第 10 类 NCERT 解决方案 - 第 1 章实数 - 练习 1.3

问题 1.证明 √5 是无理数。

解决方案：

Let √5 be a rational number.

√5 = p/q be a rational number, where p and q are co-primes and q≠0.

Then, √5q = p

=> 5q² = p² (by, squaring both the sides) ….(i)

Therefore, 5 divides p², and according to the theorem of rational number, for any prime number p which divides a² also divides a. So, we can write

p = 5k

Putting the value of p in equation (i), we get

5q² = (5k)²

5q² = 25k²

Dividing by 25,

q²/5 = k²

Similarly, we can conclude that q will be divisible by 5, and we already know that p is divisible by 5

But p and q are co-prime numbers. So there is a contradiction and it is because of the wrong assumption we made in the first place. 

√5 is not a rational number, it is irrational.

问题 2.证明 3+2√5 是无理数。

解决方案：

Let 3+2√5 be a rational number.

i.e., 3+2√5= p/q be a rational number, where p and q are co-primes and q≠0.

Subtracting 3 from both sides,

2√5=p/q-3

2√5=(p-3q)/q

Now dividing both the sides by 2, we get

√5=(p-3q)/2q

Here, p and q are integers so (p-3q)2q is a rational number. It implies, √5 should be a rational number but √5 is an irrational number and hence, there is a contradiction.

It is because of the wrong assumption.

 3+2√5 is an irrational number.

问题 3.证明以下是非理性的。

(i) 1/√2 (ii) 7√5 (iii) 6+√2

解决方案：

(i) Let us assume that 1/√2 is a rational.

 Therefore, there exist co-prime integers p and q (q ≠ 0) such that

1/√2=p/q ⇒ √2=q/p

Since p and q are integers, we get q/p is rational and so √2 is rational.

But this contradicts the fact that √2 is irrational.

This contradiction has arisen because of our incorrect assumption that 1/√2 is rational.

Hence, 1/√2 is irrational.

(ii) Let us assume that 7√5 is a rational.

Therefore, there exist co-prime integers p and q (q ≠ 0) such that

7√5=p/q ⇒ √5=q/7p

Since p and q are integers, we get q/7p is rational and so √5 is rational.

But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our incorrect assumption that 7√5 is rational.

Hence, 7√5 is irrational.

(iii) Let us assume that 6+√2 is a rational.

Therefore, there exist co-prime integers p and q (q ≠ 0) such that

6+√2 = p/q ⇒ √2 = q/p – 6

Since p and q are integers, we get q/p – 6 is rational and so √2 is rational.

But this contradicts the fact that √2 is irrational.

This contradiction has arisen because of our incorrect assumption that 6+√2 is rational.

Hence, 6+√2 is irrational.