Edexcel C3 question Watch

IgorYakov
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In the specification it says:

Preamble
Methods of proof, including proof by contradiction and disproof by
counter-example, are required. At least one question on the paper
will require the use of proof.


I've never seen a proof by contradiction/disproof by counter-example unless I have but didn't realise, can anyone give an example of when it's come up on a paper?
When it says at least one question will require proof

Is that e.g.
http://www.examsolutions.net/a-level...uary/paper.php question 2(a) or 2(b)

http://www.examsolutions.net/a-level...uary/paper.php question 2(a)?

If not which question is it?

Thank you
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Smaug123
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(Original post by IgorYakov)
In the specification it says:

Preamble
Methods of proof, including proof by contradiction and disproof by
counter-example, are required. At least one question on the paper
will require the use of proof.


I've never seen a proof by contradiction/disproof by counter-example unless I have but didn't realise, can anyone give an example of when it's come up on a paper?
When it says at least one question will require proof

Is that e.g.
http://www.examsolutions.net/a-level...uary/paper.php question 2(a) or 2(b)

http://www.examsolutions.net/a-level...uary/paper.php question 2(a)?

If not which question is it?

Thank you
Neither of those papers contains a proof-by-contradiction or a counterexample question, as far as I can tell.
The canonical example of a proof by contradiction is Euclid's proof that there are infinitely many primes, which is in the spoiler (but if you want to practise contradiction, you might like to leave this as an exercise, with the hint that "you can create a number which is not divided by any numbers in a given list, by multiplying the list together and adding 1").

Spoiler:
Show
Suppose there were not; suppose that there are only finitely many primes, p_1, \dots, p_k. Then the number p_1 \times \dots \times p_k + 1 is certainly not divisible by any of those primes, since it is one more than a multiple of any of the primes. But the number itself isn't prime (because it doesn't appear as a p_i for any i; and it doesn't have any prime factors (because no p_i divides it). This is a contradiction, because all positive integers are either prime or composite, and this number is neither of those. Hence there are, in fact, infinitely many primes.


I don't think I've ever seen it on a paper, that I remember (I took A-levels two years ago, so my memory may be faulty). It's quite hard to practise, too.
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brianeverit
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(Original post by IgorYakov)
In the specification it says:

Preamble
Methods of proof, including proof by contradiction and disproof by
counter-example, are required. At least one question on the paper
will require the use of proof.


I've never seen a proof by contradiction/disproof by counter-example unless I have but didn't realise, can anyone give an example of when it's come up on a paper?
When it says at least one question will require proof

Is that e.g.
http://www.examsolutions.net/a-level...uary/paper.php question 2(a) or 2(b)

http://www.examsolutions.net/a-level...uary/paper.php question 2(a)?

If not which question is it?

Thank you
A simple proof by contradiction is the proof that  \sqrt 2 is not a rational number.
Proof. Assume that it is i.e.  \sqrt 2=\frac{a}{b} where a and b are integers without a common factor.
Then squaring both sides  2=\frac{a^2}{b^2} \Rightarrow a^2=2b^2 so a^2  is an even number \Rightarrow a is an even number.
So put  a=2k then we have 4k^2=2b^2 \Rightarrow b^2=2k^2 \Rightarrow b^2 \mathrm{\ and\ hence\ }b are even numbers.
So a and b are both even numbers which contradicts our assumption that they had no common factor. So our initial assumption is incorrect and  \sqrt 2 cannot be expressed as a fraction.
A simple disproof by counterexample is the following.
Prove that the statement x^2>4 \Rightarrow x>2 is false.
Proof. x=-3 \Rightarrow  x^2=9>4 \mathrm{\ but\ }-3 is not >2
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IgorYakov
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(Original post by Smaug123)
Neither of those papers contains a proof-by-contradiction or a counterexample question, as far as I can tell.
The canonical example of a proof by contradiction is Euclid's proof that there are infinitely many primes, which is in the spoiler (but if you want to practise contradiction, you might like to leave this as an exercise, with the hint that "you can create a number which is not divided by any numbers in a given list, by multiplying the list together and adding 1").

