# Pythagorean Theorem and Triplets

## Pythagoras in a painting The School of Athens ( 1509 - 1510 ) by Raffaello Sanzio da Urbino ( Raphael ) - Photo from Wikipedia

## Enter Pythagoras

This Hub continues from the one on Trigonometric functions, Trigonometry to begin with , and how to find angles and sides of a right angled triangle, to the related world of the Pythagorean theorem.

From the sections on how to work out lengths in such a triangle, the other way to work out this third length is to use Pythagoras, which states that the square of the length of the hypotenuse** **is equal to the sum of the squares of the other two sides.

This means that opposite² + adjacent² = hypotenuse²,

or also, that hypotenuse² - adjacent² = opposite²

**5.7735² ** -** 5² **=** **(** 25** ÷** 3** )** **square root =** 2.887**

Which is of course the answer we had before. See, that’s all trigonometry is, finding these numbers which are just there to show what fraction of one length the other length is. These ratios are all consistent, and whichever way you work them, if you do it right, there’ll be no error.

## It's Triplets !

Taken from the previous Hub on Trigonometry, with respect to the **3** :** 4 **:** 5 **triangle, we can say again that in terms of whole numbers, there are only a select few that work, so that the sum of the squares of the lesser two equal the square of the hypotenuse. But if we were to multiply each of our lengths in the **3** :** 4 **:** 5 **triangle by the same number, we would end up with another right angled triangle.

For example, if we multiply all three sides by **two**, their lengths would become **6**, ** 8**, and **10 **inches, respectively, and **6² **+** 8² **=** 10²**, so that’s alright, but what about the internal angles ? Will they change, or stay the same ? Look back at the drawing of our triangle, then see that the side opposite our top angle is now **8**, while our hypotenuse this time is equal to **10**. If we want the sine** **again, we know now and for ever that opposite ÷ hypotenuse =** 8 **÷** 10**, which also still equals **.8**, so the ratio between these two sides has not changed, just because their lengths have, and neither therefore has the angle between them, since both sides were multiplied by the same amount. Incidentally, the area of this new triangle is now **four** times that of the previous, and I expect you to work it out.

So, if you find a basic Pythagorean Triplet, you can multiply all three sides by the same number, and you will get another right angled triangle with three whole number sides, and the same three internal angles as before.

There are even formulae for finding the three whole number sides of a right angled triangle. If you want to begin with an odd number, __square__ it, then split this __square__ into two numbers as close to each other as possible. The number you first __squared__, and the two addends of its __square__, will be the three whole number sides of your right angled triangle.

## Pythagorean Triplet Odd Number Example

For instance : odd number = **3**.** 3² **=** 9**. The two whole numbers closest to half of **nine **(** 4.5** ) =** 4** and **5**, and this of course is our lowest triplet.

Let’s try another : odd number = **7**. **7² **=** 49**. Half of **49** is **24.5**. The two whole numbers closest to this are **24** and **25**, and **7² **+** 24² **=** 25²**, so we have another triplet.

One last try : odd number = **9**. **9²** =** 81**. The two whole numbers closest to **81** are **40** and **41**, and **9² **+** 40² **=** 41²**.

## If we then even things up

So what happens if you want to begin with an even number, instead ? In this case, we are instructed to *double* the even number, and this gives us our first length. Then we square the even number, one less than this square is our second side, and one more than this square is our third.

For example : even number = **2**. **2 **× ** 2 **=** 4**

**2²** - **1 **=** 3**, and **2² **+ **1 **=** 5**. We once again have our triplet **3 **:** 4** : **5**.

Again : even number = **10**. **10 **×** 2 **=** 20**

**10² **-** 1 **=** 99**, and **10² **+** 1 ** = **101**. **20² **+** 99² **=** 101²**

## Finding a Formula for the Triplets

Both of these formulae can be expressed algebraic-ally, to see if they come full circle. Now, what we are saying with the first one to do with odd numbers, is that we have three numbers, whose relation is such that the sum of the squares of the lesser two is equal to the square of the largest one.

