Chapter 1: Rational Numbers

Solution:-

True.

Express each of the given rational numbers with 6 as the common denominator.

Now,

(2/3)= [(2×3)/ (2×2)] = (4/6)

(1)= [(1×6)/ (1×6)] = (6/6)

Then,

= 4/6 < 5/6 < 6/6

= 2/3 < 5/6 < 1

So, 5/6 lies between 2/3 and 1.

Solution:-

False.

Express each of the given rational numbers with 10 as the common denominator.

Now,

½ = [(1×5)/ (2×5)] = (5/10)

(1)= [(1×10)/ (1×10)] = (10/10)

Then,

½ is equal to 5/10

So, 5/6 does not lies between ½ and 1.

Solution:-

True.

Express each of the given rational numbers with 2 as the common denominator.

Now,

-3/1 = [(-3×2)/ (1×2)] = (-6/2)

-4/1 = [(-4×2)/ (1×2)] = (-8/2)

Then,

-8/2 > -7/2 > -6/2

-4 > -7/2 > -3

So, -7/2 is lies between -3 and -4.

Solution:-

True.

Express each of the given rational numbers with 6 as the common denominator.

Now,

1/1 = [(1×6)/ (1×6)] = (6/6)

2/1 = [(2×6)/ (1×6)] = (12/6)

Then,

6/6 < 9/6 < 12/6

1 < 9/6 < 2

So, 9/6 is lies between 1 and 2.

Solution:-

True.

Solution:-

False.

Because, the correct answer is reciprocal of a negative rational number is negative rational number. i.e. reciprocal of -3/5 is -5/3.

Solution:-

False.

The additive inverse of ½ is -½.

Solution:-

True.

Let x/y = ½ and its additive inverse c/d = -1/2

Then, (x/y) + (c/d)

= ½ + (-½ )

= ½ – ½

= 0

Solution:-

False.

Let x = 3

Then, 3 + 1 = 4

3 ≠ 4

So, it is clear that x + 1 ≠ x

Solution:-

False.

Let x/y = 2/3 and its additive inverse c/d = -2/3

Then, (x/y) – (c/d)

= (2/3) – (-2/3)

= (2/3) + (2/3)

= 4/3