# Harder A level proof questions? Watch

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Does anyone have resources? I really enjoyed these type of questions and have done all the ones in my textbook and the ones my teacher provided. By harder I mean questions which requires thinking and isn't exactly rote, like the "prove that the sum of two consecutive numbers is divisible by 4. There was a question where by considering √2^√2^√2 to prove that an irrational number when raised to an irrational power can be rational. Something similar to that.

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Does anyone have resources? I really enjoyed these type of questions and have done all the ones in my textbook and the ones my teacher provided. By harder I mean questions which requires thinking and isn't exactly rote, like the "prove that the sum of two consecutive numbers is divisible by 4. There was a question where by considering √2^√2^√2 to prove that an irrational number when raised to an irrational power can be rational. Something similar to that.

**noname900**)Does anyone have resources? I really enjoyed these type of questions and have done all the ones in my textbook and the ones my teacher provided. By harder I mean questions which requires thinking and isn't exactly rote, like the "prove that the sum of two consecutive numbers is divisible by 4. There was a question where by considering √2^√2^√2 to prove that an irrational number when raised to an irrational power can be rational. Something similar to that.

simply means divides , or in other words, is a factor of .

Last edited by RDKGames; 1 month ago

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Don't think the notation a|b is in a-level

**_gcx**)Don't think the notation a|b is in a-level

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#5

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Don't think the notation a|b is in a-level

**_gcx**)Don't think the notation a|b is in a-level

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(Original post by

You're right, I forgot to mention that in my post :s

**RDKGames**)You're right, I forgot to mention that in my post :s

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A question I wrote a while back.

simply means divides , or in other words, is a factor of .

**RDKGames**)A question I wrote a while back.

simply means divides , or in other words, is a factor of .

Let there be a finite number primes p1,p2,p3,p4...pn, where pn is the biggest prime number. Let P=p1p2p3p4...pn+1. P has to be a prime as the smallest prime factor of p1p2p3p4...pn is 2, therefore when 1 is added the p1p2p3p4...pn, P has a remainder of one when divided by its factors(which are only prime numbers) and therefore P must be a new prime number larger than pn. As there is a contradiction, there must be infinitely many primes.

B)This was really simple to the point where I think I may have misunderstood the question, but an example of this is a=4 b=2

C) b^2 has factors (b)(b), b is either prime or has smaller prime factors. If b is a prime then a must equal b and therefore b/a=1. If b isn't prime then b will have prime factors p1...pn. As we have b^2, the prime factors can be written as (p1)^2...(pn)^2, as a is a prime factor, which divides into b^2, then a will divide into b as b will have the same prime factors without an exponent. I feel like I explained this badly in how its not very mathematical/academic, but I think I kind of got my point across.

D)I have no idea. I have no idea how to carry it on from c. The best I have is that a prime number has factors 1 and itself. A rational number can be written as a fraction a/b where a,b are Integers and the fraction is in its simplest form. Since the surd cannot be simplified into a fraction with integers(only has factors 1 and itself so the surd is the most simplified form), it is irrational.

I guess this is just a start, but I would appreciate some feedback.

Last edited by noname900; 1 month ago

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A) I couldn't do this on my own, I had to search for the solution.

Let there be a finite number primes p1,p2,p3,p4...pn, where pn is the biggest prime number. Let P=p1p2p3p4...pn+1. P has to be a prime as the smallest prime factor of p1p2p3p4...pn is 2, therefore when 1 is added the p1p2p3p4...pn, P has a remainder of one when divided by its factors(which are only prime numbers) and therefore P must be a new prime number larger than pn. As there is a contradiction, there must be infinitely many primes.

**noname900**)A) I couldn't do this on my own, I had to search for the solution.

Let there be a finite number primes p1,p2,p3,p4...pn, where pn is the biggest prime number. Let P=p1p2p3p4...pn+1. P has to be a prime as the smallest prime factor of p1p2p3p4...pn is 2, therefore when 1 is added the p1p2p3p4...pn, P has a remainder of one when divided by its factors(which are only prime numbers) and therefore P must be a new prime number larger than pn. As there is a contradiction, there must be infinitely many primes.

If our list of primes is {2,3,5,7,11,13} then P = 2*3*5*7*11*13 + 1 = 30,031 which can be decomposed as 59*509. Clearly, P is not prime.

What the statement instead tells us is that since P leaves us with a remainder of 1 upon division by any prime we know, its EITHER a prime itself, or it has a prime factor that is not already in our list (59 and even 509 are both new primes from the example above).

Either way, our list of finitely many primes is incomplete hence the contradiction.

B)This was really simple to the point where I think I may have misunderstood the question, but an example of this is a=4 b=2

C) b^2 has factors (b)(b), b is either prime or has smaller prime factors. If b is a prime then a must equal b and therefore b/a=1. If b isn't prime then b will have prime factors p1...pn. As we have b^2, the prime factors can be written as (p1)^2...(pn)^2, as a is a prime factor, which divides into b^2, then a will divide into b as b will have the same prime factors without an exponent. I feel like I explained this badly in how its not very mathematical/academic, but I think I kind of got my point across.

