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Posts Tagged ‘algebra’

Why the proof 2=1 is wrong

September 17, 2009 Leave a comment

There are a number of apparent proof that 2=1. An example of one such proof is below

Suppose a=b then
a2 = ab
2a2=a2+ab
2a2-2ab = a2+ab-2ab
2a2-2ab=a2-ab
2(a2-ab) = 1(a2-ab)
2 = 1

However this is clearly not true which mean there must be a step in the above algebra which isnt valid.

The invalid step

The invalid step is  in fact the very last step in the proof, cancelling both sides by a2-ab. This is because a2-ab = 0 since we initially assumed a=b hence

ab = aa = a<sup>2</sup>

This then means that the second last line states

2 x 0 = 1 x 0

which is obviously true but you cant “cancel” both sides by 0 since any number divided by zero is meaningless and doesnt have a values. The trick involved in this proof is hiding the divide by zero which most people would recognise as invalid with a more complicated looking expression.

What this means

This example demonstrates the dangers of allowing divide by zeros to occur in your calculations even if its in the form of algebra. It also demonstrates the problems of considering infinty (x/0) as a number that can be used in calculations.

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Categories: algebra, maths Tags: , , , ,

Simultaneous Equations

September 4, 2009 Leave a comment

Simultaneous equations are two or more equations that are all related because they contain the same variables. To solve these equations you must use information from both equations to find a set of values for the variables that hold true for both equations.

You can only solve simultaneous equations if there at least as many different equations (being different meaning that the two equations cant be re-arranged or simplified into each other) as there are variables in the equations. In order to solve the equations we want to eliminate all the variables but one so that we are left with a single equation that can be solved normally.

There are several ways of solving these equations. The method of solving using substitution is shown below using an example.

Solve using substitution – with an example

Lets consider the equations
(1)- y + 3x = 12 + 4y
(2)- 3x = 7y – 6 + 4x

First of all we must simplify the equations by gathering all the similar terms together and cancelling through any common factors. So we get
(1)=> 3x = 12 + 3y => x = 4 + y
(2)=> 6 = 7y + x

We would now normally arrange one of these equations so that either x or y is the subject of the equation however in this example equation 1 naturally has x as the subject.

We can now substitute the expression for x in equation 1 (4+y) into equation 2 to get

6 = 7y + 4 +y

Again we must now simplify this to get

2 = 8y

and then we can solve it as a single equation to find
y=1/4

We can now substitute this value for y into either of the above equations to get a value of x. Substituting into 1 gives

x = 4 + 1/4
x=5/4

So that we now have our values of x and y that satisfy both of the equations

x=5/4 and y=1/4

If you have any questions or comments please them in the form below

Cartesian Equation of a Circle

June 28, 2009 2 comments

Since a circle is made up of all the points a fixed distance (its radius) from a given point (its centre) then the equation of a circle simply needs to ensure this is true. This can be done using Pythagoras’s theorem. This is because we can draw a right angled triangle with the centre of the circle at one corner and the point on the circle at the opposite corner as shown below. The radius is then the hypotenuse, the vertical side is the difference between the y co-ordinate of the point and that of the centre and the horizontal side is the difference between the x co-ordinate of the point and centre. From Pythagoras we therefore know that a circle of radius r and centre (a,b) must have a Cartesian equation

r2 = (x-a)2 + (y-b)2

Circle on Cartesian axis

Circle on Cartesian axis

However, we can expand these brackets out to get

r2 = x2 – 2ax + a2 + y2 – 2by + b2

but since a2+b2+r2, -a and -b are all constant we can let

c = a2+b2-r2,
g = -a
f = -b

to get

x2 + y2 + 2gx + 2fy + c = 0
where the circle has a centre (-g,-f) and radius √(a2+b2-c2)

Proof by Mathematical Induction

Proof by induction involves proving that if a statement is true in one case then it must also be true for the next case. Then by showing that it is true for a base case, eg 0, you can conclude that it is true for all positive integral cases, since if it is true for case 0 it must be true for case 1 and if it is true for case 1 it must also be true for case 2 etc

To do this it is normal to let Pn be the statement you are trying to prove in a general case n. Let us use the example Pn: 9n – 1 is divisible by 8 so we may right 9n-1 = 8a where a is an integer

Then we must prove that if Pk is true Pk+1 is also true. This will require using the expression Pk otherwise it isn’t induction since you are not showing that the truth of each statement can be found from the previous one.

so consider 9(k+1) – 1
This is equal to 9x9k -1=9x9k – 9+ 8=9(9k-1)+8
but from Pk we know 9k-1 = 8a so
9k+1-1 = 9(8a)+8=8(9a+1)
which is of the form 8b so is divisible by 8
Hence if Pk “9k-1 is divisible by 8″ is true then Pk+1 is also true

Finally we must prove that P1 is true and we then know that Pn is true for all k >=1 where k is an integer.

P1 is the statement 91 – 1 is divisible by eight
91-1=9-1=8 so is true

therefore Pn is true for all positive integral value of n

Equation of a Line: y = mx + c

May 3, 2009 4 comments

The equation of all straightlines can be written in the form

y=mx+c

where m is the gradient, c is the intercept on the y-axis and you are plotting xagainst y. m and c ae constants meaning their value is fixed.

graph of y=mx+c

graph of y=mx+c

The y intercept, c, is how far up, or down if its negative, he line crosses the veritcal yaxis.
The gradient, m, is how step the line is. A gradient equal to 1 means that the graph is at 45 degress to the axis, a gradient greater than 1 is steeper and a gradient less than 1 is shallower.

To find where the graph cuts the x-axis simply let y=0 and find the value of x since at the x-axis the grap has zero height so y=0.

The easiest way to sketch a graph of the form y=mx + c is to find the x and y intercepts and then draw a straight line through these points.

Find the gradient of a line

The gradient, m, of  a straight line can be found using the equation

m=change in y/change in x = (y1-y2)/(x1-x2)

Where x1 and y1 are the co-ordinates of one point on the line and x2 and y2 are the coordinate of another point on the line.


If you have any comments, questions or improvements please them as a comment below.

By David Woodford

Transformations of Graphs

January 11, 2009 Leave a comment

 

This looks at how a given graph will change when the the function is changed slightly eg how the graph y=x will change when it becomes y = 2x.

y = af(x)

The graph f(x) will "steeper" as the y value of each point is multiplied by a. It will appear like a "stretched" version of the graph y=f(x) 

graph transformation y=af(x)

 

 

y =f(x) + a

The graph f(x) will move up by the amount a as a is added to each y value. This means that the points of intersection of the graph and the y axis will increase by the amount a. The intersection of the graph and the x-axis will depend upon the function of the graph.

transformation graph f(x) + a

 

y = f(ax)

This will make the graph appear "narrower" beacuse y is taking the value of f(x) a time across. So if the value of f(x) is 3 when x =12 then the value of f(4x) = 3 when x=3 as 12/4=3.

transformation graph f(ax)

y=f(x+a)

This will shift the graph to the left by a. This is because the value of f(x) at x+a is displayed at the point x so effectively the graph occurs a earlier and therefore shifts to the left.

transformation graph y(x+a)

Categories: maths Tags: , , ,