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Can You Solve This Paradox?

By Sylvia Brewda

Plato’s dialogue the “Meno” has been one of the documents most cited, of the anti-Aristotelian faction throughout history, and is thus a most appropriate benchmark for the offensive against Eulerian, linear thinking in ourselves. In a crucial section, a young, uneducated slave-boy is guided by Socrates, from wrong opinion, to awareness of his own ignorance, to the {knowledge} of how to construct a square with exactly twice the area of a given square.

First, Socrates shows the slave-boy that his immediate, naive presumption, that a square with area equal to 8 can be constructed from one of area equal to 4 by simply doubling the lengths of the sides, from 2 to 4, is wrong. Then, the boy is shown that his guess of sides equal to 3 will not work either.

Socrates comments to his chief interlocutor in the dialogue, “Observe, Meno, the stage he has reached on the path” of coming to know.

“At the beginning he did not know the side of the square of eight feet. Nor indeed does he know it now, but then he thought he knew it, and answered boldly, as was appropriate–he felt no perplexity. Now, however, he does feel perplexed. Not only does he not know the answer, he doesn’t even think he knows…. Isn’t he in a better position now in relation to what he didn’t know?

“…|Now notice what, starting from this state of perplexity, he will discover by seeking the truth in company with me….”

Starting at that point, with nothing more than stick drawings in the sand, how can you show that this boy could {knowingly} succeed in constructing the required square? What are the {Analysis Situs} implications of this?

{Figure 1.} Here we see the original square of area equal to 4, and the method of producing a square double in area (square BUWD).

{Figure 2.} The construction undertaken here points to a simple geometric proof of the Pythagorean Theorem (which says that the square on the hypotenuse of any right triangle is equal to the sum of the squares on the other two sides, or legs.)

{Figure 3.} Fig. 3(a) shows that in a right triangle in which both legs are equal to {a}–that is, an isoceles right triangle–the area of the right triangle equals one-half of the area of the square whose side is {a.}

Figure 3(b) shows that, in a right triangle with legs of different lengths ({a} and {b}), the area of the right triangle equals one-half the area of the rectangle whose sides are {a} and {b.}

Look for the following autosorts in this article:

@s for superscripts

@ts for multi signs

? for square root signs

Also, in the use of a and b to denote sides of triangles & squares, I have marked those boldface {}, because I don’t want them to be confused with, e.g., Fig. 2(b), or something. They don’t have to be bf, but they do have to be somehow set apart.

Thanks

How Socrates Resolves the Paradox

CAN YOU SOLVE THIS PARADOX?

by Sylvia Brewda

Here we reprint the first part of this column, which appeared in the last issue of New Federalist (No. 14, April 14, p. 12), along with the second, concluding part, in which Socrates resolves the paradox.

Plato’s dialogue the “Meno” has been one of the documents most cited, of the anti-Aristotelian faction throughout history, and is thus a most appropriate benchmark for the offensive against Eulerian, linear thinking in ourselves. In a crucial section, a young, uneducated slave-boy is guided by Socrates, from wrong opinion, to awareness of his own ignorance, to the {knowledge} of how to construct a square with exactly twice the area of a given square.

First, Socrates shows the slave-boy that his immediate, naive presumption, that a square with area equal to 8 can be constructed from one of area equal to 4 by simply doubling the lengths of the sides, from 2 to 4, is wrong. Then, the boy is shown that his guess of sides equal to 3 will not work either.

Socrates comments to his chief interlocutor in the dialogue, “Observe, Meno, the stage he has reached on the path” of coming to know.

“At the beginning he did not know the side of the square of eight feet. Nor indeed does he know it now, but then he thought he knew it, and answered boldly, as was appropriate–he felt no perplexity. Now, however, he does feel perplexed. Not only does he not know the answer, he doesn’t even think he knows…. Isn’t he in a better position now in relation to what he didn’t know?

“…|Now notice what, starting from this state of perplexity, he will discover by seeking the truth in company with me….”

Starting at that point, with nothing more than stick drawings in the sand, how can you show that this boy could {knowingly} succeed in constructing the required square? What are the {Analysis Situs} implications of this?

To lead the boy to discover how to construct a square with an area twice that of the original one, Socrates draws three additional squares, each equal to the original one. He has labelled the corners of the original square, ABCD, and draws two new squares: BCUT; CDXW; and, finally, “to fill up the corner here,” CUYW (see Fig. 1). Thus, he has again drawn a square four times larger than the original one.

Then he draws one diagonal in each of the four small squares: BD, DW, WU, and UB. These diagonals form a new square, inside the larger one, and, rotated half a turn with respect to it. It is clear to the boy that, these diagonals divide each of the small squares in half. Therefore, he can see that the area inside the new square made from the diagonals is half that of the large square, since it contains half of each of its component squares.

