Chapter 1: Real Numbers

No, every positive integer cannot be of the form 4q + 2, where q is an integer.

Justification:

All the numbers of the form 4q + 2, where ‘q’ is an integer, are even numbers which are not divisible by ‘4’.

For example,

When q=1,

4q+2 = 4(1) + 2= 6.

When q=2,

4q+2 = 4(2) + 2= 10

When q=0,

4q+2 = 4(0) + 2= 2 and so on.

So, any number which is of the form 4q+2 will give only even numbers which are not multiples of 4.

Hence, every positive integer cannot be written in the form 4q+2

Yes, the statement “the product of two consecutive positive integers is divisible by 2” is true.

Justification:

Let the two consecutive positive integers = a, a + 1

According to Euclid’s division lemma,

We have,

a = bq + r, where 0 ≤ r < b

For b = 2, we have a = 2q + r, where 0 ≤ r < 2 … (i)

Substituting r = 0 in equation (i),

We get,

a = 2q, is divisible by 2.

a + 1 = 2q + 1, is not divisible by 2.

Substituting r = 1 in equation (i),

We get,

a = 2q + 1, is not divisible by 2.

a + 1 = 2q + 1+1 = 2q + 2, is divisible by 2.

Thus, we can conclude that, for 0 ≤ r < 2, one out of every two consecutive integers is divisible by 2. So, the product of the two consecutive positive numbers will also be even.

Hence, the statement “product of two consecutive positive integers is divisible by 2” is true.

Yes, the statement “the product of three consecutive positive integers is divisible by 6” is true.

Justification:

Consider the 3 consecutive numbers 2, 3, 4

(2 × 3 × 4)/6 = 24/6 = 4

Now, consider another 3 consecutive numbers 4, 5, 6

(4 × 5 × 6)/6 = 120/6 = 20

Now, consider another 3 consecutive numbers 7, 8, 9

(7 × 8 × 9)/6 = 504/6 = 84

Hence, the statement “product of three consecutive positive integers is divisible by 6” is true.

No, the square of any positive integer cannot be written in the form 3m + 2 where m is a natural number

Justification:

According to Euclid’s division lemma,

A positive integer ‘a’ can be written in the form of bq + r

a = bq + r, where b, q and r are any integers,

For b = 3

a = 3(q) + r, where, r can be an integers,

For r = 0, 1, 2, 3……….

3q + 0, 3q + 1, 3q + 2, 3q + 3……. are positive integers,

(3q)² = 9q² = 3(3q²) = 3m (where 3q² = m)

(3q+1)² = (3q+1)² = 9q²+1+6q = 3(3q²+2q) +1 = 3m + 1 (Where, m = 3q²+2q)

(3q+2)² = (3q+2)² = 9q²+4+12q = 3(3q²+4q) +4 = 3m + 4 (Where, m = 3q²+2q)

(3q+3)² = (3q+3)² = 9q²+9+18q = 3(3q²+6q) +9 = 3m + 9 (Where, m = 3q²+2q)

Hence, there is no positive integer whose square can be written in the form 3m + 2 where m is a natural number.

No.

Justification:

Consider the positive integer 3q + 1, where q is a natural number.

(3q + 1)² = 9q² + 6q + 1

= 3(3q² + 2q) + 1

= 3m + 1, (where m is an integer which is equal to 3q² + 2q.

Thus (3q + 1)² cannot be expressed in any other form apart from 3m + 1.