How to demonstrate this?We have equilateral trinagle `DeltaOAB`:`O_(((0,0))),A_(((m,n)));m,ninNN;m,n!=0` and`B_(((x,y)));x,yin(0,oo)`.Demonstrate that `B` can not have both coordinates natural numbers.

Let `R(pi/3)=((Cos(pi/3), -Sin(pi/3)),(Sin(pi/3), Cos(pi/3)))= ((1/2,-sqrt3/2),(sqrt3/2,1/2))`

be the rotation matrix which rotates `pi/3` radians.

Having a vertice at the origin, If `(m,n)` is an equilateral triangle vertice then the other vertice is located at

`((x),(y)) = R(pmpi/3)cdot((m),(n))`

Taking the counterclockwise rotation to follow we have

`((x),(y)) = R(pi/3)cdot((m),(n))=((m/2 - (sqrt[3] n)/2), ((sqrt[3] m)/2 + n/2) )`

so clearly `((x),(y))` can not be represented as an integer couple.

Note that `nm ne 0`

We use Reductio Ad Absurdum to prove the Result.

For this, suppose, to the contrary, that, there exists an equilateral

`DeltaOAB,` having vertices `O(0,0), A(m,n), B(x,y),` such that,

`m,n,x,y in NN, and, m^2+n^2!=0.`

Knowing that, Area `A_(Delta)`of an equilateral `DeltaOAB` is given by,

`A_(Delta)=sqrt3/4*OA^2=sqrt3/4(m^2+n^2)..........(ast).`

On the other hand, from Co-ordinate Geometry,

`A_(Delta)=1/2|D|,` where,

`D=|(0,0,1),(m,n,1),(x,y,1)|=|my-nx|,`

`:. A_(Delta)=1/2|my-nx|............(ast').`

`(ast), &, (ast') rArr 2|my-nx|/(m^2+n^2)=sqrt3,` which is a

contradiction, because, with, `m,n,x,y in NN, &, m^2+n^2ne0,`

the L.H.S. is a Natural No., whrereas, the R.H.S., an Irrational No.

Therefore, our supposition that all the co-ordinates of an equilateral

triangle are Natural Nos. is False.

Henc, the Proof.