# Author Archives: Gareth

# memory and answers

I attended a public lecture yesterday by Professor E.Victor Flynn on some fields within Algebraic Geometry…It was incredibly fascinating. Although maybe 20% of it went over my head, it did feel like i’d simply have to jump up to be able to reach it. So when I got home, I decided to google some of the ideas that had been expressed.

I always knew number theory was a diverse field, but I never realised HOW diverse it was. At one point, I had 6 or 7 tabs open, with each one linking to one of the others, sharing ideas and definitions – you could not read any one page completely without knowing content from the others. Most of the time, it was simply a case of definitions – when talking about an algebraic number field, you need to know about field extensions, fields and Rational numbers, and knowing the definitions of these terms allows you to understand the definition of an algebraic number field – and so this bought up some interesting questions for me.

Before I’d started looking into these ideas, I’d done my normal routine of checking 20 or so math based blogs for new content. one blog – “Godel’s lost letter and P=NP” – spoke about the importance of memorisation. Now, if you’ve read my previous blogs, you may notice that I find memorisation of theorems and definitions to be a complete waste of time.

I think now I am beginning to see the error of my ways…

without knowing definitions, we cannot hope to know other definitions that depend on the earlier ones. If an algorithm F works because algorithm G works, we need to know how G works to show F works.

On analysing these thoughts (and ideas expressed by others), I re-evaluated what I thought and why. I think one of the comments on the other blog found the real issue – ‘rote’ memorisation.

however, my thoughts are still developing on this front, and maybe at another point I shall come back to the idea. I agree – and think I have thought this way for some time, but failed to notice it – that knowing definitions and theorems etc are incredibly important. I think what I have issue with is how we learn them, and how we learn to apply them.

now, moving on to the questions from the last post:

I still don’t have much of an idea for the second question – in truth I’m a bit bored of it, so I’m just going to leave it.

but as for the first, this one appeals to me!

so: what do we know?

the number plate is only 4 digits long and contains 2 unique digits, so it’s of the form aabb or abba or abab. As the eldest child is 9 years old, it must be divisible by 9, so

9|(2a+2b)

⇒ 9|(a+b) with 0 ≤ a,b ≤ 9

⇒a+b=9

now, what else do we know? 8 children, each with a different age. the eldest is 9, which means that the other 7 children are either 1,2,3,4,5,6,7 or 8. therefore, there is either a child of 4, or a child of 8, which means the number is divisible by 4.

this tells us the number plate was one of:

9900, 1188,7272,2772, 3636,6336 or 5544.

this gives my smith an age of 00, 88, 72, 36 or 44. logic would dictate that 00 is impossible, 88 and 72 highly improbable. we shall include them for now, but we won’t include 00.

now notice that none of these possible numbers are divisible by 5, which means the children’s ages are 1,2,3,4,6,7,8 and 9. so the number must be divisible by 504=(9*8*7).

simple calculation gives:

1188 mod 504=180

7272 mod 504=216

2772 mod 504=252

3636 mod 504=108

6336 mod 504=288

5544 mod 504=0 (504*11=5544)

so the number plate was 5544, the children’s ages were 1,2,3,4,6,7,8,9 and Mr Smith is a (presumably) very tired 44-year-old.

###### Related articles

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- Further Thoughts on the Mathematical Infinite: The Coincidence of the Naturals and the Rationals (meisly.wordpress.com)

# Rambling puzzles…Number plates, shoelaces and you.

Mr Smith has 8 children, and each one is a different age. His entire family is in the car, when his eldest, aged 9, shouts out “Daddy! That number plate has only two numbers, and each number is repeated twice! And the number is divisible by each of our ages!” “You’re Correct!” Mr Smith shouts as he whips out his iPhone to check. “and look, the last two digits are my age!”

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- Mathematician Algirdas Javtokas Takes Mathematics Into Conceptual Art (prweb.com)
- Tales of Badass Mathematicians: Cardano (via Measure of Doubt) (cienciaoberta.wordpress.com)
- Mathematicians warn of damage to UK from maths funding cuts (guardian.co.uk)

# 1089 (, 495 and 6174) and all that

^{st}number)

^{nd}number)

^{rd}number)

^{th}number)

^{rd}and 4

^{th}number together.

(x-z) (y-y) (z-x)

# Do you want a mathematician or a computer?

The best one I’ve heard from this is: “what is your biggest weakness?”

“I’m dedicated to my work, so I often work straight through a lunch break”.

an awesome blog to read, and the articles make you think.

using square tiles that have an area of 1 unit squared, how can you lay thetiles such that the number of tiles used is twice the number of tiles in theperimeter?

the way people approach this is fascinating. Most of us will start writing outequations, so perimeter of a rectangle is 2 x (width+length) and the area iswidth x length, so, using w,l to denote width, length respectively,

⇒4+4l/w=l

which isn’t easily solved. It would probably be best to start throwing integersinto the equation, and see what comes out.

so

again, try values of w

8+l-1=2l

7=l

http://mindyourdecisions.com/blog/2011/09/05/monday-puzzle-paying-an-employee-in-gold/

the solution is awesome.

