"We don't need proofs, we believe what our teacher tells us." So I was told by some Sixth Form students at a boarding school somewhere in Zimbabwe a few years ago. We had gathereed one afternoon in a classroom to discuss the need to prove, while still in school, mathematical statements like Pythagoras' Theorem.
"Do you believe everything you read in the newspaper?" I replied, "Surely not! You know full well that the journalist may be biased." They had no counter-response to that, but made up for it with a fine selection of guffaws.
I usually compare proving theorems with mountain climbing. Sometimes you can just stroll, sometimes it takes more effort to reach the peak. And while there may be different ways to reach the peak, you will be unable to get there if you do not observe certain rules (like putting on hiking boots, consulting maps, or following certain paths).
And, for some of us at least, it is as much fun getting there as being at the top. You appreciate the mountain more and you have a much greater sense of achievement than if you were carried there by helicopter. You see a lot of good things on the way, you glimpse other peaks, and (perhaps above all) you develop the fitness and experience to qualify to help others to the top of the mountain, as well to climb other mountains yourself. Maybe even mountains that have never been climbed before!
Back to the paths up the mountain... in maths these are the paths of logic --- Aristotelian logic for that matter, after the ancient Greek Aristotle who lived over 2000 years. And just as you cannot always believe or fully appreciate everything people tell you, it helps to see for yourself that a2 + b2 = c2 if a, b, c are the sides of a right angled triangle (c being the hypotenuse). Sure,
32 + 42 = 52, but that could be true whether or not Pythagoras' Theorem holds in general. Ditto for the equation 52 + 122 = 132.
So how do we prove such things as mathematical theorems, lemmas, corollaries, propositions etc? Is there any foolproof method? Are there any rules one has to observe? Well, apart from logical principles, there are three main methods of proof: direct proof, proof by induction and proof by contradiction.
Let's confine this article to the easiest to explain -- direct proofs. We illustrate it using:
Pythagoras' Theorem: In any right angled triangle with sides of lengths a, b, c where c is the hypotenuse, we must have a2 + b2 = c2.
Proof: Consider a square Q of side a+b in the plane. For every side of Q place a point at a distance a from a corner in the symettrical depicted in the diagram. Call the marked points U, V, X, Y: let the corners of the square be denoted by A, B, C, D. Draw the lines UV, VX, XY, YU so that we have the situation depicted.
Now we have divided Q into five areas, the four triangles AUY, BVU, CXV, DYX and the quadrilateral Q1 = UVXY. The four triangles are congruent because they all have two sides of a, b with an included angle of 90 degrees. Consequently the angles of Q1 are all right angles. Thus Q1 is a square of side length c (the hypotenuse of each of the four triangles).
Now we establish an equation expressing the area of Q in two ways: one by the formula of a square of side a+b and one as the sum of the areas of each of the four triangles (each of area ab/2) plus the area of the inner square of side c. So we obtain (a+b)2 = 4(ab/2)+c2 or equivalently a2+2ab+b2 = 2ab + c2. Subtracting 2ab from both sides yields the desired result
a2 + b2 = c22 + b2 = c2 holds. What are these 'established facts'? For one, we used the fact that two triangles are congruent if two of their sides and the included angle are identical. We used the formula 1/2 base x height for the area of the triangle and we used the fact that the sum of the angles of a triangle is 180 degrees. And since this last detail is only true in the plane (as opposed to other surfaces like spheres) we realise that Pythagoras' Theorem may not work on all surfaces.
From this we drew the intermediate conclusion that the quadrilateral Q1 is a square of side length c. We might have been tempted to take this for granted, but we have already seen that we might have been wrong! For instance if we had been dealing with a hyperbolic surface instead of a plane. This is why we are so careful to justify all our steps - many erroneous assumptions can get unintentionally smuggled into proofs otherwise! Anyway, finally we borrowed from algebra the fact that the distributive laws (ie, x(y+z) = xy + xz and (x+y)z = xz + yz) hold in the arithmetic of numbers, and the cancellation laws (xy = xz and x not zero means that y = z). Then we arrived at our answer!
The above analysis of the proof of Pythagoras' Theorem sheds light on how to proceed in general when trying to get a direct proof. Starting from the hypothesis in general when trying to find a direct proof. Starting from the hypothesis of the theorem ("given a right angled triangle with sides a, b, c) we reach the conclusion of the theorem ("a2 + b2 = c2") by moving step-by-step. combining logical arguments with previously proved mathematical statements to motivate and justify each step. Of course, in between we may reach intermediate conclusions which are like plateaux reached as we strive to get to the mountain peak. And also along the way we get an inkling of how far the validity of our theorem may extend by noticing what we needed to prove it.
Got that? Hope so! We will cover other principles of proof in future issues. If you have had just about enough, don't worry, for this is where this article ends.
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![[Proving Pythagoras' Theorem]](proof.gif)
Now we have divided Q into five areas, the four triangles AUY, BVU, CXV, DYX and the quadrilateral Q1 = UVXY. The four triangles are congruent because they all have two sides of a, b with an included angle of 90 degrees. Consequently the angles of Q1 are all right angles. Thus Q1 is a square of side length c (the hypotenuse of each of the four triangles).
Now we establish an equation expressing the area of Q in two ways: one by the formula of a square of side a+b and one as the sum of the areas of each of the four triangles (each of area ab/2) plus the area of the inner square of side c. So we obtain (a+b)2 = 4(ab/2)+c2 or equivalently a2+2ab+b2 = 2ab + c2. Subtracting 2ab from both sides yields the desired result
a2 + b2 = c22 + b2 = c2 holds. What are these 'established facts'? For one, we used the fact that two triangles are congruent if two of their sides and the included angle are identical. We used the formula 1/2 base x height for the area of the triangle and we used the fact that the sum of the angles of a triangle is 180 degrees. And since this last detail is only true in the plane (as opposed to other surfaces like spheres) we realise that Pythagoras' Theorem may not work on all surfaces.
