Question

show tjat √2 is an irrational number​

Answer 1

Answer:

you can use 3 methods - 1st : long division method , 2nd : prove by contradiction and 3rd : by finding factors of 2.

Explanation:

1st method : √2=1.4142135.... as shown in photograph that by finding sq root of 2 the decimal representation is neither terminating nor repeating, so √2 is a irrational no.

hence, √2 is a irrational no.

3rd method : if all posible factors of a no. are even in count then no. is rational otherwise not.

for eg- for √2 we will find factors of 2 =2×1 , there is no other pair of factor 2(no. of pair is odd) so √2 is irrational.

eg. 2- for √4 we will find factors of 4,4=2×2 ,there are 2 pairs (no.of pair is even) so √4 is rational.

hence, √2is a irrational no.

2nd method : prove by contradiction

in this method we asume that any irrational no. is rational and then prove our assumed statement wrong.

let us assume that √2 is rational no. then,

√2=p/q where p and q have no common factor and q=/=0

to remove under root from 2 we multiply both side by 2 because √2 is a endless no.

2=p^2/q^2

p^2=2q^2 ...(i)

p^2 is even because anything multiplied by 2 is even so p is also even because even no. sq is always even and odd no. sq is always odd.

now let p be a integer = 2m, then

p^2=4m^2

2q^2=4m^2 ...(using i)

q^2=2m^2

now q^2 is also a even integer

so q is also a even integer

so both p and q are even integers and therefore have a common factor 2. but ., this contradicts p and q have no common factor.

hence,√2 is a irrational no.