Download Module 3 Chapter 5, Irrationals and Iterations pages 55 – 64 Popper

Module 3
Chapter 5, Irrationals and Iterations
pages 55 – 64
Popper 06 is the one in this Module
Recall:
The set of Rational numbers and the set of Irrational numbers form a partition on
the Real numbers.
i.e.
Every real number is one OR the other (NOT both or neither). The sets are
disjoint, there are no numbers in the intersection of the sets because no number
can be both) and when you union up the sets you get the Reals back.
Rational numbers are called the set Q for quotient. Each rational number is of the form
a
b
b  0 and a and b are integers.
If you choose to write a rational number as a decimal, you will find that the decimal either has an
infinite repeating part or terminates.
Irrational numbers on the other hand are in a set often called P (I is used for Integers!). They
sometimes have symbols representing the number. And if we try to put them in decimal form,
we find that the decimals are non-repeating and non-terminating. We often use symbols like 2
or  for irrational numbers.
The authors start off with an interesting discussion of how to determine if a number is irrational
by asking the question: Given the set of square numbers (a subset of the positive natural
numbers), can you find a pair of them that, when taken as a ratio leaves you with a simplified
number such as 2, 3, or 6. The answer is “no” and they discuss it pretty cleverly. On the other
hand there are pairs of squares so that one is 4 times another or 49 times another…and these,
when you take the square root of your ratio, turn out to be natural numbers.
This is by way of showing you yet another number pattern. Pattern recognition is a huge part of
mathematical behavior and is of interest to many non-mathematicians such a musicians and
doctors.
They then mosey into part of the main presentation for the chapter.
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First, they take the number 142857 and multiply it successively by the digits 1 through 7:
1 ( 142857) = 142857
2 ( 142857) = 285714
this is a permutation* of the preceding number
3 ( 142857) = 428571
another permutation
…
7 ( 142857) = 999999
oh, a pattern breaker and something we already
knew 999999 7 0
*reordering of the numbers by taking the last and placing it first – or the last 2 and placing them
first.
So, why is there a pattern in the first place…and why does it end with 7?
They cleverly divide 7 into 1 noting that
1
 .142857
7
The reciprocal of 7 is a rational number that is non-terminating BUT repeating
2
 .285714
7
And so on up to
7
which is the end of the pattern. 1 = .999999
7
Not all numbers act this way when turned into a reciprocal and divided with successively larger
numerators. But 7 is a special number.
It’s a special sort of prime known at a repetend.
Repetend:
a prime number whose reciprocal is a repeating decimal.
Not all primes are repetends! Repetends are a proper subset of the prime numbers.
2
2, for example is prime but
1
 .5 .
2
3, is prime and is a repetend:
5, is prime but
1
 .2 .
5
It is not a repetend.
1
 .3
3
It is not a repetend.
And among the repetends there is another proper subset called “full repetends”
A full repetend is a prime whose reciprocal repeats with a period one less than the repetend itself.
7 is a full repetend. The period for it’s reciprocal is 6.
1
 .142857 .
7
Let’s do a little set diagram to see these relationships:
We have the Reals, partitioned by the rationals and the irrationals, then we have repetends and
inside repetends we have full repetends. Let’s draw in the subsets and put an element or two
showing in each:
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Now let’s talk about how many repetends there are: LOTS. How do we find them – the hard
way, by actually dividing. There aren’t very many helpful theorems about repetends,
unfortunately.
There are 9 full repetends less than 100.
Now let’s spend a moment on 7 again. We know that 999999 7 0 …which I’ll rewrite as
106  1 7 0
which tells us that 106 7 1
Now recall Fermat’s Theorem from Module 2 (page 46):
If a is not divisible by p, a prime, then a P1  p 1 .
Our rewrite is another illustration of this theorem. Look at it again:
106 7 1
a is 10 and p is 7
This is an illustration of the theorem.
You would hope then, that this theorem might help us identify full
repetends!
1010 11 1 and this is true but useless…the theorem doesn’t
Using a = 10 and p = 11
guarantee that P – 1 is the SMALLEST exponent that works on a string of 9’s.
Yes, 11 divides this huge 10 digit number 9, 999, 999, 999. BUT 11 also divides 99. And it’s
reciprocal actually has a period of 2, not 10.
1
 .09
11
So Fermat’s Theorem doesn’t give us much help finding them
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On page 58, down at the bottom is a snippit of a proof that is nice to note, but not very useful
either!
“If n is a prime*, it’s period must end with a digit d which is the “complement” of n in the sense
that the product nd ends in a 9.”
*Let’s illustrate this and add one caveat. “If n is a prime and a repetend, …”
This doesn’t work for 2 or 5, or any prime that is not a repetend.
Let’s take n = 3,
1
 .3
3
Let’s take n = 7,
1
 .142857 so d is 7 and nd = 49
7
so d is 3 and nd = 9
1
 .0588235294117647 so d is 7 and nd = 7(17) = 119
17
17, btw, is the next full repetend in the list of those less than 100.
Let’s take n = 17,
Moving on the page 60 is another interesting fact that doesn’t help us much in finding the next
repetend:
The length of a period of a repetend is always a divisor of the prime less one.
Let’s illustrate this: 11 is a repetend. It has period 2 and 2 is a divisor of 11 – 1 = 10.
Another illustration. 7 is repetend it has period 6 which is a divisor of 7 – 1 = 6.
Looking a the chart, notice that 41 is repetend with period length 5 and 5 is a divisor of 41 –1 =
40.
Interesting, but not helpful!
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Popper 06 Question 1
Which of the following is NOT a property of the number 7?
A
Prime
B
Full Repetend
C
Odd
D
Repunit
E
Deficient
Still on page 60, we move to repunits. There’s a strange link between repunits and repetends
and it’s something you really wouldn’t guess at!
The chart I mentioned above lists all the primes that are repetends by period length up to 20.
Remember that some repunits with prime indices are composite, not prime, while others with
prime indices are prime.
If you take a period length that is prime (call it p) and you multiply all the repetends of that
period length you get R p . Astonishing but true.
Let’s illustrate this:
5 is prime. The 2 repetends that have period 5 are 41 and 271. When you multiply 41(271) you
get 11,111 which is R5 .
7 is prime – when you multiply 239(4649) you get R7 .
Summarizing: If we have R p a repunit with a prime number of ones and the repunit is
composite, we know that the factors are repetends with period P! Incredible but true.
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Popper 06, Question 2
21, 649 and 513, 239 are the only 2 repetends with period 11. Which of the following is NOT
true of their product (try this without your calculator!)
A
It is a rational number.
B
It is R11 .
C
It is a prime number.
D
It is deficient.
Now we shift gears hugely! Another of those detours that turn out to be important later.
Let’s review the Reals – recall the Rationals and the Irrationals form a partition on the reals.
And you’d think there would be no way to bridge the gap between these subsets, but there is a
way you just have to do it an infinite number of times.
Start with a rational number. Follow the steps of the process described below; first you’ll create
a simple irrational number. Do an infinite number of steps and the endpoint is an rational
number. You turn the simple irrational number into a rational one!
Let’s look at the process. It is called “iterated radicals”. Now an iteration is a repeated step and
a radical is the symbol that asks you to look for a root.
36
5
32
2
 6
 2
is irrational
In fact the square root of any prime is an irrational number.
So our process is to take a root of a number over and over.
Here’s the process: pick a natural number, n, and multiply n(n – 1) to get x. Then take the square
root of x. Add x to the square root and take another square root and so on. You’ll end up with a
nice irrational number.
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In symbols:
n
n(n  1)  x
x
x x
x x x
Do it an infinite number of times…and you’re back to a rational number.
Let’s look at how this works:
I’ll pick 13, a nice natural number. 13(12) = 156
156  12.489996... irrational
156  156  12.9893604... irrational
156  156  156  12.99997096...
And so on…which rational are we headed for: 13!
Amazing!
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Popper 06, Question 3
I
An iterated process is a repeated process.
II
This process takes an irrational number and turns it into a rational number in an infinite
process.
III
This process uses the formulas from page 62
Which of the above statements are true?
A
All of them are true.
B
I and II only.
C
None of them are true.
D
III only.
E
II andd III only
Now, back to partitions:
On the Natural numbers we have Evens and Odds (also known as Mod 2). The evens are a
proper subset of N and so are the odds. Evens intersect Odds is the empty set – there are no
elements in the intersection. And the union of Evens and Odds is the whole set of the Natural
numbers.
Integers Mod 3 is a partition on the integers. The disjoint subsets are [ 0 ], [ 1 ], and [ 2].
The Naturals can also be partitioned into {1}, {primes}, and {composites}.
Here’s a new partition on the Reals that was discovered in mid nineteenth century:
{algebraic numbers} and {transcendental numbers}. The definition of the Transcendental
numbers is easy: any real number that is not Algebraic is Transcendental. So let’s look at
algebraic numbers.
A real number is called algebraic if it is the solution of a polynomial with rational coefficients.
Ok, now we started the class talking about polynomials and here we are again talking about
them. Solutions to polynomials are the x values that make the polynomial equal to zero.
For example:
3x + 9 = 0
there is one solution and it is – 3.
It turns out that all integers are algebraic.
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Here’s another example:
x 2  3x  2  0
( x  1)( x  2)  0
x  1
x  2
Here’s the crucial example:
x2  2  0
x2  2
x 2
So, some irrationals are algebraic!
This is why this new partition matters! All rational numbers are algebraic and SOME, but not all
irrationals are algebraic, too. It really is a new partition on the reals.
Let’s look at this in the context of a set picture:
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The irrationals that are NOT algebraic are called transcendentals.
Dr. Liouville (French) found the first transcendental number in the middle part of the nineteenth
century. Not for the midsemester, but just for fun, let’s look at the number that he found:

