Is 0.101001000100001.....a rational or irrational number?

0.101001000100001.....is an irrational number because it cannot be put in the form m/n where m and n are integers.

To understand this we can consider how recurring decimals can be turned into the form m/n.

Firstly we consider one that repeats every 1 digit.

0.333333333333….. can be put in the required form by the following procedure:

Let x = 0.333333333333…..

Multiplying by 10 gives: 10x = 3.333333333333…..

Subtracting gives 9x = 3

Therefore x = 3/9 = 1/3

Net we consider one that repeats every 2 digits.

0.171717171717….. can be put in the required form by the following procedure:

Let x = 0. 171717171717…..

Multiplying by 100 gives: 100x = 17. 171717171717…..

Subtracting gives 99x = 17

Therefore x = 17/99

In each case we have multiplied by the length of the recursion to find the fraction.

I we look further at 0.101001000100001..... we will find that the sequence does not recur.

The digit 1 occurs at positions 1, 3, 6, 10, 15 etc. The gap between each is following an increasing series 2,3,4,5,… So we cannot find a finite number that we need to multiply by to get a rational number.

(ii) When the reciprocals 1/2,1/3,1/4,1/5,1/6,1/7,1/8......are turned into decimals, some repeat and some do not.

When turning any of these numbers into decimals we are effectively dividing into 10 or powers of 10, ignoring the decimal point.

For example ½ is: _0.5_____ 2)

1.000000 The process was a) try dividing 2 into 1. b) It wouldn’t go so we carry the 1 and divide into 10. C) The 2 went into 10 five times and so the decimal was found. Note that the 2 is a factor and so divided exactly.

The other prime factor of 10 is 5, so we could try the same thing with 1/5.

 _0.2_____ 5) 1.000000

The process was a) try dividing 5 into 1. b) It wouldn’t go so we carry the 1 and divide into 10. C) The 5 went into 10 five times and so the decimal was found.

It follows that any number that is built up from the prime factors of 10 will turn into an exact decimal.

This means any number of the form 2^m X 5^n where m and n are integers will give exact decimals e.g. 1/25 is 0.04 .

It also follows that all other numbers will not divide exactly, but will recur. We next need to know why this will happen.

First consider 1/3.

Dividing 3 into 1 won’t go, so we carry the 1 and divide into 10.

It goes 3 times with a remainder of 1. This is carried to make 10 again and so we are repeating the same division. It follows we are repeating the same process so the decimal will recur.

When dividing by 3 the remainder can’t be 0 or because that would mean the 3 had divided exactly into the number, but we know it isn’t a factor of 10 so this can’t happen. The only valid remainders are therefore 1 or 2. The decimal would therefore have to recur after 1 or 2 digits.

We can follow the same process with 1/7. This will give 0.142857142857142857…

In other words 0.142857 recurring. This has a cycle length of 6. Note that when dividing the only possible remainders are 1 to 6 so this is the maximum cycle length.

This can be seen by long division. 0 . 1 4 2 8 5 7 1

7)1 . 0 0 0 0 0 0 0 0 0 0

7

3 0

2 8

2 0

1 4

6 0

5 6

4 0

3 5

5 0

4 9

1 0 Reminder now same as step 2 so recursion has started The process can be repeated for any divisor that is not of the form 2^m X 5^n where m and n are integers.

If the numerator is not 1 then the same logic applies for any number in fully cancelled down form. For example 2/3 will recur (0.6666…) because the bottom is not of the form 2^m X 5^n where m and n are integers. 7/40 will be exact decimal (0.175) because the denominator is of the form 2^m X 5^n i.e. 40 = 2^3 X 5

Q2.describe how you would sketch the graphs of: f(x)=(x^5/5)-x^3 and two other examples of cubic functions

f(x)=(x^5/5)-x^3

i) Coordinates of the point where the graph crosses the y-axis.

At this point x=0. Therefore f(0) = 0. Point is (0,0)

ii) Coordinates of the point where the graph crosses the x-axis.

At this point f(x)=0. Therefore (x^5/5)-x^3 = 0.

Therefore x^3 ((x^2/5) -1)= 0

Implies x^3 = 0 therefore x=0 or (x^2/5) – 1 = 0 therefore x = -\/5 or \/5 Points are (0,0) when x =0; (-\/5,0) when x =-\/5; (0, \/5) when x = \/5.

iii) Local maxima and minima, points of inflection and slopes at these points.

