मराठी

Prove that sqrt(5) is an irrational number. - Mathematics

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प्रश्न

Prove that `sqrt(5)` is an irrational number. 

सिद्धांत
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उत्तर १

Let us assume, to the contrary, that `sqrt5` is irrational.

That is, we can find integers a and b (≠0) such that `sqrt5 = a/b`.

Suppose a and b have a common factor other than 1, then we can divide by the common factor and assume that a and b are coprime.

Therefore, b`sqrt5` = a

Squaring on both sides, 5b2 = a2  ...(1)

The above implies that a2 is divisible by 5 and also a is divisible by 5.

Therefore, we can write that a = 5k for some integer k.

Substituting in (1)

5b2 = (5k)2

5b2 = 25k2

b2 = 5k2

⇒ b2 is divisible by 5, which means b is also divisible by 5.

Therefore, a and b have 5 as a common factor.

This contradicts the fact that a and b are coprime.

We arrived at the contradictory statement as above since our assumption `sqrt5` is not correct.

Hence, we can conclude that `sqrt5` is an irrational number.

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उत्तर २

If possible, let `sqrt5` be rational and let its simplest form be `p/q`.

Then p and q are integers having no common factor other than 1, and q ≠ 0.

Now `sqrt5 = p/q`

⇒ 5 = `p^2/q^2`    ...[On squaring both sides]

⇒ 5q2 = p2  ...(i)

⇒ 5 divides p2 ⇒ 5 divides p  ...[∵ 5 is prime and 5 divides p2 ⇒ 5 divides p]

Let p = 5R for some integer R.

Putting p = 5R in (i), we get

5q2 = 25R2

⇒ q2 = 5R2

⇒ 5 divides q2  ...[∵ 5 divides 5R2]

⇒ 5 divides q    ...[∵ 5 divides q]

Thus, p and q share a common factor of 5. However, this contradicts the fact that p and q share no common component other than 1.

Hence, `sqrt5` is irrational.

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उत्तर ३

Let `sqrt(5)` be a rational number.

∴ `sqrt(5) = p/q`,

where p and q are co-prime integers and q ≠ 0

On squaring both the sides, we get

`5 = p^2/q^2`

⇒ p2 = 5q2

∵ p2 is divisible by 5

∴ p is divisible by 5

Let p = 5r for some positive integer r,

p2 = 25r2

∴ 5q2 = 25r2

⇒ q2 = 5r2

∵ q2 is divisible by 5

∴ q is divisible by 5.

Here, p and q are divisible by 5, which contradicts the fact that p and q are coprimes.

Hence, our assumption is false.

∴ `sqrt(5)` is an irrational number.

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Real Numbers - Exercise 1.3 [पृष्ठ १४]

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पाठ 1 Real Numbers
Exercise 1.3 | Q 1 | पृष्ठ १४
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Exercise 1B | Q 2. | पृष्ठ १२
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