Math, asked by gangarapuraj, 2 months ago

please answer this three then I will mark you as a brainliast​

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Answered by srikantadas67940
1

( 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

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