Spoiler:
Show
Suppose there were not; suppose that there are only finitely many primes, p_1, \dots, p_k. Then the number p_1 \times \dots \times p_k + 1 is certainly not divisible by any of those primes, since it is one more than a multiple of any of the primes. But the number itself isn't prime (because it doesn't appear as a p_i for any i; and it doesn't have any prime factors (because no p_i divides it). This is a contradiction, because all positive integers are either prime or composite, and this number is neither of those. Hence there are, in fact, infinitely many primes.


I don't think I've ever seen it on a paper, that I remember (I took A-levels two years ago, so my memory may be faulty). It's quite hard to practise, too.

(Original post by brianeverit)
A simple proof by contradiction is the proof that  \sqrt 2 is not a rational number.
Proof. Assume that it is i.e.  \sqrt 2=\frac{a}{b} where a and b are integers without a common factor.
Then squaring both sides  2=\frac{a^2}{b^2} \Rightarrow a^2=2b^2 so a^2  is an even number \Rightarrow a is an even number.
So put  a=2k then we have 4k^2=2b^2 \Rightarrow b^2=2k^2 \Rightarrow b^2 \mathrm{\ and\ hence\ }b are even numbers.
So a and b are both even numbers which contradicts our assumption that they had no common factor. So our initial assumption is incorrect and  \sqrt 2 cannot be expressed as a fraction.
A simple disproof by counterexample is the following.
Prove that the statement x^2>4 \Rightarrow x>2 is false.
Proof. x=-3 \Rightarrow  x^2=9>4 \mathrm{\ but\ }-3 is not >2


The first one is something I've never seen before but it makes sense, I'm not sure how to practice these and I don't know why it says at least one question will require proofs, I've not seen any at all so I don't know what type of things they expect :/
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forsparta
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(Original post by IgorYakov)
The first one is something I've never seen before but it makes sense, I'm not sure how to practice these and I don't know why it says at least one question will require proofs, I've not seen any at all so I don't know what type of things they expect :/
Out of all the C3 paper I've ever done I've never seen proof by contradiction, counterexample could be likely but as shown that is straightforward.

Proof isn't even in the C3 textbook so I wouldn't worry about it tbh,
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IgorYakov
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(Original post by forsparta)
Out of all the C3 paper I've ever done I've never seen proof by contradiction, counterexample could be likely but as shown that is straightforward.

Proof isn't even in the C3 textbook so I wouldn't worry about it tbh,

Yeah true but it's in the spec and they might throw one in there this year to give some people a shock
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brianeverit
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[QUOTE=IgorYakov;47810512]The first one is something I've never seen before but it makes sense, I'm not sure how to practice these and I don't know why it says at least one question will require proofs, I've not seen any at all so I don't know what type of things they expect :/[/QUOTE

For proof by contradiction we make use of the fact that if you start from something that is true then it is impossible (by mathematically correct steps) to finish up with something that is false.
So if we are asked to prove that A=B, we start by assuming A does not equal B and show that it leads us to a contradicion.
The simplest example I cxan come up with is
Prove that x^2+1 \geq 2x for all values of x
So we start by assuming that x^2+1<2x \Rightarrow x^2+2x+1<0
But x^2+2x+1=(x+1)^2 \geq 0 contradicting our assumption. So our assumption must be incorreect and hence x^2+1 \geq 2x
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IgorYakov
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(Original post by brianeverit)

For proof by contradiction we make use of the fact that if you start from something that is true then it is impossible (by mathematically correct steps) to finish up with something that is false.
So if we are asked to prove that A=B, we start by assuming A does not equal B and show that it leads us to a contradicion.
The simplest example I cxan come up with is
Prove that x^2+1 \geq 2x for all values of x
So we start by assuming that x^2+1<2x \Rightarrow x^2+2x+1<0
But x^2+2x+1=(x+1)^2 \geq 0 contradicting our assumption. So our assumption must be incorreect and hence x^2+1 \geq 2x
Aah right, thanks a lot
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brianeverit
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(Original post by IgorYakov)
Aah right, thanks a lot
No problem.
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