# Maths Fairy

**Much ado about not having a clue.**

**A NUMBER (or an x, y or z representing a number) may be engaged to the cm, the gram, the apple, the monkey, the Rand etc. but it is never married to any of them.**

**It is like its ‘manipulations’ (addition, multiplication, squaring, etc.) called MATHS a thing of its own but of course gets applied for use in Physics and Engineering and most likely all activities one can think of.**

**Maths with its numbers is a (an indispensable!) universal Tool, a thing of its own? There is not even (really) a multiplication sign between 5 and cm of 5cm, which really represents/means 5 of (rather than times) a thing/unit called centimetre; same with 5apple = 5 of a thing called apple. We tend to think of 5 times 1 of a sheep in the same way as buying 5 times ½ of a sheep in a butcher shop, hence the multiplication sign stuck in our minds even whilst there is still an of between ½ and sheep.**

**Point to make: You can modify a number you choose as you like as long as the basic value of what you get remains the same. Then you ‘engage’ it to whatever measuring unit or object you wish to make sense of what you wish to use the combination for. They can have a fling (see underlined below) but they are not married!**

**Length: 64cm**^{1} = 8^{2} cm^{1} = 4^{3} cm^{1};

^{1}= 8

^{2}cm

^{1}= 4

^{3}cm

^{1};

**Also: 64**^{1} of cm’s but write (64cm)

^{1}of cm’s but write (64cm)

**Area: 64 cm**^{2} = 8^{2} cm^{2} = 4^{3} cm^{2};

^{2}= 8

^{2}cm

^{2}= 4

^{3}cm

^{2};

**Also: 8**^{2} of square cm’s but write (8cm)^{2}

^{2}of square cm’s but write (8cm)

^{2}

**Volume: 64 cm**^{3} = 8^{2 }cm^{3} = 4^{3 }cm^{3};

^{3}= 8

^{2 }cm

^{3}= 4

^{3 }cm

^{3};

**Also: 4**^{3} of cubic cm’s but write (4cm)^{3}

^{3}of cubic cm’s but write (4cm)

^{3}

###### Pythagoras states with regards to the three sides of the rectangular triangle: square of length of side + square of length of side = square of length of hypotenuse in number of them times units such as

###### cm^{2}

**He might have stuck to the numbers only and have said: If the proportion of the rectangular sides is 3:4 (maybe he used the length of a thumb to determine this?) you find by squaring, adding and square-rooting these numbers the hypotenuse to join them in the proportion 3:4:5. Then if you wish you are welcome to think of a unit to engage with these numbers which of course in this case everyone knows must indicate length, and when squared, area.**

**By the way or furthermore: In this case, Pythagoras could, and you can, attach little squares to the sides of his/your triangle as a drawn proof, of his theorem. In a real world Pythagoras could go with numbers and units from triangle sides to squares shown on these sides. In the next paragraph you find De Gua could not get his figures to likewise match any shape!**

**De Gua says about the trirectangular tetrahedron’s four faces:**

**area**^{2} of face + area^{2} of face + area^{2} of face = area^{2}

^{2}of face + area

^{2}of face + area

^{2}of face = area

^{2}

**of base triangle. He might have prioritised the numerical values and ignored the unit parts of the areas (perhaps using his personalized ruler based on the length of his nose for measurement and an area equation to find it) finding say as on the Definitions page of Swannie’s Arty Maths, the numbers 5,2 and 8,0 and 11,2 and 14,7 **

**(all to 1’**^{st} decimal only) which each one squared in turn gives

^{st}decimal only) which each one squared in turn gives

**27 + 64 + 125 = 216,**

**which is actually the full/ story he had or wanted or needed to tell us.**

**Retaining the units with the numbers he however ended up with, selecting one above,**

**64^inch**^{4} or (2,83 inch)^{4} etc.

^{4}or (2,83 inch)

^{4}etc.

