Question

Prove that√3 is an irrational number hence prove 3-√5 is also an irrational number?

Answer 1

let us assume that √3 is a rational number. and hence let's represent √3 in the form of p/q where p and q are vo primes.
therefore,
√3=p/q
squaring on both sides,
3= p^2/q^2
3q^2=p^2
Thus, 3 divides p^2 and if a number divides the square of a number, it also divides it's square root. hence 3 divides p
3a = p where a is any number which when multiplied to 3 gives p.
3a=p
sq. on both sides,
9a^2=p^2
substitute this in first equation
3q^2=9a^2
q^2=3a^2
thus, 3 divides q^2 and hence it also divided q.

Thus means 3 is a common factor between p and q which contradicts the fact that o and q must be co primes. hence our assumption is wrong and hence √3 is an irrational number.

prove √5 irrational in the similar way as shown up

then assume that 3-√5 is irrational and can be represented as p/q form where p and q are co primes.

3-√5= p/q
-√5=p/3q
√5= -p/3q
√5= -(integer)/(integer)(integer)
according to this, √5 is rational which contradicts our proof that √5 is irrational

hence our assumption that 3-√5 is rational is wrong

hence 3-√5 is irrational.

hope this helps!

Answer 2

Step-by-step explanation:

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