Differentiating f(x)=(x^5/5)-x^3 gives:

f’(x) = x^4 – 3x^2

(f’(x) is alternative notation for dy/dx)

Stationary points occur when f’(x) = 0 i.e. when slope = 0,

therefore: x^4 – 3x^2 = 0

therefore: x^2(x^2 – 3) = 0

therefore: x^2 = 0 so x = 0

or (x^2 – 3) = 0 so x = -\/3 or \/3

Need to find f’’(x) to classify points as max, min or point of inflection. (This can also be done by putting small values in f(x) either side of x= 0, x = -\/3 and x = \/3).

 As f’(x) = x^4 – 3x^2 by differentiating again f’’ (x) = 4x^3 – 6x

x = 0

When x = 0, f’’ (0) = 0 i.e. value of second differential when x=0.

When second differential = 0 this is usually a point of inflection, but need to check that f’’ (x) changes sign passing through the point.

f’’ (-0.1) = 0.596 and f’’ (0.1) = - 0.596 so condition is satisfied and the conclusion is that it is a point of inflection.

x = -\/3

When x = -\/3, f’’ (-\/3) = -10.39 i.e. NEGATIVE indicating a MAXIMUM

x = \/3

When x = \/3, f’’ (\/3) = 10.39 i.e. POSITIVE indicating a MINIMUM

iv) Where the function is increasing and where the function is decreasing. 

When x < -\/3 function is increasing (positive slope f’ (x) > 0)

When x =-\/3 function is level (zero slope f’ (x) = 0)  

When 0<x<v3 function is decreasing (negative slope f' (x)

When -\/3<0 x="0"><0> \/3 function is increasing (positive slope f’ (x) > 0)

cubic functions

Note: the standard definition of a cubic function is one in which the highest power is 3. The above wasn’t a cubic function as the highest power was 5. The following two are cubic.

EXTRA FUNCTION 1

f(x)=x^3-3x

 i) Coordinates of the point where the graph crosses the y-axis.

At this point x=0. Therefore f(0) = 0. Point is (0,0)

ii) Coordinates of the point where the graph crosses the x-axis.

At this point f(x)=0. Therefore x^3-3x = 0.

Therefore x (x^2 - 3)= 0

Implies x = 0 therefore x=0 or x^2 – 3 = 0 therefore x = -\/3 or \/3

Points are (0,0) when x =0; (-\/3,0) when x =-\/3; (0, \/3) when x = \/3.

iii) Local maxima and minima, points of inflection and slopes at these points.

Differentiating f(x)=x^3-3x gives:

f’(x) = 3x^2 – 3

Stationary points occur when f’(x) = 0 i.e. when slope = 0,

therefore: 3x^2 – 3= 0

therefore: 3(x^2 – 1) = 0

therefore: x^2 = 1 so x = -1 or 1

Need to find f’’(x) to classify points as max, min or point of inflection.

As f’(x) = 3x^2 – 3 by differentiating again f’’ (x) = 6x

x =-1

When x = -1 f’’ (-1) = -6 i.e. NEGATIVE indicating a MAXIMUM

x = 1

When x = 1 f’’ (1) = 6 i.e. POSITIVE indicating a MINIMUM

iv) Where the function is increasing and where the function is decreasing.  

When x < -1function is increasing (positive slope f’ (x) > 0)

When x =-1 function is level (zero slope f’ (x) = 0)

When -1<1 x="1"><1>1function is increasing (positive slope f’ (x) > 0)

EXTRA FUNCTION 2

f(x)=x^3

i) Coordinates of the point where the graph crosses the y-axis.

At this point x=0. Therefore f(0) = 0. Point is (0,0)

ii) Coordinates of the point where the graph crosses the x-axis.

At this point f(x)=0. Therefore x^3= 0.

Therefore x = 0

Points is (0,0) when x =0.

iii) Local maxima and minima, points of inflection and slopes at these points.

Differentiating f(x)=x^3 gives:

f’(x) = 3x^2

Stationary points occur when f’(x) = 0 i.e. when slope = 0,

therefore: 3x^2 = 0

therefore: x = 0 is the only solution

Need to find f’’(x) to classify points as max, min or point of inflection.

As f’(x) = 3x^2 by differentiating again f’’ (x) = 6x

x = 0

When x = 0, f’’ (0) = 0 i.e. value of second differential when x=0.

When second differential = 0 this is usually a point of inflection, but need to check that f’’ (x) changes sign passing through the point.

f’’ (-0.1) = -0.001 and f’’ (0.1) = 0. 001 so condition is satisfied and the conclusion is that it is a point of inflection.

iv) Where the function is increasing and where the function is decreasing.

When x < 0 function is increasing (positive slope f’ (x) > 0)

When x = 0 function is level (zero slope f’ (x) = 0)

When x >1function is increasing (positive slope f’ (x) > 0)

At no point does this function decrease.