In the first row (0 + 1)/(1 + 1) a third of the quantities below is (2)/(3), set in order these three quantities 1 (2)/(9) (2)/(3). The first is 1, equal to the sum above, the last is (2)/(3), equal to the subtriple of the sum below. The middlemost is (2)/(9) subtriple to the last quantity (2)/(3). The excess of the proportion of 1 to (2)/(3) above the subtriple proportion of (2)/(9) to (2)/(3) is the proportion of 1 to (2)/(9) that is of 9 to 2, that is, of 18 to 4. Secondly, in the second row, which is (0 + 1 + 4)/(4 + 4 + 4), a third of the sum below is 4, the sum above is 5. Set in order these quantities, 1, 5, 4, 12. There the proportion of 15 to 12 is the proportion of 5 to 4. The proportion of 4 to 12 is subtriple; the excess is the proportion of 15 to 4, which is less than the proportion of 18 to 4, as it ought to be; but not less by the proportion of (1)/(6) to (1)/(12) as you would have it. Thirdly, in the third row, which is (0 + 1 + 4 + 9)/(9 + 9 + 9 + 9). A third of the sum below is 12, the sum above is 14. Set in order these quantities, 42, 4, 12. There the proportion of 42 to 12 is the same with that of 14 to 4. And the proportion of 4 to 12 subtriple, less than the former excess of 15 to 4. And so it goes on decreasing all the way in this manner, 18 to 4, 15 to 4, 14 to 4, &c. which differs very much from your 1 to 6, 1 to 12, 1 to 18, &c. and the cause of your mistake is this: you call the twelfth part of twelve (1)/(12), and the eighteenth part of thirty-six you call (1)/(18), and so of the rest. But what need of all those equations in symbols, to show that the proportion decreases; is there any man can doubt, but that the proportion of 1 to 2 is greater than that of 5 to 12, or that of 5 to 12 greater than that of 14 to 36, and so on continually forwards; or could you have fallen into this error, unless you had taken, as you have done in very many places of your _Elenchus_, the fractions (1)/(6) and (1)/(12), &c. which are the quotients of 1 divided by 6 and 12, for the very proportions of 1 to 6 and 1 to 12. But notwithstanding the excess of the proportions of the increasing quantities, to subtriple proportion decrease, still, as the number of terms increaseth, and that what proportions soever I shall assign, the decrement will in time (in time, I say, without proceeding _in infinitum_) produce a less, yet it does not follow that the row of increasing quantities shall ever be equal to the third part of the row of so many equals to the last or greatest. For it is not, I hope, a paradox to you, that in two rows of quantities the proportion of the excesses may decrease, and yet the excesses themselves increase, and do perpetually.