# Squaring the Circle/Fallacious Proof/Edward J. Goodwin

## Fallacious Proof of Squaring the Circle

To quadrate the circle is to find the side of a square whose perimeter equals that of the given circle; rectification of the circle requires to find a right line equal to the circumference of the given circle. The square on a line equal to the arc of $90^\circ$ fulfills both of the said requirements.
It is impossible to quadrate the circle by taking the diameter as the linear unit, because the square root of the product of the diameter by the quadrant of the circumference produces the side of a square which equals $9$ when the quadrant equals $8$.
It is not mathematically consistent that it should take the side of a square whose perimeter equals that of a greater circle to measure the space contained within the limits of a less circle.
Were this true, it would requiire a piece of tire iron $18$ feet to bind a wagon wheel $16$ feet in circumference.
This new measure of the circle has happily brought to light the ratio of the chord and arc of $90^\circ$, which is as $7:8$; and also the ratio of the diagonal and one side of a square, which is as $10:7$. These two ratios show the numerical relation of diameter to circumference to be as $\frac 5 4:4$.
Authorities will please note that while the finite ratio ($\frac 5 4:4$) represents the area of the circle to be more than the orthodox ratio, yet the ratio ($3.1416$) represents the area of a circle whose circumference equals $4$ two $\%$ greater than the finite ratio ($\frac 5 4:4$), as will be seen by comparing the terms of their respective proportions, stated as follows: $1 : 3.20 :: 1.25 : 4$, $1 : 3.1416 :: 1.2732 : 4$.
It will be observed that the product of the extremes is equal to the product of the means in the first statement, while they fail to agree in the second proportion. Furthermore, the square on a line equal to the arc of $90^\circ$ shows very clearly that the ratio of the circle is the same in principle as that of the square. For example, if we multiply the perimeter of a square (the sum of its sides) by $\frac 1 4$ of one side the product equals the sum of two sides by $\frac 1 2$ of one side, which equals the square on one side.
Again, the number required to express the units of length in $\frac 1 4$ of a right line, is the square root of the number representing the squares of the linear unit bounded by it in the form of a square whose ratio is as $1:4$.
These properties of the ratio of the square apply to the circle without an exception, as is further sustained by the following formula to express the numerical measure of both circle and square.
Let $C$ represent the circumference of a circle whose quadrant is unity, $Q$ $\frac 1 2$ the quadrant, and $C Q^2$ will apply to the numerical measure of a circle and a square.
We are now able to get the true and finite dimensions of a circle by the exact ratio $\frac 5 4:4$, and have simply to divide the circumference by $4$ and square the quotient to compute the area.
-- Edward J. Goodwin

## Resolution

To quadrate the circle is to find the side of a square whose perimeter equals that of the given circle ...

From Perimeter of Circle, this would produce a square whose area is $\left({\dfrac {\pi r} 2}\right)^2 = \dfrac {\pi^2 r^2} 4$.

If this were so, then $\pi$ would equal $4$.

... rectification of the circle requires to find a right line equal to the circumference of the given circle.

Accurate enough, if one assumes that a right line is the same thing as a (straight) line segment.

The square on a line equal to the arc of $90^\circ$ fulfills both of the said requirements.

False -- as pointed out above, this would produce a square of area $\dfrac {\pi^2 r^2} 4$, which is considerably greater than that of the circle. Trivially, of course, it does indeed have the correct perimeter.

The circle as described by Goodwin has a diameter of $10$ and, as he states it, a circumference actually of $32$, delivering a value of $\pi$ of $3.2$, not $3.1416$.

The chord of $90^\circ$ has length stated as $7$, and not what it actually is, approximately $7.0710$.

Further mistakes are equally egregious.

$\blacksquare$