RD Sharma Solutions Class 10 Real Numbers Exercise 1.1

RD Sharma Class 10 Solutions Chapter 1 Ex 1.1 PDF Free Download

Exercise 1.1

Q.1: If a and b are two odd positive integers such that a > b, then prove that one of the two numbers \(\frac{ a + b }{ 2 }\) and \(\frac{ a – b }{ 2 }\) is odd and the other is even.

 

Sol:

Consider “a” and “b” to be any two odd positive integers in a way that a > b.

We know that any positive integer is of the form q, 2q + 1

Now, let a = 2q + 1 and b = 2m + 1 (here “q:, “m” are whole numbers)

\(\frac{ a + b }{ 2 }\) = \(\frac{ (2q + 1)  + (2m + 1)  }{ 2 }\)

\(\frac{ a + b }{ 2 }\) = \(\frac{ 2 ((q + m)  + 1)  }{ 2 }\)

\(\frac{ a + b }{ 2 }\) = (q + m + 1)  which is a positive integer.

Also,

\(\frac{ a – b }{ 2 }\) = \(\frac{ (2q + 1)  – (2m + 1)  }{ 2 }\)

\(\frac{ a – b }{ 2 }\) = \(\frac{ 2 (q + m)   }{ 2 }\)

\(\frac{ a – b }{ 2 }\) = (q – m)

Since a > b

⇒ 2q + 1 > 2m + 1

2q > 2m

q > m

So, \(\frac{ a – b }{ 2 }\) = (q – m)  > 0

∴ \(\frac{ a – b }{ 2 }\) is a positive integer.

Now. it has to be proved that one of the two numbers \(\frac{ a + b }{ 2 }\) and \(\frac{ a – b }{ 2 }\) is an odd and other is an even number.

 

Consider , \(\frac{ a + b }{ 2 }\) – \(\frac{ a – b }{ 2 }\) = \(\frac{ (a + b)  – (a – b)  }{ 2 }\) = \(\frac{  2b }{ 2 }\) = b

 

We already know that \(\frac{ a + b }{ 2 }\) and \(\frac{ a – b }{ 2 }\) are positive integers.

It is also known that the difference of two integers is an odd number if one of them is odd and another is even. (Also, the difference between two odd and two even integers is even)

Thus, it is proved that if a and b are two odd positive integers is even.

So, it is proved that if a and b are two odd positive integers such that a > b then one of the two number \(\frac{ a + b }{ 2 }\) and \(\frac{ a – b }{ 2 }\) is odd and the other is even.

 

Q.2:  Prove that the product of two consecutive positive integers is divisible by 2.

 

Sol:

Consider n and (n – 1) be any two consecutive positive integers.

Now, their product will be n (n – 1) = n2 – n

It is known that every positive integer is of the form 2q or 2q + 1 for some integer q.

Now,

let n = 2q

So, n2 – n = (2q) 2 – (2q)

n2 – n = (2q) 2 – (2q)

n2 – n = 4q2 – 2q

n2 – n = 2q (2q – 1)

n2 – n = 2r [where r = q (2q – 1)]

⇒ n– n is even and divisible by 2

 

Let n = 2q + 1

So, n2 – n = (2q + 1) 2 – (2q + 1)

⇒ n2 – n = (2q + 1)   (2q + 1) – 1)

⇒ n2 – n = (2q + 1)   (2q)

⇒ n2 – n = 2r [r = q (2q + 1)]

⇒ n2 – n is even and divisible by 2

Therefore, the product of two consecutive integers is divisible by 2 (hence proved).

 

Q.3: Prove that the product of three consecutive positive integers is divisible by 6.

 

Sol:

Consider “n” as any positive integer.

We know, positive integers are in the form of 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5.

Now, if n = 6q,

⇒ n (n + 1)   (n + 2)  = 6q (6q + 1)   (6q + 2), which is divisible by 6

Now, if n = 6q + 1

⇒ n (n + 1)   (n + 2)  = (6q + 1)   (6q + 2)   (6q + 3)

i.e. n (n + 1)   (n + 2)  = 6 (6q + 1)   (3q + 1)   (2q + 1)  Which is divisible by 6

Now, if n = 6q + 2

⇒ n (n + 1)   (n + 2)  = (6q + 2)   (6q + 3)   (6q + 4)

i.e. n (n + 1)   (n + 2)  = 12 (3q + 1)   (2q + 1)   (2q + 3), which is divisible by 6.

In a similar way, others can also be proved.

So, the product of three consecutive positive integers is divisible by 6 (hence proved).

 

Q.4: For any positive integer n, prove that n3 – n divisible by 6.

 

Sol:

Consider “n” to be a positive integer.

Now, n3 — n = (n – 1)   (n)   (n + I)

We know, any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5

If n = 6q,

Then, (n —1) n (n + 1)  = (6q —1)  6q (6q + 1)

This is divisible by 6

Now, if n = 6q + 1,

Then, (n —1) n (n + 1) = (6q)   (6q + 1)   (6q + 2)

This is also divisible by 6.

Now agian if n = 6q + 2,

Then, (n – 1) n (n + 1) = (6q + 1)   (6q + 2)   (6q + 3)

(n – 1) n (n + 1) = 6 (6q + 1)   (3q + 1)   (2q + 1)

Which is again divisible by 6.

Similarly, we can prove others.

So, it is proved that for any positive integer n, n3— n is divisible by 6.

 

Q.5: Prove that if a positive integer is of form 6q + 5, then it is of the form 3q + 2 for some integer q, but not conversely.

