please answer this three then I will mark you as a brainliast
Answers
( 1 ) Answer:
let us assume that √2 be a rational number
so , √2 = a/b ( where a and b are co prime and b≠0)
now, squaring on both side :-
2= a^2/b^2
b^2 = a^2/2
since, a^2 is divisible by 2
so, a also divisible by 2
NOW, let a=2m
so, b^2=4m^2/2
b^2 = 2m^2
m^2 = b^2/2
since, b^2 is divisible by 2
so, b is also divisible by 2.
SINCE, 2 is the common factor of both a and b
but this contradicts our assumption that a and b are co prime ...
therefore our assumption is incorrect that √2 is a rational number
so, √2 is an irrational number ...
PROVED
(2) answer :-
If the number 4n , for any natural number , were to end with the digit zero then it would be divisible by both 2 and 5 i.e. it would contain both 2 and 5 as it's prime factors
but this is not possible because 4^n = (2^2)n ; 4n only contain 2 as it's prime factor ...
that's why there are no any number for which 4n end with the digit zero