Question

Prove that√3 is an irrational number and hence prove that 2+√3 is also an irrational number.

Answer 1

Let us assume that √3 is a rational number

So,

√3=a/b

by cross cut the common numbers we got

√3=p/q

so here p and q co-prime

q√3=p

sq. both the sides

3q²=p²..........(1)

As 3 divides p²

so it is also divides p

now

p=3c

for c is any integer

now by sq.both the sides

p²=9c²

from (1)

3q²=9c²

q²=3c²

As 3 divides q²

so it is also divides q

Now 3 is divided by p and q both but we assume that they are co-prime

this had arise a condradiction because our intial assumption is wrong that √3 is rational so √3 is irrational

In second part

take

2+√3=p/q

√3=(p/q)+2

√3=(p+2q)/q

by cross cuit the common numbers we got

√3=a/b

and then follow the above steps

and if it is low number question then you cqan directly state that √3 is irrational

                    

HOPE ITS HELPFUL

Answer 2

Answer:

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