Question

Show that root 5 are not rational number by contradiction method

Answer 1

Answer:

Let us assume √5 is a rational number which can be written in the form p/q, where p&q are integers, q≠0, both p&q have common factor 1.

Step-by-step explanation:

√5=p/q

Squaring both the sides

5=$p^{2}$/$q^{2}$

Therefore, $p^{2}=5q^{2}+0$

5 is a factor of $p^{2}$

5 is a factor of $q^2$

Therefore, p is the multiple of 5

p=5m (where m is any integer)

Squaring both sides

$p^{2}$=25m^2

5$q^2$=25m^2....(from 1)

$q^2$=5m^2+0

5 is the factor of $q^2$

5 is the factor of q

Therefore, 5 is the common factor of p&q which contradicts our assumption

Therefore, √5 is irrational

Answer 2

Answer:

Let us assume √5 is a rational number which can be written in the form p/q, where p&q are integers, q≠0, both p&q have common factor 1.

Step-by-step explanation:

√5=p/q

Squaring both the sides

5=p^{2}p

2

/q^{2}q

2

Therefore, p^{2}=5q^{2}+0p

2

=5q

2

+0

5 is a factor of p^{2}p

2

5 is a factor of q^2q

2

Therefore, p is the multiple of 5

p=5m (where m is any integer)

Squaring both sides

p^{2}p

2

=25m^2

5q^2q

2

=25m^2....(from 1)

q^2q

2

=5m^2+0

5 is the factor of q^2q

2

5 is the factor of q

Therefore, 5 is the common factor of p&q which contradicts our assumption

Therefore, √5 is irrational