# The Elements of a New Method of Reasoning in Geometry

Elements

of

a New Method of Reasoning

in

Geometry:

applied to the rectification of the circle.

by Thomas Taylor

*–Nec altum sapiamus, nec ultra sobrium, sed veritatem in charitate colamus.*

Bacon.

London:

Printed for the Author; and Sold by J. Denis and Son, № 2, New Bridge-Street, Black-Friars. MDCCLXXX.

[Price Two Shillings and Six-pence.]

Page 7. Line 5. *Instead of* “the X Arch,” *read* “the Arch,”

ADVERTISEMENT

The Author of the following small Tract is not ashamed to confess, that it has been the employment of his leisure hours for a considerable time. If he has failed in the execution, he can, however, safely affirm he has not been wanting in the most earnest endeavours towards the completion of his purpose. He considered that the object of his search and enquiry, although arduous, was at the same time glorious, and that the Discovery of Truth is always a sufficient recompense for the difficulty attending its Investigation. As he cannor, by the most impartial scrutiny, detect any false Reasoning in his Demonstrations, he flatters himself they will not be found altogether destitute of support, nor wholly unworthy the {iv} approbation of the Public. However, sensible of his own weakness, he would not too confidently presume on success, since in this case the desire is not sufficient to obtain.

In short, animated by a sincere Love of Truth, he flatters himself the integrity of his Intentions will in some measure atone for his want of greater genius and abilities.

Indeed, as an elegant Writer observes, since Mediocrity is now become a Proection, he has probably obtained that protection in spite of himself. He only adds (with the same Author) and would with the Reader to remember, that while Error sinks into the abyss of forgetfulness, Truth alone swims over the vast extent of ages.

## The Elements, &c.

### Lemma.

IF there be any three Quantities, X, Y, Z, the former of which X, is greater than either of the other two, Y, Z, then it cannot be said *Z* is greater than *Y*, by the excess of X above Y; but if it is at all greater, is must be by some Quantity less than the excess of *X* above Y.

For if it is denied, *Z* shall be equal to *X*, which is absurd.

*Fig*. I.

If from the same Centre *s*, any two Arches *tR*, *a<sub>x/sub>*, are described as described as in the Figure, and with the Radius *s a*, of the greatest *a<sub>x/sub>*, {2} the Arch *t z* is drawn touching the Arch *R t*, in the Point t*,* then the Arch *R t*, shall not be greater than the Arch *t z*.

From the Point z, draw the Line zς parallel to the Line s a; then , as is well known, the Arch ρa shall be equal to the Arch z t. Suppose the Arch χρ to be divided by a continual Bisection into an infinite number of equal Parts, at the points ο, ο, ο, &c. and let ox represent one of such parts: From ο draw the right line ο ο, parallel to a s, and with a radius equal to a s describe the arch λ ο and from the centre s, the arch λ s. Now let (M) represent the infinitely small quantity χ ο, then if the arch as exceed the Arch λ ο, by the quantity M, it shall be equal (from the Lemma) to the Arch which is absurd. The same consequences will evidently follow, if a Quantity less than χ ο is assigned, and so the Arch λ s is not greater than the Arch λ ο.

Again, Suppose the Arch λ ο, to exceed the Arch mλ by the like Quantity (M), then if the Arch in mχ exceed mλ, by (M), it shall be equal to the Arch λ ο. But the Arch a s, is not greater than λ ο from the Demonstration, and so in this case is not greater than the Arch mχ, which is evidently absurd. In like manner proceeding infinitely, we shall at last prove, that the Arch t R is not greater than the Arch t z. Q.E.D. {3}

*Fig*. 1. It must be carefully observed that this method of Reasoning will not take place, if the Angle a s π is assumed greater than a right Angle, (suppose the Angle a s G). For since it is always required to the Demonstration that an infinitely small Part of the Arch a G must be taken, and a Line drawn parallel to s a. If from the Point F, which terminates the Quadrant a F, an in finitely small Arch F o is taken, and a Line drawn parallel to a s, the Line so drawn shall be a Tangent to the Arch F G, and so fall without the Curve.–In like manner, any Line drawn from a Point below F, parallel to a s, shall fall without the Curve towards G; and therefore in this case the preceding Demonstration can not take place.

*Fig*. 1. No. II. It may not, perhaps, be unnecessary to observe, that after the same manner as in this Proposition, it is easy to prove that in the Quadrant a K, if any right line φ o is drawn parallel to x a, then any other right Line φρ ſhall not be greater than any right line φ o.

*Fig*. 1. No. III. If Concentric and Tangent Arches are described as in the Figure terminated by the Quadrantal Arch πs, {3} then from this Proposition, (if we suppose the concentric Arches from s towards π continually increase) any Concentric Arch x y shall not be greater than it's Tangent Arch δs.

Hence, (the same Things remaining as in Cor. II.) if mθ is described exceeding Δδ by (M), and if mλ = mθ, then all the concentric Arches above Δo shall equal their correspondent Tangent Arches, described with the Radius sς as per Scheme.

For, suppose X, to exceed mθ by (M), then if mλ = mθ, the Arch xy shall be greater than mθ, and not greater than xς (Pr. I.). Therefore xy çannot be less than xso and so must be equal to it. Q.E.D.

It were easy to infer a Variety of other Conclusions from this Proposition, but as they will naturally present themselves to the Reader's observation, without much enquiry, they are omitted in this place.