We shall call the first one, our odd number, which ends up being our shortest length, **x**. Now we find our next number, the middle length, by squaring** x**, and as it works out, subtracting **one**, then halving. This is what in effect happens when we find the two numbers closest to the half of **x²**. As an example, when we squared ** nine** we got **81**. If we subtract **one**, we get **80**, and then if we halve this, we end up with **40**, which is what we did have. To get the third length, which turns out to be our hypotenuse, we add **one** to **x²**, then halve. This gives us **82 **÷** 2**, which equals **41**, which, as we saw when we did this one before, was indeed our third value. Algebraically, what we do ends up looking like this :

This is because all three values we are squaring are the **three** numbers we find by applying the directions we are given. You can see that the polynomials within the braces would end up being **40** and 41 respectively, if ** x** was equal to **nine**, and the squaring is what would be done to show their Pythagorean relationship.

Now to simplify this whole equation, we would first need to rewrite it like this :

Which, if you work it out, you will see it is true. The difference between both these polynomials is indeed **4x²**. We don’t even need to plug any **x **value into the above equation to work it out, since this is as much a proof, which is all I intended to show here, although it can still become a formula if that is what we also need to use. By similar methods I shall prove that the process for even numbers used to determine whole number sides of Pythagorean triangles is also mathematically accurate.

## Looking next at Even Numbers

In the case of the even numbers, our even number this time is *not* one of the sides, but rather, twice what its value is. If again we say our even number equals **x**, then the first side we work out, which happens to end up being our middle length side, is equal to **2x**, our shortest side is **x²** -** 1**, and our hypotenuse is **x² **+** 1**. So, the relation between these three lengths with respect to them being whole number sides of a right angled triangle, is that :

(** 2x **)**² **+ (** x² **-** 1 **)**² **= (** x² **+** 1** )**²**, because again, the unsquared values within parentheses are our triangle sides, and they are then squared to show that the two values on the left squared equal the value of the square of the one on the right, which is our hypotenuse. Thusfore :

which is exactly what we had with the formula using an odd number, so it is no surprise both formulae should be related. Now note that these two formulae are to be used to find the *base* Pythagorean triplets ; that is, ones like **3 **:** 4** : **5**, and **5 **:** 12 **:** 13**, where all three of the numbers have no common factors. That only occurs when we multiply each number within a triplet by the same factor, but then those are **not** base triplets.

## Concluding Remarks about the Triplet Formula

Thus, the Pythagorean Theorem gives all basic triplets of positive integers, **a**, **b **and **c**, having no common factors, and satisfy the equation **a² **+** b² **=** c²**, so that **a **=** 2mn**, **b **= ** m² **- **n²**, and **c **=** m² **+** n²**, and one, but not both, of **a **or **b **is even, and **m **and **n** are positive integers, where **m **>** n **>** 0**. For example :

**3 **:** 4 **:** 5 **: if **a ** = **2mn**, well the only number here divisible by **two **is **four**, so **4 **=** 2mn**, which means that **mn **= ** 2**. Now, the only two whole numbers which multiply to make **2** are in fact **1** and **2**, where of course **2 **×** 1 **=** 2**, and seeing that of these, **m **is greater than **n**, **m ** = **2**, and **n **=** 1**. This works out, because **b**, which is **3**, equals **m² **- **n²**, or **2² **-** 1²**, so that **c**, which is **5**, ends up being **m² **+** n²**, or **2² **+ **1²**.

## Where's the Proof, Prof ?

Of course, it’s one thing for us to *tell* you that Pythagoras’ Theorem works, but why should you be asked to accept it, without proof ? Sure, we could go through example after example of Pythagorean triangles that *do* support the statement that **a² **+ **b² **= **c²**, but all you would need to see is one exception to this rule, to blow the whole idea to pieces.

You can see the triangle on the left corresponds to that at right. The one on the right also has along its edges the __square__ __areas__, such that the largest of the three squares is equal in __area__ to both of the other two put together, that is, *if * Pythagoras’ Theorem is correct. What we do to prove it, is to rearrange the above diagram in the following way :

## Stephen Grover Cleveland, 22nd and 24th United States President, once also hid a serious dental operation from the American Public - photo courtesy Wikipedia

## Explaining the Diagram above

Here the largest area, that, according to Pythagoras, equals both the others, has been rearranged with four right angled triangles, each the same as the centre one of our first diagram above.

