS = RHS

Hence the distributivity of multiplication over addition holds true for rational numbers

(i)

=

=

(ii)

=

(iii)

=

=

(i)

=

(ii)

=

(iii)

=

=

(iv)

=

=

=

Every non zero rational number is invertible, but only ± 1 are invertible integers.

The value is

=

=

=

The only rational number which is equal to its reciprocal is ± 1

The number of trips made by the driver on particular day are 5 ( 8-10, 10-12, 12-2, 2-4, 4-6)

Number of passenger keeps on decreasing with each trip so total passengers carried on that day

= 30 + 29 + 28 + 27 + 26

= 140

There are no rational numbers between 0 and 1, of the form as after the division the rational number obtained would always be greater than 1.

Let n = 0

Then which is an integer

Let n = 1

Let n = -1

which is an integer

Let n = -3

which is an integer

Let n = -4

, which is an integer

The values are

(2 × 3) + 4 = 6 + 4 = 10

3 + ( 4 × 5) = 3 + 20 = 23

2 + (4 × 5) = 2 + 20 = 22

(2 × 3 + 4) × 5 = (6 + 4 ) × 5

= 10 × 5 = 50

Now,

=

=

Now note that a common factor of p + 2q and q is also a factor of (p + 2q)−2⋅q = p, therefore must be 1.

Also a common factor of p + 2q and p + q is a factor of both (p + 2q) − ( p + q) = q and 2⋅(p + q) − ( p + 2q) = p hence must be 1.

Hence it is in its lowest form.

Let n = 1

Then which is in its lowest form.

Let n = 3

which is again in its lowest form

Hence for every value of n natural number, is in its lowest form.

Let n = 0

= (0 + 3)(0-1)

= (3)(-1)

= -3 which is also an integer

Let n = -1

= (-1 + 3)(-1-1)

= (2)(-2)

= -4 which is also an integer

Let n = 1

= (1 + 3)(1-1)

= (4)(0)

= 0, which is also an integer

Therefore, for any integer value of n the equation will result an integer.