Question

prove that √5 is irrational​

Answer 1

➠ AnSwer :-

To prove :

• √5 is irrational number

Let us assume that √5 is rational

Then √5 = a/b

(a and b are co primes, with only 1 common factor and b≠0)

⇒ √5 = a/b

(cross multiply)

⇒ a = √5b

⇒ a² = 5b² -------> α

⇒ 5/a²

(by theorem if p divides q then p can also divide q²)

⇒ 5/a ----> 1

⇒ a = 5c

(squaring on both sides)

⇒ a² = 25c² ----> β

From equations α and β

⇒ 5b² = 25c²

⇒ b² = 5c²

⇒ 5/b²

(again by theorem)

⇒ 5/b-------> 2

we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong.

This contradiction arises because we assumed that √5 is a rational number

∴ our assumption is wrong

∴ √5 is irrational number

Answer 2

${\huge{\star{\underline{\underline{\mathtt{\red{Question}}}}}}}{\huge{\star}}$

✳Prove that $\sqrt{5}$ is irrational⭐

$\huge\underline\matbb{\red♡A\pink{N}\green{S}\orange{W}\purple{E}\blue{R:}}$

➡Suppose,$\sqrt{5}$ is a rational number

Then, $\sqrt{5}$ can be expressed in the form $\frac{a}{b}$ , where a and b are integers which have no common factor other than one and b is not equal to 0

➡therefore, $\sqrt{5}$ = $\frac{α}{b}$

☆On squaring both sides we get.

➡5= $\frac{a²}{b²}$

➡a²= 5b².............(I)

⭐Here 5 divides a²

5 divides a....... (By their am).... (II)

➡So, we can take a=5m

a²= 25m²(on squaring both sides)

➡On putting the value of a² in equation(i) we get

25m²=5b²

5m²=b²

⭐Here, 5 divides b²

5divides b........ (by theoram) (iii)

➡Thus, from equation (II) 5 divides a and from equation (iii) 5 divides b. It means 5 is a common factor of a and b

➡This contradicts that there is no common factor of a and b.

➡this contradiction arises by our wrong assumption.

⚡Hence, $\sqrt{5}$ is irrational number.

$\mathcal{\large{\fbox{Verified}}}$

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