Now, the side of the whole square at left = **a** +** b**, therefore the area of this whole square, which also includes the diamond shaped square area ** c²**, will be (** a** + **b** )**²**, which, when expanded, = **a²** + **2ab** + **b²**. Now the area of each of the four small triangles around the area ** c²** = **½( a** × **b)**, so the area of all ** 4** triangles will be **four** times that, or = **2ab**. This tells us that (** a** + **b** )**²**- **2ab** = **c²**, since if we took the area of the four triangles away, that’s what we would have. Now, remember that ( **a** + **b** )**²** = **a²** + **2ab** +** b²**, so, if **a²** + **2ab** + **b²** - **2ab** =** c²**, this means, since the **2ab’s** cancel out, that **a²** + **b²** = **c²**, which is indeed what we wanted to prove.

In addition, the truth of Pythagoras’ Theorem can be proven in a more graphical way, and to do so is no mean feat, since it was Grover Cleveland, future President of the United States, who himself gave a proof of this very thing in his time.

We see if we rearrange our diagram from just above

## Final explanation and conclusion

We recognise that the four triangles that in the diagram at left are bordering the diamond, are in the second diagram put into two groups of two, making up the same __area__ as before, so that, the remaining __area__, which in the first diagram is represented by **c²**, is in the second shown to be equal also to **a² **+** b²**, proving then that **a² **+** b²** = **c²**. *Q**uod **E**rat **D**emonstrandum**. *

## Disclaimer

As much as some of this Hub contains certain Mathematical knowledge accessible in the public domain, and not subject to any Copyright, other information has been drawn from textbooks which themselves are Copyright, but only in the sense of how they deliver the information which itself is shared and sometimes ancient Mathematical knowledge. Other information has also been found on Wikipedia ( Copyright 2013, the Wikimedia Foundation ), which is a good source of information. Part of this is also my own discovery, but may also independently have been found by others. The Where's the Proof, Prof ? Title could be seen as inspired by the 1984 Wendy's Commercial using Clara Peller ( 1902 - 1987 ) to ask, Where's the Beef ? Written by Cliff Freeman at Dancer Fitzgerald Sample Advertising.

Some of the illustrations in this particular Hub are my own, and have primarily been done using Microsoft™ Paintbox, edited from illustrations done using Microsoft™ Word. Others are, as noted, from Wikipedia. Any quote or part of this material which seems to belong to any other author should be treated as such, and I claim no ownership of anything I did not myself invent or discover, nor of any obvious copyright, trade mark, or registered trade mark.

The Adventure continues in the next Hubs on Geometry, and these are : Things to do with Shapes, Pyramids - How to find their Height and Volume, How to find the Area of Regular Polygons,

If You are curious, then do not hesitate to take a good look at the other Hubs, The Maths They Never Taught Us - Part One, The Maths They Never Taught Us - Part Two , The Maths They Never Taught Us - Part Three, The Very Next Step - Squares and the Power of Two , And then there were Three - a Study on Cubes, Moving on to Higher Powers - a First look at Exponents, The Power of Many More - more on the Use of Exponents, Mathematics - the Science of Patterns , More on the Patterns of Maths , Mathematics of Cricket , The Shape of Things to Come , Trigonometry to begin with The Wonder and Amusement of Triangles - Part One, The Wonder and Amusement of Triangles - Part Two, the Law of Missing Lengths, The Wonder and Amusement of Triangles – Part Three : the Sine Rule, and The Wonder and Amusement of Triangles - Part Four : the Cosine Rule.

Also, feel free to check out my non Maths Hubs :

Bartholomew Webb , They Came and The Great New Zealand Flag

There will also be many More to come on a wide variety of Subjects.

Just take a good look at it, and note how interesting it all is, then see if you can come up with anything else along the same lines. As usual, I would appreciate any comments, feedback and suggestions which would be given due credit, or indeed have a go and publishing Your ideas Yourselves, but firstly, by all means, add Your comments - it's a free Country.

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