My name is Liam and I'm currently studying Mathematics at Queen's University, Belfast.

I am a good people person and have great enthusiasm for Maths in particular.

I have previously tutored students preparing for GCSE and A-Level exams and felt great satisfaction in helping those students understand the subject in any way that I could.

In my experience, to do well in Maths regular practice and repetition is essential. Although receiving tuition for the subject is beneficial, in order to truly excel the student must be independent and have a mature approach to the subject; practicing past paper questions regularly in their own time.

My name is Liam and I'm currently studying Mathematics at Queen's University, Belfast.

I am a good people person and have great enthusiasm for Maths in particular.

I have previously tutored students preparing for GCSE and A-Level exams and felt great satisfaction in helping those students understand the subject in any way that I could.

In my experience, to do well in Maths regular practice and repetition is essential. Although receiving tuition for the subject is beneficial, in order to truly excel the student must be independent and have a mature approach to the subject; practicing past paper questions regularly in their own time.

No DBS Check

It's worth remembering FOIL: First, Outside, Inside, Last.

Say we have the following:

*(a+b)(c+d), where a,b,c and d are positive integers.*

Using FOIL, we have:

*(First) + (Outside) + (Inside) + (Last)*

ac + ad + bc + bd

__Examples:__

Expand the following brackets:

1. (2x+3)(4x+6) = **(****F) **8x^{2 }+ **(O) **12x + **(I) **12x + **(L) **18

= 8x^{2 }+ 24x + 18

2. (6x-1)(2x^{2}+4) = (6x)(2x^{2}) + 4(6x) + (-1)(2x^{2}) + (-1)(4)

= 12x^{3 }+ 24x - 2x^{2 }- 4

3. (x-1)(2x+4)(3x-6) *(Note: For this example, first expand the first two brackets, then expand the result with the third bracket, as follows):*

= (2x^{2}+2x-4)(3x-6) *(To expand this, multiply the first term in the 1st bracket by the 1st term and 2nd term in the 2nd, the 2nd term in the 1st by the 1st and 2nd term in the 2nd bracket, the 3rd term in the 1st bracket by the 1st and 2nd terms in the 2nd bracket, as follows):*

= 6x^{3 }- 12x^{2 }+ 6x^{2 }- 12x - 12x + 24

= 6x^{3 }- 6x^{2 }- 24x + 24

^{ }

It's worth remembering FOIL: First, Outside, Inside, Last.

Say we have the following:

*(a+b)(c+d), where a,b,c and d are positive integers.*

Using FOIL, we have:

*(First) + (Outside) + (Inside) + (Last)*

ac + ad + bc + bd

__Examples:__

Expand the following brackets:

1. (2x+3)(4x+6) = **(****F) **8x^{2 }+ **(O) **12x + **(I) **12x + **(L) **18

= 8x^{2 }+ 24x + 18

2. (6x-1)(2x^{2}+4) = (6x)(2x^{2}) + 4(6x) + (-1)(2x^{2}) + (-1)(4)

= 12x^{3 }+ 24x - 2x^{2 }- 4

3. (x-1)(2x+4)(3x-6) *(Note: For this example, first expand the first two brackets, then expand the result with the third bracket, as follows):*

= (2x^{2}+2x-4)(3x-6) *(To expand this, multiply the first term in the 1st bracket by the 1st term and 2nd term in the 2nd, the 2nd term in the 1st by the 1st and 2nd term in the 2nd bracket, the 3rd term in the 1st bracket by the 1st and 2nd terms in the 2nd bracket, as follows):*

= 6x^{3 }- 12x^{2 }+ 6x^{2 }- 12x - 12x + 24

= 6x^{3 }- 6x^{2 }- 24x + 24

^{ }

**[NOTE: An irreducible fraction (or fraction in lowest terms) is a fraction in which the numerator and denominator are integers that have no other common divisors than 1.]**

Given that p is a prime, positive integer and not a square number, we know that √p is irrational. Let's prove this:

__Proof:__*We shall use Proof by Contradiction;*

Let's suppose towards a **contradiction** that √p is in fact rational.

This implies that there exists two non-negative integers, call them **m and n**, such that:

√p = m / n , where n is not zero and m/n is an __ irreducible fraction__.

<=> p = m^{2} / n^{2 } *(Squared both sides)*

<=> pn^{2 }= m^{2}....(**#)** *(Multiplied through equation by n ^{2})*

From eqn(**#)**, it follows that, since p is prime, p | m^{2 }(** i.e. p divides m^{2}) **which thus implies also that p | m.

This means that there exists some natural number, call it k, such that:

pk = m....(**##**).

Now, sub eqn(**##) **into eqn(**#**), we thus obtain:

pn^{2} = p^{2}k^{2}

<=> n^{2 }= pk^{2 }*(Divided through by p)*

This last implies that p | n^{2}, which further implies:

p | n.

