[NumTh Expository] “Co-prime” as more basic than “Prime”

Introduction

     While mucking around with another problem, I used the above to argue and solve it.  If  a  is a factor (divisor) of  bk,  and  a  and  b  are co-prime (they do not have any common factor), then “obviously”  a  must divide  k.  I wanted to prove this by just using this definition of  a  and  b  being co-prime

          d | a  and  d | b   Þ   d | 1   Þ   d = 1  (since 1 | d  also)

The concept of being co-prime (or relatively prime) is more basic than the concept of prime numbers.  The prime number property

          p | ab  Þ   p | a  or  p | b

should be built on top of this common sense observation, instead of the other way round.  In advanced modern mathematics, the above prime number property is the definition of prime numbers.   By the way,  hcf  stands for highest common factor, which Americans call “greatest common divisor” or gcd.

     It turns out that “common sense” facts are harder to prove.  Furthermore, I wanted to tie one hand behind my back and still see if I could still do it.  I came up with three proofs.  In what follows, the notation  A | B | C  means  A | B  and  B | C  and by transitivity  A | C  and can be read as  “A  divides  B,  which divides  C”.

One more try ...

     I think the above proofs are correct as they stand, but I still have not achieved my objective of using just the concept of divisibility.  Proof #3 is short and sweet, but this borrows from the theory of Linear Diophantine Equations.  Looking back at proof  2a,  I decided to replace “prime number” with “smallest non-trivial divisor” and had a go at it again.  The observant reader will realise that the smallest non-trivial divisor is a prime number in disguise but “Shhhhhhh!  Don’t tell other people, OK?” 

Final Remark

     The key idea that makes this proof work is that the hcf is guaranteed to exist, and  hcf(A, B) always divides  AB  since it divides both  A  and  B.  Finally, I got the proof with the flavour that I wanted.  So how does the prime number property   p | ab  Þ   p | a  or  p | b    follow from the above theorem?  Easy: If  p  is a prime that does not divide  a,  then  hcf(p, a) = 1  i.e.  p  and  a  are coprime.  Then  p | b.  ©

[S2_20150513IXS] The Power of Repeated Chunks

Question

Introduction

     This is a question on indices (taught in secondary 2 or 3) that looks daunting.  There is a way to unravel it.  Look at the equation carefully.  Can you see anything that looks like it is repeated?  What about that small little “+1”?  How do you deal with it?

Reminders and Tips

    

Solution

Summary

     The key to solving this question is to apply the laws of indices and recognising repeated chunks.  This allows one to see the overall structure of the expression and this often leads to a simplification of the expression involved.  It is good to recognise the powers of small prime numbers.  Like atoms, prime numbers are the building blocks of all whole numbers.  Using prime bases allows us to compare indices more easily.

Heuristics Used

H04. Look for pattern(s)   e.g. repeated chunks

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (usually Sec 2)

* “Elementary” Mathematics or “E Maths”

* Additional Mathematics (revision)

* other syllabuses that involve indices

[OlymLSec_20150507BDP] A Partition of Unity via Factorisation

Question

Introduction

     A premier school in Singapore is said to have set a question similar to this question in a regular school test.  I suspect it was meant to select pupils to represent the school for mathematics Olympiad.
     I present two solutions: one using the idea of prime factor, the other one using the possible factorisation of some number into two factors.

Solution 1

Solution 2

Remark

     The Factorisation Grid helps us to perform the factorisation.  It is presented in the form that I learned it, in which the cross terms are written in the middle.  I prefer it this way.  Nowadays, the textbooks and the schools put these terms on a column on the right instead of in the middle.

     The word “unity” in mathematics is just a fancy word for the number  1.  It is a “partition” because the number is being split into two separate parts (fractions).  The “partition of unity” concept can be extended, and is useful in advanced mathematics including signal processing, weighted averages, spline functions and topology.

Suitable Levels

* Lower Secondary Mathematics Olympiad

* any student who loves a challenge in algebra or number theory

[AM_20150412SQS] Perfect Squares Lurking Absurdly

Question

Introduction

     Surds are expressions involving roots like square roots.  They are usually irrational numbers.  If you try to put them into a ratio of integers, you are absurd!  Many students (and teachers?) are not sure of how to put square roots of numbers in a simple form.  The trick is to use square numbers or perfect squares.  These are squares of whole numbers.  For example:-

     1² = 1 ´ 1 = 1                      Ö1 = 1

     2² = 2 ´ 2 = 4                      Ö4 = 2

     3² = 3 ´ 3 = 9                      Ö9 = 3

     4² = 4 ´ 4 = 16                  Ö16 = 4

     5² = 5 ´ 5 = 25                  Ö25 = 5

A number like  4  can be represented by a real square whose sides have length  2  units.  Note also that if you take the square root of a perfect square, you always get a nice whole number.

     How do you deal with numbers that are not perfect squares?  You factor out as many perfect squares as possible.  This would eventually lead to surds with small numbers, which are more manageable.  Here are some examples:-

With this weapon in our hands, let us kick some butt.

Solution

Moral of the Story

     Using perfect squares reduces your square roots to surds involving square roots of prime numbers, which are easier to combine or cancel.  This gives a short and sweet solution.  In mathematics, always try to do things by the cleanest way (if you can).

Heuristics Used

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels
* GCE ‘O’ Level Additional Mathematics
* revision for GCE ‘A’ Level H2 Mathematics
* revision for IB Mathematics HL / SL
* other syllabuses that involve surds
* precocious kids who always want to learn more