And thus, the new square, with the diagonal of the first as its side, is {known} to have twice the area of the first.

Clearly, this discovery has been made without the use of any asserted authority. Instead, it is the innate characteristic of the human mind, its own {Analysis Situs,} that allowed it to know the truth of what has been drawn out of it by the Socratic method employed.

This is the power of this brief section of the “Meno” dialogue; this is the reason why constructive geometry is of such importance in true education. Johannes Kepler, the great Renaissance scientist, wrote about this passage in his great work, <cf2>“Harmonice Mundi”<cf1> (“Harmonies of the World”):

“And, indeed, this was Plato’s judgment concerning mathematical things: that the {human mind} is, from itself, fully informed about all, species or figures, axioms, and conclusions about these things; truly, when the mind seems to be instructed, this [process] is nothing other than to be reminded by diagrams, which can be grasped by the senses, of those things which the mind must know through itself. This he represents with singular art in the dialogues, introducing a boy who, being questioned by a teacher, answers everything that is asked.”

Another construction can be generated which solves the problem, and also points to a simple geometric proof of the famous Pythagorean Theorem–namely, that the square on the hypotenuse of {any} triangle containing a right angle is equal to the sum of the squares on the other two sides (the hypotenuse is the side of a right triangle opposite the right angle).

In this construction, a square of side {a} is first cut by diagonals, forming four equal triangles on the four sides. Then, an equal square is constructed on each of the sides of the first square, and each of these new squares is also divided by its diagonals into four equal parts [see Fig. 2(a)]. Next, erase from the sand the three triangular parts of each of these new squares whose bases do not correspond with those of the original square [see Fig. 2(b)]. Now, a new square is seen, one which is made up of the original square plus the four remaining triangles outside that original square, each of which equals one-quarter of the original. That is:

4|@ts|[{a}@s2/4]|+|{a}@s2|=|2{a}@s2

What is the length of the side of this new square? Each side is made up of two parts, each equal to half the diagonal of the original square. Let us call this part {b.} The sides of the outer square are equal to the diagonal of the original one, that is, 2{b.}

Consider the four right triangles which are made by the diagonals meeting inside the original square, {a}@s2. Each has legs equal to {b} and hypotenuse equal to {a} [one is shown as the shaded area in Fig. 2(c)]. The construction has already generated the squares on the legs of each such triangle [a pair are shown marked by hatching in Fig. 2(c)]. For each pair of such squares, the triangular sections left outside the original square correspond exactly to the area within it left uncovered by those sections of the same small squares which are inside it. Thus, the area of the original square, whose side is the diagonal of the small square, is twice that of each small square, or the sum of the squares on the legs of the small right triangle.

However, this construction also leads us further. We can see from Fig. 3 that the area of any right triangle is equal to one-half the product of the two legs. Here, in our construction, this means that the areas of the small triangles are each = ({b}|@ts|{b})/2, and the area of the original square is equal to four of these. That is:

{a}@s2|=|4|@ts|({b}@s2/2)|=|2|@ts|{b}@s2

From this, if the symbol ?{a}@s2 indicates the side of a square of area {a}@s2, clearly {a,} the side of the original square, can be denoted as:

{a}|=|?(2|@ts|{b}@s2)|=|{b}|@ts|@sr(2)

In future articles, we will discover that this relationship means that the two magnitudes, {a} and {b,} are of two completely different types, such that neither can be measured by the other.

The last construction, plus the material presented up to this point, provides the basis for a geometric proof of the Pythagorean Theorem. The starting point is consideration of the effect of changing the lengths of the legs of the small triangles, both on the angles of such triangles, and on the geometry of their fitting together.

How {do} you know that the sum of the squares on the sides of a right triangle is equal to the square on the hypotenuse?

Look for the following autosorts in this article:

@s for superscripts

@ts for multi signs

@sr for square root signs

There are also + and = signs, but I left them uncoded–no need to code.

Also, in the use of a, b, and c to denote sides of triangles & squares, I have marked those boldface {}, because I don’t want them to be confused with, e.g., Fig. 2(b), or something. They don’t have to be bf, but they do have to be somehow set apart.

Thanks

CAN YOU SOLVE THIS PARADOX?

by Sylvia Brewda

As we said in last week’s column, a construction can be generated which solves the problem of doubling the square, and also points to a simple geometric proof of the famous Pythagorean Theorem, which says that the square on the hypotenuse of {any} triangle containing a right angle is equal to the sum of the squares on the other two sides. (The hypotenuse of a right triangle is the side opposite the right angle.)

