MAM1000 UCT lecture notes – Mathemafrica http://www.mathemafrica.org All about maths in Africa Mon, 20 Apr 2020 06:44:50 +0000 en-US hourly 1 https://wordpress.org/?v=4.4.2 Proof by Contradiction http://www.mathemafrica.org/?p=13399 http://www.mathemafrica.org/?p=13399#respond Mon, 17 Apr 2017 21:35:39 +0000 http://www.mathemafrica.org/?p=13399 The concept of proof by contradiction refers to taking a statement and assuming the opposite is true. When assuming the opposite is true we begin to further examine the our ‘opposite’ statement and reach to a conclusion which doesn’t add up or in simple terms is absurd.

Take the case:

Statement: There are infinite number of prime numbers.

Using the concept of proof by contradiction, we will assume the opposite is true.

If an integer (2) divides an integer (6) we say that 2 divides 6 or 2|6. In a more general sense we can say that if any integer ‘a’ divides any other integer ‘b’ then a|b.

Prime numbers: it is an integer (n ≥ 2) that has exactly two positive factors (1 and itself).

eg. 2, 3, 5 …

Composite numbers: it is an integer (n ≥ 2) that has more than two positive factors.

eg. 4, 6, 8 …

Fundamental Theorem of Arithmetic: Every integer n ≥ 2 has a unique (exactly one) prime factorization.…

]]> The concept of proof by contradiction refers to taking a statement and assuming the opposite is true. When assuming the opposite is true we begin to further examine the our ‘opposite’ statement and reach to a conclusion which doesn’t add up or in simple terms is absurd.

Take the case:

Statement: There are infinite number of prime numbers.

Using the concept of proof by contradiction, we will assume the opposite is true.

If an integer (2) divides an integer (6) we say that 2 divides 6 or 2|6. In a more general sense we can say that if any integer ‘a’ divides any other integer ‘b’ then a|b.

Prime numbers: it is an integer (n ≥ 2) that has exactly two positive factors (1 and itself).

eg. 2, 3, 5 …

Composite numbers: it is an integer (n ≥ 2) that has more than two positive factors.

eg. 4, 6, 8 …

Fundamental Theorem of Arithmetic: Every integer n ≥ 2 has a unique (exactly one) prime factorization.…

]]> http://www.mathemafrica.org/?feed=rss2&p=13399 0 The mean value theorem for integrals – proof http://www.mathemafrica.org/?p=11091 http://www.mathemafrica.org/?p=11091#respond Fri, 07 Aug 2015 09:23:19 +0000 http://www.mathemafrica.org/?p=11091 I left out one line at the beginning, which was pointed out in class.

We start with some function, f(t) which is continuous on [a,b]. We define:

F(x)=\int_a^x f(t)dt

By the fundamental theorem of calculus, this is continuous of [a,b] and differentiable on (a,b).

Now, we use the mean value theorem for derivatives (somewhere in an interval, a function has a gradient which is equal to its average gradient) which say that if g is differentiable on (a,b) and continuous on [a,b] then there exists some c between a and b such that g'(c)=\frac{g(b)-g(a)}{b-a}.

Now simply using our function F(x) in the mean value theorem for derivatives, we have that:

F'(c)=\frac{F(b)-F(a)}{b-a}

But we know that F'(c)=f(c). We also have defined F(x) above, so we can plug in a and b to get:

f(c)=\frac{\int_a^b f(t) dt-\int_a^a f(t) dt}{b-a}

The second term on the right hand side is zero as the two limits are the same, so we have:

f(c)=\frac{\int_a^b f(t) dt}{b-a}

But we know that \int_a^b f(t) dt=\int_a^b f(x) dx as t and x are just dummy variables, and so we have proved that:

f(c)=\frac{\int_a^b f(x) dx}{b-a}

This finishes the proof.…

]]> I left out one line at the beginning, which was pointed out in class.

We start with some function, f(t) which is continuous on [a,b]. We define:

F(x)=\int_a^x f(t)dt

By the fundamental theorem of calculus, this is continuous of [a,b] and differentiable on (a,b).

Now, we use the mean value theorem for derivatives (somewhere in an interval, a function has a gradient which is equal to its average gradient) which say that if g is differentiable on (a,b) and continuous on [a,b] then there exists some c between a and b such that g'(c)=\frac{g(b)-g(a)}{b-a}.

Now simply using our function F(x) in the mean value theorem for derivatives, we have that:

F'(c)=\frac{F(b)-F(a)}{b-a}

But we know that F'(c)=f(c). We also have defined F(x) above, so we can plug in a and b to get:

f(c)=\frac{\int_a^b f(t) dt-\int_a^a f(t) dt}{b-a}

The second term on the right hand side is zero as the two limits are the same, so we have:

f(c)=\frac{\int_a^b f(t) dt}{b-a}

But we know that \int_a^b f(t) dt=\int_a^b f(x) dx as t and x are just dummy variables, and so we have proved that:

f(c)=\frac{\int_a^b f(x) dx}{b-a}

This finishes the proof.…

]]> http://www.mathemafrica.org/?feed=rss2&p=11091 0