To find the first apotome.

Set out a rational straight line, and let *BG* be commensurable in length with *A*. Then *BG* is also rational.

Set out two square numbers *DE* and *EF*, and let their difference *FD* not be square. Then *ED* does not have to *DF* the ratio which a square number has to a square number.

Let it be contrived that *ED* is to *DF* as the square on *BG* is to the square on *GC*. Then the square on *BG* is commensurable with the square on *GC*.

But the square on *BG* is rational, therefore the square on *GC* is also rational. Therefore *GC* is also rational.

Since *ED* does not have to *DF* the ratio which a square number has to a square number, therefore neither has the square on *BG* to the square on *GC* the ratio which a square number has to a square number. Therefore *BG* is incommensurable in length with *GC*.

And both are rational, therefore *BG* and *GC* are rational straight lines commensurable in square only. Therefore *BC* is an apotome.

I say next that it is also a first apotome.

Let the square on *H* be that by which the square on *BG* is greater than the square on *GC*.

Now since *ED* is to *FD* as the square on *BG* is to the square on *GC*, therefore, in conversion, as *DE* is to *EF* as the square on *GB* is to the square on *H*.

But *DE* has to *EF* the ratio which a square number has to a square number, for each is square, therefore the square on *GB* also has to the square on *H* the ratio which a square number has to a square number. Therefore *BG* is commensurable in length with *H*.

And the square on *BG* is greater than the square on *GC* by the square on *H*, therefore the square on *BG* is greater than the square on *GC* by the square on a straight line commensurable in length with *BG*.

And the whole *BG* is commensurable in length with the rational straight line *A* set out.

Therefore *BC* is a first apotome. Therefore the first apotome *BC* has been found.

Q.E.F.