Here, the square of side {a} is first cut by diagonals, forming four equal triangles on the four sides. Then, an equal square is constructed on each of the sides of the first square, and each of these new squares is also divided by its diagonals into four equal parts [see Fig 1(a)]. Next, erase from the sand the three triangular parts of each of these new squares whose bases do not correspond with those of the original square [see Fig. 1(b)]. Now a new square is seen, which is made up of the original one plus the four remaining triangles outside it, each of which equals one-quarter of the original. That is:

4|@ts|{[a}@s2/4]|+|{a}@s2|=|2{a}@s2

What is the length of the side of this new square? Each side is made up of two parts, each equal to half the diagonal of the original square. Let us call this part {b.} The sides of the outer square are equal to the diagonal of the original one, that is, 2|@ts|{b.}

Consider the four right triangles which are made by the diagonals meeting inside the original square, {a}@s2. Each has legs equal to {b} and hypotenuse equal to {a} [one is shown as the shaded area in Fig. 1(c)]. The construction has already generated the squares on the legs of each such triangle [(a pair are shown hatched in Fig. 1(c)].

This construction provides the basis for a geometric proof of the Pythagorean Theorem. The starting point is consideration of the effects of changing the lengths of the legs of the small triangles, on both the angles of such triangles, and on the geometry of their fitting together inside the square.

– * * * * –

There are many proofs of the famous Pythagorean Theorem, but the following requires no further information than the evident fact that any right triangle can be considered as half a rectangle, constructed with sides equal to the legs of the triangle and cut along one diagonal. From this construction, it is clear that the two angles in each such triangle which are not right angles must fit together to form one right angle, since they can be generated by cutting the right angles in the corners of the rectangle.

To start, consider the figure of the square we just constructed [see Fig. 1(b)]. Clearly, this is a square whose side is the hypotenuse of the right triangle created by the intersection of its diagonals, and the square contains four of these triangles. What changes if these four right triangles with {equal} legs, are replaced by, again, four copies of the particular right triangle we are investigating, but now it is a triangle with legs of different lengths, {a} and {b,} and hypotenuse of length {c?} The square is still the square on the hypotenuse of this particular triangle, but now the copies have to be placed so that each corner of the square is filled by the meeting of the two different acute angles, to equal a right angle. That means that the triangles have to be placed with the short leg of one coinciding with part of the longer leg of the next [see Fig. 2(a)].

Each of these four right triangles is also half of a rectangle, and when the rectangles are drawn in, a new square is created outside the first, with each side equal to the sum of the sides of the rectangle, or the legs of the triangle, {a}|+|{b} [see Fig. 2(b)]. From this we can see that {c}@s2, the square on the hypotenuse, is equal to the square on the sum of the legs, {(a}|+|{b)}@s2, less the four triangular sections of the rectangles which are outside the square on {c,} each of which is equal to the triangle we started with. These four triangles are each half of a rectangle with sides {a} and {b,} and thus add up to two such rectangles. This can be denoted as:

{c}@s2|=|{(a}|+|{b)}@s2|@ms|2|@ts|{(a}|@ts|{b)}

Now, we must investigate the size of the square on {(a}|+|{b),} that is with a side made by adding the two, unequal, legs of the triangle. It can be divided to include one square with side {a} and, in the diagonally opposite corner, another with side {b} (see Fig. 3). Since the side of the large square is {(a}|+|{b),} the area that remains is that of two rectangles, each with sides {a} and {b.} Therefore, the square on {(a}|+|{b)} is equal to sum of the squares on {a} and on {b,} plus these two rectangles:

{(a}|+|{b)}@s2|=|{a}@s2|+|{b}@s2|+|2|@ts|{(a}|@ts|{b)}

Thus, these two rectangles are equal to {both,} the difference between the square on {(a}|+|{b)} and the square on {c, and}, the difference between that same square on {(a}|+|{b)} and the sum of the two squares, on {a} and on {b.}

{(a}|+|{b)}@s2|=|{c}@s2|+|2|@ts|{(a}|@ts|{b)}

{(a}|+|{b)}@s2|=|{a}@s2|+|{b}@s2|+|2|@ts|{(a}|@ts|{b)}

We now {know} that the square on the hypotenuse, {c,} of any right triangle, is equal to the squares on the two legs, {a} and {b:}

{c}@s2|=|{a}@s2|+|{b}@s2

Further, the hypotenuse, {c,} will be equal to the side of the square which is the sum of the squares on the two legs, or, if we use the symbol @sr{c}@s2 for the side of the square with area {c}@s2, we can write:

{c}|=|@sr{(a}@s2|+|{b}@s2)

Simple? Yes, but necessary. Only when this basic theorem of the relation between two independent directions of action has been established on a solid basis, can we develop the experimental aspect of scientific progress, the process of measuring the effects of an additional dimension. Now, secure in the knowledge of this simple relationship, we can begin to measure the universe.