Saleem A. A Level Physics tutor, A Level Maths tutor, GCSE Maths tuto...

Saleem A.

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Mathematics (Masters) - Oxford, St Catherine's College University

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About me

I've always excelled at Maths from a very young age and due to hard work and commitment, I'm now studying at the University of Oxford. My siblings weren't similarly gifted in their mathematical abilites so I've tutored them in the past. I've also tutored my neighbours on occasion when they felt as though they were struggling. I believe I'd make a great tutor because I'm charismatic and have a thorough understanding of Maths. I find enjoyment in passing on my knowledge and seeing my students succeed which drives me to help my students fulfil their potential. I place emphasis on understanding material rather than learning it, you don't need a tutor to help you memorize content, a tutor's job is to improve your understanding of it and that's what I aim to do.

I've always excelled at Maths from a very young age and due to hard work and commitment, I'm now studying at the University of Oxford. My siblings weren't similarly gifted in their mathematical abilites so I've tutored them in the past. I've also tutored my neighbours on occasion when they felt as though they were struggling. I believe I'd make a great tutor because I'm charismatic and have a thorough understanding of Maths. I find enjoyment in passing on my knowledge and seeing my students succeed which drives me to help my students fulfil their potential. I place emphasis on understanding material rather than learning it, you don't need a tutor to help you memorize content, a tutor's job is to improve your understanding of it and that's what I aim to do.

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Qualifications

SubjectQualificationGrade
MathsA-level (A2)A*
Further MathsA-level (A2)A*
PhysicsA-level (A2)A
ComputingA-level (A2)B

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
PhysicsA Level£20 /hr
MathsGCSE£18 /hr
-Oxbridge Preparation-Mentoring£22 /hr
-Personal Statements-Mentoring£22 /hr

Questions Saleem has answered

How do you prove that (3^n)-1 is always a multiple of 2?

Proof by Induction: The method of Proof by Induction is a simple but very powerful technique. It involves 3 steps:

1) Showing a claim is true for a basic integer value of n (e.g. 0 or 1)

2) Assuming the claim is true for n=k where k is an arbitrary integer

3) Using this assumption to show that the claim is true for n=k+1

The reason why these 3 steps prove the claim is because you've shown 2 things: You've shown that the claim's true for 0 (or 1). You've also shown that is the claim is true for n=k, then it's also true for n=k+1. And so if the claim is true for n=0 (or 1), then the claim is true for n=1(or 2). Then if the claim is true for n=1 (or 2) then it's true for n=2 (or 3) and so on. So the claim is true for all integers n greater than or equal to 0.

 

1) In this particular case, we'll start with n=0:

(3^0)-1 = 1-1 = 0 = 2 x 0 and so the claim holds.

2) Now let's assume the claim is true for n=k. That is, (3^k)-1 is a multiple of 2. So (3^k)-1 = 2c for some integer c.

3) Now let's look at n=k+1:

(3^(k+1))-1 = 3 x (3^k) - 1

Using our assumption, (3^k)-1 = 2c, so that (3^k)=2c+1

Now we have 3 x (2c + 1) - 1 = 6c + 3 -1 = 6c +2 = 2(3c + 1)

(3^(k+1))-1 = 2(3c + 1) and since (3c + 1) is an integer, we have proven the claim.

Proof by Induction: The method of Proof by Induction is a simple but very powerful technique. It involves 3 steps:

1) Showing a claim is true for a basic integer value of n (e.g. 0 or 1)

2) Assuming the claim is true for n=k where k is an arbitrary integer

3) Using this assumption to show that the claim is true for n=k+1

The reason why these 3 steps prove the claim is because you've shown 2 things: You've shown that the claim's true for 0 (or 1). You've also shown that is the claim is true for n=k, then it's also true for n=k+1. And so if the claim is true for n=0 (or 1), then the claim is true for n=1(or 2). Then if the claim is true for n=1 (or 2) then it's true for n=2 (or 3) and so on. So the claim is true for all integers n greater than or equal to 0.

 

1) In this particular case, we'll start with n=0:

(3^0)-1 = 1-1 = 0 = 2 x 0 and so the claim holds.

2) Now let's assume the claim is true for n=k. That is, (3^k)-1 is a multiple of 2. So (3^k)-1 = 2c for some integer c.

3) Now let's look at n=k+1:

(3^(k+1))-1 = 3 x (3^k) - 1

Using our assumption, (3^k)-1 = 2c, so that (3^k)=2c+1

Now we have 3 x (2c + 1) - 1 = 6c + 3 -1 = 6c +2 = 2(3c + 1)

(3^(k+1))-1 = 2(3c + 1) and since (3c + 1) is an integer, we have proven the claim.

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3 years ago

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