V3 = 1. 7 3 2 0 5 0

Let us assume that root 3 is rational.therefore, It cab be express in the form a/b there a and b are co-prime. Now,root5 = a/broot5b=a(root5b)sq=a Squaretherefore, a square is a multiple of 5a is also a multiple of 5let a=5c for any ctherefore,=) 5b square= 5c square.=) bsquare= 5c squaretherefore b is also multiple of 5therefore a and b has a common factor 5 which is contradiction as a and b are coprimetherefore root5 is irrational

v3= 1.7 3 2 0 5 0 that might work

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