 

Sol:

Consider n = 6q + 5

Now, any positive integer “n” is of the form of 3k or 3k + 1, 3k + 2

If q = 3k,

Then, n = 6q + 5

Here, q = 3k

⇒ n = 18k + 5

Or, n = 3 (6k + 1) + 2

n = 3m + 2 (where m = (6k + 1))

If q = 3k+ 1,

Then, n = (6q + 5)

n = (6 (3k + 1) + 5)   (q = 3k + 1)

n = 18k + 6 + 5

n = 18k + 11

n = 3 (6k + 3) + 2

n = 3m + 2 (where m = (6k + 3))

If q = 3k + 2,

Then, n = (6q + 5)

n = (6 (3k + 2) + 5)   (q = 3k + 2)

n = 18k + 12 + 5

n = 18k + 17

n = 3 (6k + 5) + 2

n = 3m + 2 (where m = (6k + 5))

For the converse prove:

Let a number, say, 8 which is the form 3q + 2 i.e. 3 x 2 + 2. Now, 8 cannot be written in the form 6q + 5. So, the converse is not true.

 

Q.6: Prove that square of any positive integer of the form 5q + 1 is of the same form.

 

Sol:

Here, n = 5q + 1

So, n= (5q + 1) 2

⇒ n= (5q) 2 + 2 (1) (5q) + 12 = 25q2 + 10q + 1

⇒ n2 = 5m + 1 (where m = (5q2 + 2q))

∴ ninteger is of the form 5m + 1.

 

Q.7: Prove that the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.

 

Sol:

Since positive integer n is of the form of 3q , 3q + 1 and 3q + 2

If n = 3q

n= (3q) 2

n= 9q2

n= 3 (3q) 2

n= 3m (m = 3q) 2

If n = 3q + 1

Then, n2 = (3q + 1) 2

n2 = (3q) 2 + 6q + 1

n2 = 9q2 + 6q + 1

n2 = 3q (3q + 1) + 1

n= 3m +1 (where m = (3q + 2) )

If n = 3q + 2

Then, n2 = (3q + 2) 2 = (3q) + 12q + 4

n2 = 9q2 + 12q + 4

n2 = 3 (3q + 4q + 1) + 1

n= 3m + 1 (where q = (3q + 4q + 1))

Hence, ninteger is of the form 3m, 3m + 1 but not of the form 3m +2.

 

Q.8: Prove that the Square of any positive integer is of the form 4q or 4q + 1 for some integer q.

 

Sol:

As positive integer n is of the form of 2q or 2q + 1

If n = 2q

Then, n= (2q) 2

n2 = 4q2

n2 = 4m (where m = q2)

If n = 2q + 1

Then, n2 = (2q + 1) 2

n2 = (2q) + 4q + 1

n2 = 4q2 + 4q + 1

n2 = 4q (q + 1) + 1

n2 = 4q + 1 (where m = q (q + 1))

So, it is proved that the square of any positive integer is of the form 4q or 4q + 1, for some integer q.

 

Q.9:  Prove that the Square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.

 

Sol:

Since positive integer n is of the form of 5q or 5q + 1, 5q + 4.

If n = 5q

Then. n= (5q) 2

n2 = 25q2

n= 5 (5q)

n2 = 5m (Where m = 5q)

If n = 5q + 1

Then, n2 = (5q +1) 2

n2 = (5q) + 10q + 1

n2 = 25q2 + 10q + 1

n2 = 5q (5q + 2) + 1

It n2 = 5q (5q +2) + 1

n2 = 5m +1 (where m = q (5q + 2))

If n = 5q + 2

Then, n= (5q + 2) 2

n2 = (5q) + 20q + 4

n2 = 25q2 + 20q + 4

n2 = 5q (5q + 4) + 4

n2 = 5m + 4 (where m = q (5q + 4))

If n = 5q + 4

Then, n= (5q + 4) 2

n2 = (5q) + 40q + 16

n2 = 25q2 + 40q + 16

n2 = 5 (5q2 + 8q + 3) + 1

n2 = 5m + 1 (where m = 5q2 + 8q + 3 )

Hence it is proved that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.

 

Q.10:  Show that the Square of odd integer is of the form 8q + 1, for some integer q.

 

Sol:

To Prove: the square of any positive integer is of the form 8q + 1 for some integer q.

Proof: Since any positive integer n is of the form 4m + 1 and 4m + 3

If n = m + 1

Then,

n2 = (4m + 1) 2

n2 = (4m) + 8m + 1

n2 = 16m2 + 8m + 1

n2 = 8m (2m + 1) + 1

n2 + 8q +1 (where q = m (2m + 1))

If n = 4m + 3

Then, n2 = (4m + 3)2

n2 = (4m)+ 24m + 9

n2 = 16m2 + 24m + 9

n2 = 8 (2m2 + 3m + 1) + 1

n2 = 8q + 1 (where q = (2m2 + 3m + 1))

Hence, ninteger is of the form 8q + 1, for some integer q.

 

Q.11:  Show that any positive odd integer is of the form 6q +1 or 6q + 3 or 6q + 5, where q is some integer.

 

Sol:

Consider a two positive integers ‘a’ and b = 6.

Now, by division algorithm there exists integers q and r such that

a = 6q + r, 0 ≤ r < 6

⇒ a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4

We know, 6q or 6q + 2 or 6q + 4 are even positive integers.

∴ a = 6q + 1 or 6q + 3 or 6q + 5 

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