Majors

PROPOSITION 39

If 2 straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial, be added together, the whole straight line is irrational : and let it be called major.

For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 33 ], be added together; I say that AC is irrational.

For, since the rectangle AB, BC is medial, twice the rectangle AB, BC is also medial. [X. 6 and 23, Por.]

But the sum of the squares on AB, BC is rational; therefore twice the rectangle AB, BC is incommensurable with the sum of the squares on AB, BC, so that the squares on AB, BC together with twice the rectangle AB, BC that is, the square on AC, is also incommensurable with the sum of the squares on AB, BC; [X. 16 ] therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 ]

let it be called major. Q. E. D.

PROPOSITION 40

If two straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational, be added together, the whole straight line is irrational; and let it be called the side of a rational plus a medial area. For let two straight lines AB, BC incommensurable in square, and fulfilling the given conditions [X. 34 ], be added together; I say that AC is irrational.

For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC; so that the square on AC is also incommensurable with twice the rectangle AB, BC. [X. 16 ]

But twice the rectangle AB, BC is rational; therefore the square on AC is irrational.

Therefore AC is irrational. [X. Def. 4 ]

And let it be called the side of a rational plus a medial area. Q. E. D.

PROPOSITION 41

If 2 straight lines incommensurable in square which make the sum of the squares on them medial, and the rectangle contained by them medial and also incommensurable with the sum of the squares on them, be added together, the whole straight line is irrational; and let it be called the side of the sum of two medial areas.

For let two straight lines AB, BC incommensurable in square and satisfying the given conditions [X. 35 ] be added together; I say that AC is irrational.

Let a rational straight line DE be set out, and let there be applied to DE the rectangle DF equal to the squares on AB, BC, and the rectangle GH equal to twice the rectangle AB, BC; therefore the whole DH is equal to the square on AC. [II. 4 ]

Now, since the sum of the squares on AB, BC is medial, and is equal to DF, therefore DF is also medial.

And it is applied to the rational straight line DE; therefore DG is rational and incommensurable in length with DE. [X. 22 ]

For the same reason GK is also rational and incommensurable in length with GF, that is, DE.

And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, DF is incommensurable with GH; so that DG is also incommensurable with GK. [VI. 1 , X. 11 ]

And they are rational; therefore DG, GK are rational straight lines commensurable in square only; therefore DK is irrational and what is called binomial. [X. 36 ]

But DE is rational; therefore DH is irrational, and the side of the square which is equal to it is irrational. [X. Def. 4 ]

But AC is the side of the square equal to HD; therefore AC is irrational.

And let it be called the side of the sum of two medial areas. Q. E. D.

LEMMA. And that the aforesaid irrational straight lines are divided only in one way into the straight lines of which they are the sum and which produce the types in question, we will now prove after premising the following lemma. Let the straight line AB be set out, let the whole be cut into unequal parts at each of the points C, D, and let AC be supposed greater than DB; I say that the squares on AC, CB are greater than the squares on AD, DB.

For let AB be bisected at E.

Then, since AC is greater than DB, let DC be subtracted from each; therefore the remainder AD is greater than the remainder CB.

But AE is equal to EB; therefore DE is less than EC; therefore the points C, D are not equidistant from the point of bisection.

And, since the rectangle AC, CB together with the square on EC is equal to the square on EB, [II. 5 ] and, further, the rectangle AD, DB together with the square on DE is equal to the square on EB, [id.] therefore the rectangle AC, CB together with the square on EC is equal to the rectangle AD, DB together with the square on DE.

And of these the square on DE is less than the square on EC; therefore the remainder, the rectangle AC, CB, is also less than the rectangle AD, DB, so that twice the rectangle AC, CB is also less than twice the rectangle AD, DB.

Therefore also the remainder, the sum of the squares on AC, CB, is greater than the sum of the squares on AD, DB. Q. E. D. 1

1 3. and which produce the types in question. The Greek is ποιουσῶν τὰ προκείμενα εἴδη, and I have taken εἴδη to mean “types (of irrational straight lines),” though the expression might perhaps mean “satisfying the conditions in question.”

PROPOSITION 42. A binomial straight line is divided into its terms at one point only . Let AB be a binomial straight line divided into its terms at C; therefore AC, CB are rational straight lines commensurable in square only. [X. 36 ] I say that AB is not divided at another point into two rational straight lines commensurable in square only.

For, if possible, let it be divided at D also, so that AD, DB are also rational straight lines commensurable in square only.

It is then manifest that AC is not the same with DB.

For, if possible, let it be so.

Then AD will also be the same as CB, and, as AC is to CB, so will BD be to DA; thus AB will be divided at D also in the same way as by the division at C: which is contrary to the hypothesis.

Therefore AC is not the same with DB.

For this reason also the points C, D are not equidistant from the point of bisection.

Therefore that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, because both the squares on AC, CB together with twice the rectangle AC, CB, and the squares on AD, DB together with twice the rectangle AD, DB, are equal to the square on AB. [II. 4 ]

But the squares on AC, CB differ from the squares on AD, DB by a rational area, for both are rational; therefore twice the rectangle AD, DB also differs from twice the rectangle AC, CB by a rational area, though they are medial [X. 21 ]: which is absurd, for a medial area does not exceed a medial by a rational area. [x. 26 ]

Therefore a binomial straight line is not divided at different points; therefore it is divided at one point only. Q. E. D.