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Degree: Mathematics (Masters)  Queen's, Belfast University
CRB/DBS Standard 
No 

CRB/DBS Enhanced 
No 
It's worth rememberingÂ 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:
Â (First) +Â (Outside) + (Inside) + (Last)
Â Â ac Â Â + Â Â Â ad Â Â Â Â + Â Â bc Â Â Â + Â Â bd
Examples:
Expand the following brackets:
1. (2x+3)(4x+6) =Â (F)Â 8x^{2Â }+Â (O)Â 12x +Â (I)Â 12x +Â (L)Â 18
Â Â Â Â Â Â Â Â Â Â Â Â Â = 8x^{2Â }+ 24x + 18
2. (6x1)(2x^{2}+4) = (6x)(2x^{2}) + 4(6x) + (1)(2x^{2}) + (1)(4)
Â Â Â Â Â Â Â Â Â Â Â Â Â = 12x^{3Â }+ 24x  2x^{2Â } 4
3. (x1)(2x+4)(3x6) Â (Note: For this example, first expand the first two brackets, then expand the result with the third bracket, as follows):
= (2x^{2}+2x4)(3x6) Â Â (To expand this, multiply the first term in the 1st bracket by the 1st term and 2nd termÂ in the 2nd, the 2nd term in the 1stÂ 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^{3Â } 12x^{2Â }+ 6x^{2Â } 12x  12x + 24
= 6x^{3Â } 6x^{2Â } 24x + 24
Â Â Â Â Â Â Â Â Â Â Â Â Â Â
^{Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â }
see more[NOTE:Â AnÂ irreducible fractionÂ (orÂ fractionÂ in lowest terms)Â is aÂ fractionÂ 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 thatÂ âˆšp is irrational. Let's prove this:
Proof:Â We shall use Proof by Contradiction;
Let's suppose towards aÂ contradictionÂ thatÂ âˆšp is in fact rational.
This implies that there exists two nonnegative integers, call themÂ m andÂ n, such that:
âˆšp = m / n , where n is not zero and m/n is anÂ irreducible fraction.
<=> p = m^{2}Â / n^{2Â }Â (Squared both sides)
<=> pn^{2Â }= m^{2}....(#)Â Â (Multiplied through equation by n^{2})
From eqn(#), it follows that, since p is prime, p  m^{2Â }(i.e. p divides m^{2})Â which thus implies also that p  m.Â
This means that there exists some natural number, call it k, such that:
pk = m....(##).
Now, sub eqn(##)Â into eqn(#), we thus obtain:
pn^{2}Â = p^{2}k^{2}
<=> n^{2Â }= pk^{2 Â Â }(Divided through by p)
This last implies that p  n^{2}, which further implies:
p  n.
BUTÂ 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 thatÂ âˆšp is in fact irrational, as required.Â
see more[Recall:Â the numerator of a fraction is the top number; the denominator refers to the bottom number.Â
A surd is an irrational number, e.g.Â âˆš3,Â âˆš5, etc.]
Given the following fraction:
(a+âˆšb)/(câˆšb), where a,b and c are nonnegative integers and b is not a square number.
We can see that the denominator (câˆšb) is irrational. To rationalise the denominator we take the following steps:
1. Multiply BOTH the numerator and the denominator of our fraction by (c+âˆšb) in order to eliminate the irrational surd in the denominator.Â
Note:Â we perform this multiplication to both the numerator and denominator in order to preserve the value of the original fractionÂ .
2. We now have for our numerator: (a+âˆšb)(c+âˆšb), and for our denominator: (câˆšb)(c+âˆšb).Â
Expand these brackets, thus we obtain the following fraction:
(ac+(a+c)âˆšb+b) / (c^{2Â } b)
Clearly we have succeeded inÂ rationalising our denominator (whilst still maintaining the value of our original fraction) since (c^{2}b)Â is clearly a rational number, as required.
Example:
Write (5+7âˆš3)/(5âˆš3) in the form a+bâˆš3, where a and b are rational.
Soln: We carry out the steps stated above;
Multiply numerator and denominator by (5+âˆš3), in doing so eliminating the irrational surd from our denominator.Â We thus obtain:
{(5+7âˆš3)(5+âˆš3)} /Â {(5âˆš3)(5+âˆš3)} Â (expand brackets)
=(46+12âˆš3) / (25+5âˆš35âˆš3Â Â 3)
=(46+12âˆš3) / (22)
=23/11+(6/11)âˆš3.
Clearly, from our orginal hypothesis, a=23/11, b=6/11Â are both rational numbers, thus we are done.
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