EMAT 6600

Problem Solving


 

 

 


Square root of 2

Problem:

Prove that the Square Root of 2

is an irrational number.

The proof of 2 is very well known. It is however a powerful proof.

Proof: Assume that 2 is rational. That is 2 can be written in the form of a fraction such as a/b, where a and b have no factor other than 1 in common.

sqrt(2) = (a/b)

2 = (a(^2)/(b^2))

2(b^2) =( a^2)

( b^2) = ((a^2) /2) indicates that (a^2) is even, for it is divisible by two

the square of an even number is even

consider 2k, where k is an integer number

((2k)^2) = 4(k^2)

= 2(2(k^2)) which is even

the square of 2k + 1, on the other hand, which is an odd number, is odd

((2k + 1)^2) = 4(k^2) + 4k + 1

= 2(2(k^2) + 2k) + 1 which is odd

So, let a = 2n then (a^2) = 4(n^2) for an integer number n

thus,

(b^2) = (4(n^2) /2)

(n^2) = ((b^2)/2) which implies that b is an even number and is divisible by 2

we have a contradiction, for our assumption was that a and b have no factor in common other than 1.

This proof holds for every non-perfect square positive integer. That is because every integer can be written in form of a/b in its lowest term.

Consider

is irrational

Proof:

sqrt(15) = (a/b)

15 = (a(^2)/(b^2))

(15b^2) =( a^2)

( b^2) = (( a^2) /15) indicates that (a^2) is divisible by 15

 

So, let a = 15n then (a^2) = 225(n^2) for an integer number n

thus,

(b^2) = ((225n^2) /15)

(n^2) = ((b^2)/(15^2)) = (b/15) which implies that b is divisible by 15

we have a contradiction, for our assumption was that a and b have no factor in common other than 1


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