Question

prove that root 3 is a irrational​

Answer 1

Answer:

Answers

ALTAF11

ALTAF11Maths AryaBhatta

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

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

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

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:Let us assume root 3 be rational, root 3 = a/ b,b not equal to 0,b€z,(a,b) = 1, root 3b= a, square on both sides 3b^2 =a^2[ equation one], 3 divides a^2 ,3 divides a( 3 is prime)[equation two]. For some integer C .  3*c=a, square on both sides 9c^2=a^2,3c^2=b^2[ from equation one](9c^2=3b^2), then 3 divides b^2 ,3 divides b( 3 is prime)[equation three] From two and three 3 is the common factor of a and b. Therefore, our assumption is  wrong and root is a irrational