This is all well and good…except for a little thing called the internet. Throughthe use of google, wolframalpha and Wikipedia, that 90% of the course can bedone by someone with no previous knowledge of the course. I would also guess that they can do it moreaccurately, and faster.

But enough of that. It’s fairly obvious I have crap marks and think I’m abetter mathematician than those marks show, but that’s my opinion. I likesolving problems, and this is a taster of the next post:

what do the numbers 1089 and 495 have in common?

# problematic views, futuring maths.

This view point is note entirely correct. Whilst there is a lot of logic in mathematics and it is certainly helpful to be able to think logically, it is not all there is to it. Mathematics is – by necessity – a very creative science. To stretch mathematics to new areas, you need to be creative in your approach. To attempt a proof of something that no one else has done, you need to be creative.

This being said, there are rules and logical steps that you follow – just like everywhere else. Music is governed by certain laws of physics. Art is governed by visual laws. Poetry and literature are governed by laws of the medium it is written in – even poets like e.e. cummings ascribe to certain laws, albeit they enjoy breaking most of them. Mathematics is the same.

A few people I have told this to have been sceptical of my views. They remember their mathematics at school as being very strict and formal. Very few were privileged enough to be shown the joys of creating new mathematics, or taught to think rather then remember. To people who view mathematics as being a logic and rule based science with no deviations, I suggest you look up some recreational math authors (or contact me for some who I’ve read).

You can’t start counting and eventually get there – infinity is easily viewed as the biggest number you can think of, plus 1. You can’t ever reach it. Your only way of examining it is to simplify it, or to creep up on it and surprise it.

assume you want to show that there is a set of consecutive composite numbers of any size.

(so the number n is not prime, nor is n+1, n+2,…for the next s numbers.)

here’s the trick: if you take a number n, and you times it by n-1,n-2…2 then the number you get is called the factorial of n, and is shown as n!. so n!=n*(n-1)*(n-2)*…*2

now this is obviously composite. We can also show that n!+2 is composite as n! has a factor of 2, as does 2, so it is composite. We can do this for any factor of n! – so for all numbers from 2 to n-1, we can add it n! and get a composite number. Assume we choose k. then n!=n*(n-1)*…*k*…*2, and n!+k=k( (n*(n-1)*…*(k+1)*(k-1)*…*2+1), so k is a factor.

this means that all numbers from n!+2 up to n!+n are composite. So to find a set of size s, take the s numbers after (s+1)!+1 – they’ll all be composite. (quick question, are there any sets of size 2? Why?)

I am not saying here that computers will not be able to prove number theory problems. There is a role for computers in proofs. Sometimes we can reduce a problem down to a set of numbers – so if it satisfies these numbers, then it satisfies all numbers. This is the way the four colour theorem in Graph theory was solved. I wouldn’t be surprised if the Collatz conjecture is similar. but I believe that no computer will ever be able to take a conjecture and prove it from scratch. I may be mistaken, I do think I recall a book that said that Turing proved that computers won’t be able to find solutions to all problems, which leaves the possibility that some problems will be proved by computers, but my current knowledge does not know of any that have been done by a computer.

there is a separate blog post about this, but to reiterate: take any number n. if it’s odd, multiply it by 3, and add 1. If it’s even, divide it by 2. Repeat for the new number. Keep doing that, and eventually you’ll reach 1 – or at least, that’s the conjecture (which means it hasn’t been proved).

Assume that, a long time ago, there was a village in the middle of nowhere with a very detailed language. Now a huge fight erupted, and because of the fight, many people in the village decided to leave and start new villages. These people all formed their own groups which then left the village and headed off in their own direction, never to travel back to the original village.

now over the following thousands of years, this happened many times. Occasionally, splinter groups from one village would join up with another splinter group, forming a new village, but people never travelled to an existing village.

# Collatz’ Conjecture

^{k}, where k is any natural number, so 4=2

^{2}, 8=2

^{3},256=2

^{8}). So if a sequence ever reaches a power of 2, it will reach 1.

^{2}, we already knew it would lead to 1, so we ignore it for now. But 3 is a ‘new’ number, and the interesting thing is that 3 multiplied by a power of 2 will also eventually lead to 1!

^{k}. so it will look like this:

Now, as stated above, the number 5 leads into 16, so we can view this as a branch with a root of 5, which links into 16. Our tree will now look like this:

^{k}, which is also 2

^{k}× 1. If we look at each branch, you will notice that each one can become a new tree of it’s own, with the main trunk having a root number, and then consisting of all natural numbers of the form 2

^{k}× the root.

^{k}× the root, and has branches extending out of it at some points.

^{k}× the root. The roots will then be referred to as the p

^{th}root, meaning that this root leads into the p-1 branch, and we will define it as root(p). the branches will then be branch(p), so that branch(p) is the sequence of numbers 2

^{k}× root(p).

^{3}× 5, and 5 is a member of root(2).