**Units to the fourth power! * He could not follow Pythagoras’s example (as mentioned in the previous paragraph ) to give us a drawing or a model as a further proof because he did not have a clue as to what something with dimensions**

**2,83 inch x 2,83 inch x 2,83 inch x 2,83 inch**

**would look like!* De Gua thus did fine going with his numbers to a next level but starting with triangles and **

**cm**^{2} landed in a never/nether world (where for Swannie’s maths competency level a Devil rules!) with

^{2}landed in a never/nether world (where for Swannie’s maths competency level a Devil rules!) with

**inch**^{4} and a none existing shape.

^{4}and a none existing shape.

**Now, Swannie clearly states in the definition of the First Theorem that use is made of the numerical value of the area of a face, once used for the area of the base of a prism and once, for the height of the prism; such as an 8 for the area and an 8 for the height giving 64 as its volume’s number.**

**Same done for the other faces. Number wise this is fully supported by De Guas’s Theorem as shown above and validates the First Theorem.**

**The chosen inch units, as inch**^{3} units (cubed inches = a little block of something, can just be space, measuring an inch by an inch by an inch) now “engaged” to the number, gives a volume for the prism of

^{3}units (cubed inches = a little block of something, can just be space, measuring an inch by an inch by an inch) now “engaged” to the number, gives a volume for the prism of

**64**^{1} inch^{3}, or 4^{3} inch^{3}, a real property in our world.

^{1}inch

^{3}, or 4

^{3}inch

^{3}, a real property in our world.

**Thus for Plato’s equation the cube root of the number of this prisms volume is 4.**

**For the 27, 64, and 216 prisms you have 3, 5 and 6, the other root values for Swannies **

**3**^{3} + 4^{3} + 5^{3} = 6^{3}, all in inch^{3}, stainless steel model.

^{3}+ 4

^{3}+ 5

^{3}= 6

^{3}, all in inch

^{3}, stainless steel model.

###### By** Swannie’s Theorem the partially (the sensible part of it) included De Gua’s Theorem can now be visualised!**

**Until now for many centuries De Gua’s (and let’s not forget De Tinseau the actual originator) Theorem has been pretty useless, hanging around in an equation.**

**But not anymore – it led, (with skilled? Manipulation – **see later), to Swannie’s Theorem which can at least be used for building little models!**

**The spirits of these two Frenchmen can now rejoice, their brainchild has found a haven in this real world, for all to see! (But they probably achieved much more than just this.) **

**Now to come clear: In no way was it necessary for the Maths Fairy to assist with the above, that was, once Swannie knew what to do. But she does exist and so does the (Maths?) Devil!**

**Trying to find a volume equivalent to Pythagoras’s “triangle with its attached squares” theorem the trirectangular tetrahedron seemed to Swannie the obvious choice to start off with, but he also contemplated other shapes, so paged through a number of text books for a clue.**

**He was then, 2005, off computers, only using 1983 Multimate for real old time letters to a friend or family member, having had enough of watching screens whilst being a member of the working class.**

**Then he was most excited to find in James Steward’s Calculus, International Students edition 5e a problem set therein, on page 858, no.3 about a trirectangular tetrahedron:**

**“Suppose the tetrahedron **** in the figure has a trirectangular vertex S.**

**Let A, B and C be the ***areas* of the three faces that meet at S, and let D be the *area* of the opposite face PQR.

*areas*of the three faces that meet at S, and let D be the

*area*of the opposite face PQR.

**Using the result of problem one, or otherwise show that**

**D**^{2} = A^{2} + B^{2} + C^{2}

^{2}= A

^{2}+ B

^{2}+ C

^{2}

**This is a ***three*** *-dimensional version of the Pythagorean Theorem. ” (*Italic* lettering by Swannie)

*three****-dimensional version of the Pythagorean Theorem. ” (

*Italic*lettering by Swannie)

**The above model gives a top view of such a tetrahedron, trirectangular, showing faces A, B, C; and D as a reflection.**

**Excitement turned into four months of extreme frustration – The trirectangular tetrahedron was tied up/seemed married to a unit to the fourth power, some advanced multidimensional maths way beyond Swannie’s comprehension and a shape for which you could try to invoke the addition of the parallel world some folks believe exist as a twin to our one.**

