20 replies to this topic

[Mr Walker]

Sherapy, on 20 March 2013 - 04:35 PM, said:

MW, I would add that you don't have to believe in the properties/axioms of arithmetic and algebra, you have to know/learn them, it's in knowing them-- you apply them; it is in the application they prove themselves.

You would use the natural number axiom(basic assumptions we hold to be true/that have been proven to be true) which allows for one to logically induce that all natural numbers follow in an order.(The variable n represents all natural numbers in this case.) Then you would use addition property, then the property of equality and so on and so on.

You are representing the fact that 1 +1 = 2 by using the symbols(language)/rules/laws of Mathematics.

It is not unlike--Bonjour means hello regardless if it's in English or French it's just a different representation of the same thing. It still holds true that both are ways(facts)of saying hello.

Precisely.  Th e latter part of what I posted said exactly this. But  one still needs to chose to believe (or disbelieve) anything not known by personal experience. So a person unfaniliar with the theory and practices of algebra MUST accept its accuracy in faith or belief.

  The two types of proof are the same, but one is more accessible to everyone than the other. Mathematics and language are both human constructs representing physical truths or realities. MAthematics indeed IS a form of language.

PS i did pre university double maths in a time when there were no calculators or computers. We learned first as young children how to do mental arithmetic,  such as our times tables and simple and long division and mulitlication, in our heads, and second how to use slide rules and logarithmic tables for calculations. Thus we had to have the knowledge and discipline to understand the very basic and more complex building blocks of mathemeatics. We had to memorise, and hold in our heads, all the mathematical and  geometric formulae/processes used in maths, and  work  out complex questions in our heads, or on paper, in a few minutes under test conditions. At the end of our final year we had two,  state set and assessed, three hour exams, one on each maths subject, in order to assess our state ranking.  Maths wasn't my strong suite but I achieved in the top 25% of the state's students in both exams.

[monk 56]

Some like visual illusions, so if you still like toys, get yourself a hologram chamber ha ha!

Neat visual mathematics!

Edited by monk 56, 22 March 2013 - 03:13 PM.

[freetoroam]

Setton, on 17 March 2013 - 12:08 AM, said:

Yes.

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

yes, but this does not make a cup of tea?

[Frank Merton]

freetoroam, on 22 March 2013 - 03:26 PM, said:

yes, but this does not make a cup of tea?

Give them a while.

[monk 56]

Hi Freetoroam & Frank Merton,

Including me, thats 3 that are waiting for a cup of tea, one more and we get a T(ea) Square Ha Ha!

http://en.wikipedia.org/wiki/T-square

My math interest is Mandelbrot Set, Fractals and PHI, different strokes for different folks!

[Roy Perry]

God first

thanks you all my friends

with love and a holy kiss of friendship friom Roy