10
k
 101  102  ... 
k 1
1
1
1



10 100 1000
11000100...001000...0001000...0001...
The 1’s are appearing at the factorials!
1!
2!
3! = 6 place is a 1
4! = 24th digit is a 1
5! = the 120th digit is a 1
…
And so on…non-terminating and non-repeating!
In 1873
the euler number e is proved transcendental.
In 1882
pi is transcendental
Then, any natural number to an irrational power is transcendental! Aka: an irrational power of a
natural number.
So, what are these?
5
3
R3
2
Who would have thought up THAT number?
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Popper 06, Question 4
7 is an algebraic number.
A
true
B
false
Now back to transcendental numbers. It is a fact that you can add 2 carefully chosen
transcendental numbers and get an algebraic number. This is a situation that is rather strange
because if you add or multiply 2 rational numbers, you get another rational number!
Let’s look
    0
Now, it’s true that adding 2 arbitrary transcendentals will probably stay transcendental:


2

3
2
for example.
But if you’re careful! You can add or multiply your way right out of the set – no interated
process needed at all!
You can also do the trick of moving out of transcendentals to algebraic with carefully chosen
powers:
(5 2 )
2
 52  25
And doing it NOT carefully will leave you in the transcendentals.
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Popper 06, Question 5
Which of the following is NOT an irrational power of a natural number?
A
(1)
B
5
C
R5 3
2
3
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