EWD 1231

The arithmetic and geometric means once more

In the following,  x, y, c  are positive.

In EWD1140, I used

(0)         (x+y)² = (x−y)² + 4 · x · y

to argue that

(1)         x,y = c, x+y−c       ,

which does not change x+y, increases x · y provided c lies between the initial values of x and y.

In EWD1171, I used (0) to argue that

(2)         x,y := c, x · y/c       ,

which does not change x · y, decreases x+y provided c lies between the initial values of x and y.

In both cases the use of (0) came a little bit as a rabbit and the link between the condition on c and the decrease of the distance between x and y remained informal. Last Thursday, when I asked for an expression that contained both x+y and x · y, my [student] An Thai Nguyen suggested that we look at (c−x)·(c−y), and this expression indeed plays a central role in the derivations from which all rabbits have been removed.

*         *         *

We want to change x,y such that (i) their sum is not changed, and (ii) their product is increased. Any assignment satisfying (i) can be written like (1); in order to satisfy (ii) we now observe for any c

        “(1) increases x · y
=         {program semantics, (1)}
        x · y < c ·(x+y−c)
=         {algebra}
        (c−x)·(c−y) < 0      .

We now consider the change of x,y such that (iii) their product is not changed, and (iv) their sum is decreased. Any assignment satisfying (iii) can be written like (2); in order to satisfy (iv) as well, we observe for any positive c

        “(2) increases x+y
=         {program semantics, (2)}
        c + x · y/c < x+y
=         { c > 0 }
        c² + x · y < c ·(x+y)
=         { algebra }
        (c−x)·(c−y) < 0      .

So, in both cases, the completely forced calculations lead in exactly the same form to the conclusion that c should lie between the initial values of x and y. The secret is that Nguyen’s expression can be rewritten as

        x · y − c ·(x+y−c)

and as

        (c² + x · y) − (c · x + c · y)       ,

i.e. the difference of two products with equal sums of their factors, and the difference of two sums with equal products of their addenda. I was surprised.

Austin, 10 February 1996

prof.dr. Edsger W.Dijkstra
Department of Computer Sciences
The University of Texas at Austin
Austin, TX 78712-1188, USA