Question

prove that rote 5 is an irrational
​

Answer 1

Step-by-step explanation:

Let assume that √5 is a rational number.

Then,

√5 is written in the form of p/q

It means,

√5 = p/q [ where p and q are integers and q ≠ 0 ]

Now,

→ √5 = p/q

Squaring both sides :-

→ (√5)² = (p/q)²

→ 5 = p²/q²

→ 5q² = p² ..... (i)

From here, we can see 5 divides p and 5 also divides p².

Now,

Let p = 5c ( where, c is some integer )

→ 5q² = (5c)²

→ 5q² = 25c²

→ q² = 5c² ..... (ii)

From here, we can see 5 divides q and 5 also divides q².

Thus,

5 is a common factor of p and q.

So,

Our assumptions is wrong

Hence,

√5 is an irrational number.

Answer 2

Answer:

$\sf \pink{Let \: assume \: that \: √5 \: is \: a \: rational \: number. }$

$\bf \red{Then, }$

√5 is written in the form of p/q

It means,

√5 = p/q [ where p and q are integers and q ≠ 0 ]

$\bf \gray{Now, }$

→ √5 = p/q

Squaring both sides :-

→ (√5)² = (p/q)²

→ 5 = p²/q²

→ 5q² = p² ..... (i)

From here, we can see 5 divides p and 5 also divides p².

$\huge \sf \blue{Now, }$

Let p = 5c ( where, c is some integer )

→ 5q² = (5c)²

→ 5q² = 25c²

→ q² = 5c² ..... (ii)

From here, we can see 5 divides q and 5 also divides q².

Thus,

5 is a common factor of p and q.

So,

Our assumptions is wrong

$\tt \pink{Hence, }$

√5 is an irrational number.