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Proofs of Irrationality

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of Squareroot of 2


Irrational numbers are numbers that cannot be written in the form of `p/q`, where p and q are integers and q ≠ 0. Example `sqrt3,sqrt2`, π.


1)Theorem: let p be a prime number. If P divides `a^2` the p divides a, where a is a positive integer.

Proof: If `a^2/p` true then `a/p` is also ture.

a=`(p_1p_2...p_n)` as per fundamental theorem, this means

`a^2`= `(p_1p_2..p_n)^2`

= `p_1^2 p_2^2..p_n^2`

given: p divides `a^2`

that means p must belong to pi ,where i lies between 1 to n

Thus, is a factor of a also.

2)Theorem: `sqrt2` is irrational

Proof: Let's assume `sqrt2` is rational. If `sqrt2` is rational the `sqrt2= a/b`, where a and b are co prime.

`sqrt2= a/b`

b `sqrt2`= a

`2b^2`= `a^2` That means 2 divides `a^2`. And hence 2 is prime number it also divides a.

2×k=`a^2` (k is constant)

2×k= a .......eq1

2`b^2`= 2`k^2`= 4`k^2`

`b^2`= 2`k^2`, Therefore 2 divides `b^2`, and since 2 is prime  number it also divides b.

That means 2 is factor of a,b but according to our assumption a and b are co prime which

means they dont have common factors. Here our assumption that `sqrt2` is rational is

proven incorrect, thus sqrt is a irrational number.


If you would like to contribute notes or other learning material, please submit them using the button below. | Real Numbers part 8 (Irrational Numbers)

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