**Thus convinced by the (Maths?) Devil that numbers get married to units Swannie spent four frustrating months making all sorts of little maths shapes out of cardboard, post toasty boxes and bought sheets, glue and Artist paint, imagining himself to be an Artist in order to deal with despondency.**

**Then the Maths Fairy paid him a visited, waved her wand, broke the Devils spell and showed him the way out!**

**The Theorem with its shapes and their volumes could be defined – a true three dimensional follow up to Pythagoras! The Maths Fairy had to be credited somehow without then relating or even having the above true story available.**

**So Swannie credited her with ‘throwing away part of some unknown unit, one cm of**

**cm**^{4} to get a useful cm^{3}

^{4}to get a useful cm

^{3}

**which in away is what practically happened!**

**The comment earlier mentioned is now much appreciated as it led Swannie to give thought to a simple, but subtle subject, aided by ingrained views, before this story could be generated.**

**He surely now also has a better understanding of how it all came about! **

**By the way the above somehow turned up this nonsense rhyme:-**

**Numbers divorced, or rather apart, from units, sometimes they tend to hide, being all on their own, but never from those who like to play with them, must be very alone; like making them add and divide.**

**The noun and verb that follows are relevant (please, also check in your Dictionary and Thesaurus):**

**Theorem: A Mathematical Statement established by means of a proof.**

**** Manipulate:**

**More positive wording: ***To use with skill.*

*To use with skill.*

**More negative wording: ***To use deviously to own advantage. *

*To use deviously to own advantage.*

**Yes, Swannie did manipulate Maths to find his theorem and he really likes to think it was with skill, and yes it was to his own advantage, it surely boosted his ego, but as to whether the maths was used deviously he won’t admit.**

**Whether the theorem is perhaps seen as trivial and seems to be of no practical use other than making little models, may be so, but Swannie has no doubt that it is valid.**

**He also better feel good about it as it is the only original (to his knowledge) discovery he has ever made! (Except, perhaps also his unique treatment of piles with cold water). **

**Thus he immensely appreciates it if any one bothers to comment and/or suggests an improvement. **

***** Note: The three above in ***italic* is, I would say, incorrect because

*italic*is, I would say, incorrect because

**area**^{2}, as was shown, is length^{4}.

^{2}, as was shown, is length

^{4}.

**As “James Stewart” used De Gua’s number with **

**unit**^{4} theorem he should have said:

^{4}theorem he should have said:

**This is a fourth-dimensional, version of the Pythagorean Theorem, or did he anticipate something like Swannie’s contribution? Perhaps just meant: taking it to a next level.**

**In Brief:**

**What follows would seem to wipe away most of the above directed at the First Theorem and make the theorem real elementary and straight forward.**

**This approach obviously initially bypassed Swannie; being simply an ‘abbreviated’ definition of De Gua’s theorem: **

*The sum of the squares of the numerical values (only!) of the squares of the areas of the right angle faces of a trirectangular tetrahedron add up to the square of the numerical value (only!) of the area of the base of the tetrahedron.*

*The sum of the squares of the numerical values (only!) of the squares of the areas of the right angle faces of a trirectangular tetrahedron add up to the square of the numerical value (only!) of the area of the base of the tetrahedron.*

**???**

**Any problem with simply ignoring the units (?)**

**So say a face has an area of**

**2,645…**^{2}**cm**^{2}, that is 7cm^{2}**,**

^{2}

^{2}, that is 7cm

^{2}

** ignore the unit and square the 7 only, that is 49 = 7 x 7**

**— ****use the one 7 for the base of a prism on the face and now give it a **

**cm**^{2} unit l, that is a base area of 7cm^{2}

^{2}unit l, that is a base area of 7cm

^{2}

**— then use the other 7 for the height of the prism and give it a cm unit, that is a height of 7cm.**

**So you have a volume 7**^{2} cm^{3} for the prism = 49cm^{3}.

^{2}cm

^{3}for the prism = 49cm

^{3}.

*Thus are the units initially ignored as defined above*** and lastly attached as required. **

*Thus are the units initially ignored as defined above*