Question

prove that √3 is an irrational no​

Answer 1

Answer:

• Let √3 be a rational number
• Hence, Let √3=p/q {where p&q are integers and q not equal to 0}
• √3q=p
• (√3q)^2=p
• 3q^2=p
• 3 divides p^2--------(eq 1)
• 3 divides p
• P=3K for some integer
• 3q^2=(3k)^2
• 3q^2=9k^2
• q^2 =3k^2
• 3 divides q ---------(eq 2)

From eq 1 and eq 2 we know that √3 is not rational but irrational.

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Answer 2

Answer

Let us assume that √3 is a rational number

√3  = p/q     (where p and q are integers and co- prime to each other)

So

√3  = p/q

=>  $\sqrt{3} q = p$

Squaring Both sides

$(\sqrt{3} q)^2 = p^2$

=>  $3q^2 = p^2 ........ ( i )$

So

If 3 is the factor of p²

then, 3 is also a factor of p ............. ( ii )

=> Let p = 3r { where r is any integer }

Squaring both sides

$p^2 = (3r)^2$

=> $p^2 = 9r^2$

Now

Putting the value of p² in equation ( i )  We get

$3q^2 = p^2$

=> $3q^2 = 9r^2$

=> $q^2= 3r^2$

So

If 3 is factor of q²

then, 3 is also factor of q

Since

P & Q have one common factor 3

So our assumption was totally wrong

Hence

√3 is an irrational number (PROVED)