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The Maths Fairy
Much appreciated comment on the presentation of the theorems included a remark that the doings of a Fairy shall surely not be acceptable to any serious practitioner of the ultra pure (snobbish?) discipline called Maths. This led to presenting the following boring alternative to the fairy waving her wand:
If you decide not to wade through all of 1. below then at least read 2., a most elementary 'give away' of the simple nature of the First Theorem.
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 centimeter; 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 ; Also: 64^1 of cm's but write (64cm).
Area: 64 cm^2 = 8^2 cm^2 = 4^3 cm^2 ; Also: 8^2 of square cm's but write (8cm)^2.
Volume:64 cm^3 = 8^2 cm^3 = 4^3 cm^3; Also: 4^3 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
By the way or furthermore:
De Gua says about the trirectangular tetrahedron's four faces: area ^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 Swannies Website the numbers 5,2 and 8,0 and 11,2 and 14,7 (all to 1 st dec.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 (2,83inch)^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
