------------------------------------------------------------

Ask NRICH: Onwards and Upwards: A Level Mathematics

------------------------------------------------------------

Posted by QED963 on Tuesday, 11 June, 2013 - 07:39 am:

Hi I am a 17 year old who is taking the A Level Syllabus As of now, I am
studying Michael Spivak's text on Calculus and have done
the questions up until Chapter 1 Question 13. I know Calculus although not
rigorously and not the way I would have liked to be taught. (Limits were not defined mathematically;
intuition had to guide us most of the time.)

The following is a list of questions I have about mathematics: 1. Numbers and
its meaning.

1.1: Proof that root 2 is an irrational and how to generalize it. I have made
this as rigorous as possible although some will surely say that I have
overworked the problem. Proof that root 2 is irrational; the first axiom I
shall make and you must accept before we proceed is that root 2 is a number. (I
know that this is an obvious 'fact' but it is a fact which cannot be 'proved'
mathematically). Now if you have accepted this, I am about to make another
postulate. (A postulate is an 'axiom' which is not as evident.) Postulate: All
numbers are complex numbers ( in fact we don't know if this is the full
classification of numbers, there may very well be a bigger classification which
we are just unaware of.) If you have accepted this two rather obvious facts,
then we can proceed. Otherwise, re-read the above statements and convince
yourself of its truth.

Listing the axiom and postulate in order,

1. Root 2 is a number

2. All numbers are complex numbers.

Using this two facts and elementary inference, we can conclude that root 2 is a
complex number. The question as to
whether it is real or imaginary has to be determined. Most of us would accept
based on intuition that root 2 is real. But here, there is a more rigorous way
to prove such a statement. Let us first turn to the definition of the imaginary
number. Imaginary number: An imaginary number is a number that when squared gives
a negative result. (This is a rather loose definition, but for our
purpose, it will serve us well.) Now let us compare root 2 with the definition
of an imaginary number. Squaring root 2 gives a positive result which is not the
definition of an imaginary number. Hence, root 2 is real. I would like you to
re-read what I have just done. I have proved that root 2 is real by proving
that root 2 is not imaginary. ( i.e, if there are two
cases and we eliminate one, then the other case must be true.)

This method will naturally suggest the way to prove that root 2 is irrational. (
i.e, we prove that root 2 is irrational by proving
that it is not rational) Before we can show all this, you must understand that
a real number can be broken down into either a rational or an irrational
number.

We once again start off with the definition of a rational number. Rational
Number: A rational number is a real number that can be expressed as a fraction
p/q where p and q are integers and q is not equal to 0. The rest is simple
mathematics. We simply suppose that root 2 can be expressed as a fraction p/q
where p/q is a fraction in its simplest form. I must stress the last portion of
the assumption for this is what leads to the proof.

The formal proof: Proof by contradiction (Also known in Latin as Reducto Ad Absurdum).

Suppose root 2 = p/q where p/q is a fraction is in it's
simplest form.

Squaring both sides: 2 = p^2/q^2; 2q^2 = p^2

This implies that q^2 is even. Thus, q is even.

Therefore, we can substitute 2k for q where k is an element of the integer set.

Then 2q^2 = 4k^2 Dividing by 2 throughout. q^2 = 2k^2 this implies that q is
even.

But, p/q is supposed to be a fraction in its simplest form. This contains a
contradiction which can only mean one of two things:

1. My working was wrong

2. The original hypothesis was wrong.

Checking, we see that 1. Is out of the equation. That must mean that out
original hypothesis is wrong. This implies that root 2 is not rational, which
means to say that root 2 is irrational. (Q.E.D.)

NB: (Q.E.D. is Latin for 'which was to be shown') and the proof is complete.

Now I would like to generalize this proof as much as possible:

We see that root 3 can be proved to be an irrational using the same method
described above. Now I want to generalize this proof and prove that root x is
irrational where x is an element of positive real numbers and x not equivalent
to n^2 where n is an element of rational numbers.

This proof however is still not good enough. What about the generalization nth
root of a is an irrational where a is an element of real numbers (for odd nth)
and a not equal to m^n for m
an element of rational number?

Also what does a number, b, raised to the number of an irrational mean?

For instance, x^root 2 = ??

I would do it as follows;

Let x^root 2 = y

Raising to the power of root 2 to both sides, we obtain :

x^2 = y^ root 2

But this is (in hindsight) plain stupid.

I have therefore accepted the fact that we can only approximate x^root

2 (in passing, I am tempted to ask how would one prove that a^root
2 is an irrational.)

2. Differentiation

2.1: Differentiating e^x w.r.t x from first
principals. d/dx (e^x) = lim
h-> 0 (e^x.e^h - e^x)/(h)

This can be factored into e^x lim
h->0 (e^h -1)/(h).

Now this is then equal to 1.

My question is this; suppose you replace e with say a number, a. Then, why does
lim h->0 (a^x -1)/(h)
not equal to 1. What makes it so different?

I would also like to add that I do NOT want an answer that says that this is
experimentally proven. That would be a disgusting and frankly speaking
dishonest answer.

Thank you.

------------------------------------------------------------

DO NOT REPLY TO THIS MESSAGE!

Use this link to go directly to the discussion:

https://nrich.maths.org/cgi-bin/discus/show.cgi?27/155790