How to define ‘Greater Than’

September 20, 2010 in General, MS 2001 | Leave a comment

How can a statement like “5 is greater than 4” be quantified? Or is it just obvious? If we know anything about mathematics we know that there is no way we can assume something as obvious, there must be an axiomatical contruct that puts a rigorous meaning on “5 is greater than 4“.

The first attempt would be to say that 5=4+1 so 5 is “1 more” than 4 so must be bigger. This translates to 5-4=1: “5 is greater than 4 because 5-4 is positive“. Careful! 4=5+(-1) so 4-5=-1: “4-5 is negative“. But what does positive and negative mean? Easy? Positive is greater than zero… At this point a stronger construct is needed:

Definition: Call a set P\subset \mathbb{R} positive if for all a,b\in P

  1. a+b\in P
  2. ab\in P
  3. Given x\in \mathbb{R} either x\in P, -x\in P or x=0

If we think carefully, this definition concurs exactly with that of the naive notion of positive. So we can say that “5 is greater than 4 because 5-4 is positive.”

Definition: Given a,b\in\mathbb{R},  a is said to be greater than b, a>b, if a-b is positive.

MS 2001: General Information

September 20, 2010 in MS 2001 | Leave a comment

Lecturer: Mr. J.P. McCarthy

Office: Mathematics Research, Western Gateway Building
Meetings by appointment via email only.

Email: jippo@campus.ie

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An inductive proof of the Factor Theorem

September 8, 2010 in Leaving Cert, MS 2001 | 2 comments

At Leaving Cert you are only required to prove the Factor Theorem for cubics. This is a more general proof.

The Factor Theorem is an important theorem in the factorisation of polynomials. When (x-k) is a factor of a polynomial p(x) then p(x)=(x-k)q(x) for some polynomial q(x) and clearly k is a root. In fact the converse is also true. Most proofs rely on the division algorithm and the remainder theorem; here a proof using strong induction on the degree of the polynomial is used. See Hungerford, T.W., (1997), Abstract Algebra: An Introduction. Brooks-Cole: U.S.A for the standard proof (this reference also describes strong induction).

Factor Theorem
k\in\mathbb{C} is a root of a polynomial p(x) if and only if (x-k) is a factor: p(x)=(x-k)q(x) where q(x) is a polynomial.

Proof:
Suppose deg p(x)=1. Then p(x)=ax+b and p(k)=ak+b=0\Rightarrow k=-b/a. Clearly p(x)=(x-(-b/a))a and q(x)=a is a polynomial (of degree 0).

Suppose for all polynomials of degree less than or equal to n-1, that k a root implies (x-k) a factor.  Let p(x)=\sum_ia_ix^i be a polynomial of degree n and assume k\in\mathbb{C} a root. Now

p(x)-p(k)=\sum_{i=0}^na_ix^i-\sum_{i=0}^na_ik^i

\underset{p(k)=0}{\Rightarrow} p(x)=\sum_{i=1}^n a_i(x^i-k^i)

Now each x^i-k^i for i=1,\dots,n-1 is a polynomial of degree less than or equal to n-1, with a root x=k, hence (x-k) is a factor of each. Thence x^i-k^i=(x-k)q_i(x) where q_i(x) is a polynomial:

p(x)=\sum_{i=1}^{n-1}a_i(x-k)q_i(x)+a_n(x^n-k^n)

Now x^n-k^n=(x-k)x^{n-1}+kx^{n-1}-k^n but kx^{n-1}-k^n is a polynomial of degree n-1 with root k and hence (x-k) is a factor:

x^n-k^n=(x-k)x^{n-1}+(x-k)h(x)

Hence

p(x)=(x-k)\lbrack x^{n-1}+h(x)+\sum_{i=1}^{n-1}a_iq_i(x)\rbrack

 

Hence (x-k) a factor \bullet

The Basics of Linear Algebra

July 10, 2010 in MA 2055 | 1 comment

An essence of mathematics is to cut through the gloss of each topic and exhibit the core features of the construct.

We all encountered vectors as a teenager. We may have been told that a vector is a physical quantity with magnitude in direction. This is to distinguish it from a scalar – which has magnitude alone. It begs the question – what has direction without magnitude? Anyway we can think of those familiar arrow vectors in the plane, think of all possible magnitudes and directions, and think of them as some big, infinite set V. One basic thing about these vectors is that we can add them together – by the parallelogram law and more than that when we add them together we get another vector. So there is a way to add vectors u, v together, and also u+v\in V. This means + is a closed binary relation. Furthermore it is doesn’t matter in what order we add the vectors: u+v=v+u, the addition is commutative. Also we can take a vector v and double it, or treble it or indeed multiply it by any real number k\in \mathbb{R} – and again we’ll get another vector kv\in V (with 1v=v). We don’t care about multiplying two vectors – at the moment. There is special zero vector, \mathbf{0}, such that for any vector v, v+\mathbf{0}=v. Every vector v has a negative, namely -v, such that v+(-v)=\mathbf{0}. Also the addition and scalar multiplication satisfies some nice algebraic laws; those of  associativity and distributivity.

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MS 2001: Useful Links

July 10, 2010 in MS 2001 | Leave a comment

Stephen Wills MS 2001 homepage: http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/MS2001.html

Notes from there: http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/MS2001_notes.pdf

Past Exam Papers: http://booleweb.ucc.ie/ExamPapers/maths_studies.html