The document RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.

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** Exercise - 1.5**

We know that √5 is an irrational number

Here we see that √5 is a rational number which is a contradiction.

Here we see that √2 is a rational number which is a contradiction as we know that √2 is an irrational number

Hence 6 + √2 is an irrational number**(iii)** Let us assume that 3 - √5 is rational. Then, there exist positive co primes a and b such that

Here we see that √5 is a rational number which is a contradiction as we know that √5 is an irrational number

Hence 3 - √5 is an irrational number.**Question: 2****Prove that the following numbers are irrationals.****(i) 2√7****(ii) 3/(2√5)****(iii) 4 + √2****(iv) 5√2****Solution:****(i) **Let us assume that 2√7 is rational. Then, there exist positive co primes a and b such that

√7 is rational number which is a contradiction

Hence 2√7 is an irrational number

**(ii)** Let us assume that 3/(2√5) is rational. Then, there exist positive co primes a and b such that

√5 is rational number which is a contradiction

Hence 3/(2√5) is irrational.

**(iii)** Let us assume that 3/(2√5) is rational. Then, there exist positive co primes a and b such that

√2 is rational number which is a contradiction

Hence 4+ √2 is irrational.

**(iv) **Let us assume that 5√2 is rational. Then, there exist positive co primes a and b such that

√2 is rational number which is a contradiction

Hence 5√2 is irrational**Question: 3****Show that 2 - √3 is an irrational number.****Solution:**

Let us assume that 2 - √3 is rational. Then, there exist positive co primes a and b such that

Here we see that √3 is a rational number which is a contradiction

Hence 2- √3 is irrational**Question: 4****Show that 3 + √2 is an irrational number.****Solution:**

Let us assume that 3 + √2 is rational. Then, there exist positive co primes a and b such that

Here we see that √2 is a irrational number which is a contradiction

Hence 3 + √2 is irrational**Question: 5****Prove that 4 - 5√2 is an irrational number.****Solution:**

Let us assume that 4 - 5√2 is rational. Then, there exist positive co primes a and b such that

This contradicts the fact that √2 is an irrational number

Hence 4 – 5√2 is irrational**Question: 6****Show that 5 - 2√3 is an irrational number.****Solution:**

Let us assume that 5 -2√3 is rational. Then, there exist positive co primes a and b such that

This contradicts the fact that √3 is an irrational number

Hence 5 – 2√3 is irrational**Question: 7****Prove that 2√3 - 1 is an irrational number.****Solution:**

Let us assume that 2√3 – 1 is rational. Then, there exist positive co primes a and b such that

This contradicts the fact that √3 is an irrational number

Hence 5 - 2√3 is irrational**Question: 8****Prove that 2 - 3√5 is an irrational number.****Solution:**

Let us assume that 2 - 3√5 is rational. Then, there exist positive co primes a and b such that

This contradicts the fact that √5 is an irrational number

Hence 2 - 3√5 is irrational**Question: 9****Prove that √5 + √3 is irrational.****Solution:**

Let us assume that √5 + √3 is rational. Then, there exist positive co primes a and b such that

Here we see that √3 is a rational number which is a contradiction as we know that √3 is an irrational number

Hence √5 + √3 is an irrational number**Question: 10****Prove that √3 + √4 is irrational.****Solution:**

Let us assume that √3 + √4 is rational. Then, there exist positive co primes a and b such that

Here we see that √3 is a rational number which is a contradiction as we know that √3 is an irrational number

Hence √3 + √4 is an irrational number**Question: 11****Prove that for any prime positive integer p, √p is an irrational number.****Solution:**

Let us assume that √p is rational. Then, there exist positive co primes a and b such that

⇒ pb^{2} = a^{2}

⇒ p|a^{2} ⇒ p|a^{2}

⇒ p|a ⇒ p|a

⇒ a = pc for some positive integer c ⇒ a = pc for some positive integer c

b^{2}p = a^{2}

⇒ b^{2}p = p^{2}c^{2} ( ∵ a = pc)

⇒ p|b^{2} (since p|c^{2}p) ⇒ p|b^{2}(since p|c^{2}p)

⇒ p|b ⇒ p|a and p|b

This contradicts the fact that a and b are co primes

Hence √p is irrational**Question: 12****If p, q are prime positive integers, prove that √p + √q is an irrational number.****Solution:**

Let us assume that √p + √q is rational. Then, there exist positive co primes a and b such that

Here we see that √q is a rational number which is a contradiction as we know that √q is an irrational number

Hence √p + √q is an irrational number** Exercise - 1.6**

**Question: 1****Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.****(i) 23/8****(ii) 125/441****(iii) 35/50**** (iv) 77/210****(v) ****Solution:****(i) **The given number is 23/8

Here, 8 = 2^{3} and 2 is not a factor of 23.

So, the given number is in its simplest form.

Now, 8 = 2^{3} is of the form 2^{m} × 5^{n}, where m = 3 and n = 0.

So, the given number has a terminating decimal expansion.**(ii) **The given number is 125/441

Here, 441 = 3^{2} × 7^{2} and none of 3 and 7 is a factor of 125.

So, the given number is in its simplest form.

Now, 441 = 3^{2} × 7^{2} is not of the form 2^{m} × 5^{n}

So, the given number has a non-terminating repeating decimal expansion.**(iii)** The given number is 35/50 and HCF (35, 50) = 5.

Here, 7/10 is in its simplest form.

Now, 10 = 2 × 5 is of the form 2^{m} × 5^{n}, where in = 1 and n = 1.

So, the given number has a terminating decimal expansion.**(iv)** The given number is 77/210 and HCF (77, 210) = 7.

Here, 11/30 is in its simplest form. 30

Now, 30 = 2 × 3 × 5 is not of the form 2^{m} × 5^{n}.

So, the given number has a non-terminating repeating decimal expansion.**(v) **The given number is Clearly, none of 2, 5 and 7 is a factor of 129.

So, the given number is in its simplest form.**Question: 2****Write down the decimal expansions of the following rational numbers by writing their denominators in the form of 2 ^{m} × 5^{n}, where m, and n, are the non- negative integers.**

(v)

Solution.

Clearly, 8 = 2

So, the given number has terminating decimal expansion.

Clearly, 125 = 5

So, the given number has terminating decimal expansion.

Clearly, 80 = 2

So, the given number has terminating decimal expansion.

Clearly, 625 = 5

So, the given number has terminating decimal expansion?

So, the given number has terminating decimal expansion?

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