__BUT__ since p divides both m and n, this contradicts the fact that m and n were chosen to be irreducible, so our original assumption was incorrect.

**It thus follows that **√**p is in fact irrational, as required. **

**[NOTE: An irreducible fraction (or fraction in lowest terms) is a fraction in which the numerator and denominator are integers that have no other common divisors than 1.]**

Given that p is a prime, positive integer and not a square number, we know that √p is irrational. Let's prove this:

__Proof:__*We shall use Proof by Contradiction;*

Let's suppose towards a **contradiction** that √p is in fact rational.

This implies that there exists two non-negative integers, call them **m and n**, such that:

√p = m / n , where n is not zero and m/n is an __ irreducible fraction__.

<=> p = m^{2} / n^{2 } *(Squared both sides)*

<=> pn^{2 }= m^{2}....(**#)** *(Multiplied through equation by n ^{2})*

From eqn(**#)**, it follows that, since p is prime, p | m^{2 }(** i.e. p divides m^{2}) **which thus implies also that p | m.

This means that there exists some natural number, call it k, such that:

pk = m....(**##**).

Now, sub eqn(**##) **into eqn(**#**), we thus obtain:

pn^{2} = p^{2}k^{2}

<=> n^{2 }= pk^{2 }*(Divided through by p)*

This last implies that p | n^{2}, which further implies:

p | n.

__BUT__ since p divides both m and n, this contradicts the fact that m and n were chosen to be irreducible, so our original assumption was incorrect.

**It thus follows that **√**p is in fact irrational, as required. **

[__ Recall:__ the numerator of a fraction is the top number; the denominator refers to the bottom number.

A surd is an irrational number, e.g. √3, √5, etc.]

Given the following fraction:

(a+√b)/(c-√b), where a,b and c are non-negative integers and b is not a square number.

We can see that the denominator (c-√b) is irrational. To rationalise the denominator we take the following steps:

1. Multiply BOTH the numerator and the denominator of our fraction by (c+√b) in order to eliminate the irrational surd in the denominator.

__ Note:__ we perform this multiplication to both the numerator and denominator in order to preserve the value of the original fraction .

2. We now have for our numerator: (a+√b)(c+√b), and for our denominator: (c-√b)(c+√b).

Expand these brackets, thus we obtain the following fraction:

(ac+(a+c)√b+b) / (c^{2 }- b)

*Clearly we have succeeded in rationalising our denominator (whilst still maintaining the value of our original fraction) since (c ^{2}-b) is clearly a rational number, as required.*

**Example:**

**Write (5+7****√3)/(5-**√**3) in the form a+b**√**3, where a and b are rational.**

*Soln: We carry out the steps stated above;*

*Multiply numerator and denominator by (5+*√3), in doing so eliminating the irrational surd from our denominator. *We thus obtain:*

{(5+7√3)(5+√3)} / {(5-√3)(5+√3)} **(expand brackets)**

=(46+12√3) / (25+~~5√3~~-~~5√3~~ - 3)

=(46+12√3) / (22)

=23/11+(6/11)√3.

Clearly, from our orginal hypothesis, a=23/11, b=6/11 are both rational numbers, thus we are done.

[__ Recall:__ the numerator of a fraction is the top number; the denominator refers to the bottom number.

A surd is an irrational number, e.g. √3, √5, etc.]

Given the following fraction:

(a+√b)/(c-√b), where a,b and c are non-negative integers and b is not a square number.

We can see that the denominator (c-√b) is irrational. To rationalise the denominator we take the following steps:

1. Multiply BOTH the numerator and the denominator of our fraction by (c+√b) in order to eliminate the irrational surd in the denominator.

__ Note:__ we perform this multiplication to both the numerator and denominator in order to preserve the value of the original fraction .

2. We now have for our numerator: (a+√b)(c+√b), and for our denominator: (c-√b)(c+√b).

Expand these brackets, thus we obtain the following fraction:

(ac+(a+c)√b+b) / (c^{2 }- b)

*Clearly we have succeeded in rationalising our denominator (whilst still maintaining the value of our original fraction) since (c ^{2}-b) is clearly a rational number, as required.*

**Example:**

**Write (5+7****√3)/(5-**√**3) in the form a+b**√**3, where a and b are rational.**

*Soln: We carry out the steps stated above;*

*Multiply numerator and denominator by (5+*√3), in doing so eliminating the irrational surd from our denominator. *We thus obtain:*

{(5+7√3)(5+√3)} / {(5-√3)(5+√3)} **(expand brackets)**

=(46+12√3) / (25+~~5√3~~-~~5√3~~ - 3)

=(46+12√3) / (22)

=23/11+(6/11)√3.

Clearly, from our orginal hypothesis, a=23/11, b=6/11 are both rational numbers, thus we are done.