Page:Carroll - Euclid and His Modern Rivals.djvu/212

174 to both members the part MTSRQD, the result will be MDQRSTM less than MPQRSTM.' Is that result proved?

Nie. No.

Min. Is it true?

Nie. Not necessarily so.

Min. Perhaps it is a lapsus pennœ. Try to amend it.



Nie. If we add to MD the part MTSRQD, we do get MDQRSTM, it is true: but, if we add it to MPQD, we get QD twice over; that is, we get MPQRSTM together with twice QD.

Min. How does that addition suit the rest of the proof?

Nie. It ruins it: all depends on our proving the perimeter MDQRSTM less than the perimeter MPQRSTM, which this method has failed to do—as of course all methods must, the thing not being capable of proof.

Min. Then the whole proof breaks down entirely?

Nie. We cannot deny it.

Min. Let us turn to the next author.