D)I have no idea. I have no idea how to carry it on from c. The best I have is that a prime number has factors 1 and itself. A rational number can be written as a fraction a/b where a,b are Integers and the fraction is in its simplest form. Since the surd cannot be simplified into a fraction with integers(only has factors 1 and itself so the surd is the most simplified form), it is irrational.

Suppose is rational. Hence where and . The fraction is of course irreducible, as usual.

So . Notice that . Can you proceed from here?

Last edited by RDKGames; 1 month ago

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#9

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A) I couldn't do this on my own, I had to search for the solution.

Let there be a finite number primes p1,p2,p3,p4...pn, where pn is the biggest prime number. Let P=p1p2p3p4...pn+1. P has to be a prime as the smallest prime factor of p1p2p3p4...pn is 2, therefore when 1 is added the p1p2p3p4...pn, P has a remainder of one when divided by its factors(which are only prime numbers) and therefore P must be a new prime number larger than pn. As there is a contradiction, there must be infinitely many primes.

B)This was really simple to the point where I think I may have misunderstood the question, but an example of this is a=4 b=2

C) b^2 has factors (b)(b), b is either prime or has smaller prime factors. If b is a prime then a must equal b and therefore b/a=1. If b isn't prime then b will have prime factors p1...pn. As we have b^2, the prime factors can be written as (p1)^2...(pn)^2, as a is a prime factor, which divides into b^2, then a will divide into b as b will have the same prime factors without an exponent. I feel like I explained this badly in how its not very mathematical/academic, but I think I kind of got my point across.

D)I have no idea. I have no idea how to carry it on from c. The best I have is that a prime number has factors 1 and itself. A rational number can be written as a fraction a/b where a,b are Integers and the fraction is in its simplest form. Since the surd cannot be simplified into a fraction with integers(only has factors 1 and itself so the surd is the most simplified form), it is irrational.

I guess this is just a start, but I would appreciate some feedback.

**noname900**)A) I couldn't do this on my own, I had to search for the solution.

Let there be a finite number primes p1,p2,p3,p4...pn, where pn is the biggest prime number. Let P=p1p2p3p4...pn+1. P has to be a prime as the smallest prime factor of p1p2p3p4...pn is 2, therefore when 1 is added the p1p2p3p4...pn, P has a remainder of one when divided by its factors(which are only prime numbers) and therefore P must be a new prime number larger than pn. As there is a contradiction, there must be infinitely many primes.

B)This was really simple to the point where I think I may have misunderstood the question, but an example of this is a=4 b=2

C) b^2 has factors (b)(b), b is either prime or has smaller prime factors. If b is a prime then a must equal b and therefore b/a=1. If b isn't prime then b will have prime factors p1...pn. As we have b^2, the prime factors can be written as (p1)^2...(pn)^2, as a is a prime factor, which divides into b^2, then a will divide into b as b will have the same prime factors without an exponent. I feel like I explained this badly in how its not very mathematical/academic, but I think I kind of got my point across.

D)I have no idea. I have no idea how to carry it on from c. The best I have is that a prime number has factors 1 and itself. A rational number can be written as a fraction a/b where a,b are Integers and the fraction is in its simplest form. Since the surd cannot be simplified into a fraction with integers(only has factors 1 and itself so the surd is the most simplified form), it is irrational.

I guess this is just a start, but I would appreciate some feedback.

C) Very serious mistake here. What about ? Then b^2

**cannot**be written in the form where are prime for all i. The real prime factorisation of b is where are distinct primes and are natural numbers for all i.

"as a is a prime factor, which divides into b^2, then a will divide into b as b will have the same prime factors without an exponent" - after correcting the mistake, write this as proper maths.

If a is prime, and , and , then for some . Note that , which has a factor of , so .

D) Start with the idea that where a is prime and m, n are coprime natural numbers. Use part C to show that m is divisible by a, and then use the fact that m is divisible by a to show that n is divisible by a, which contradicts the coprime condition. It's very similar to irrationality of root2 proof.

If I come across as harsh I apologise, I think it's important to be as cautious as possible when writing proofs (and if I've written something wrong here I would hope that somebody can point it out to me as well).

Last edited by I hate maths!; 1 month ago

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#10

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Pretty much. As you say, . Since it means can be any product combination of these factors and their squares, however we know is prime, therefore it is EXACTLY equal to one of the . Hence it definitely divides since is a factor.

**RDKGames**)Pretty much. As you say, . Since it means can be any product combination of these factors and their squares, however we know is prime, therefore it is EXACTLY equal to one of the . Hence it definitely divides since is a factor.

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#11

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C) Very serious mistake here. What about ? Then b^2

**I hate maths!**)C) Very serious mistake here. What about ? Then b^2

**cannot**be written in the form where are prime for all i.
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#12

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Pretty much, altough you've fell into the common misconception trap of saying P must be prime. It's not.

If our list of primes is {2,3,5,7,11,13} then P = 2*3*5*7*11*13 + 1 = 30,031 which can be decomposed as 59*509. Clearly, P is not prime.

What the statement instead tells us is that since P leaves us with a remainder of 1 upon division by any prime we know, its EITHER a prime itself, or it has a prime factor that is not already in our list (59 and even 509 are both new primes from the example above).

Either way, our list of finitely many primes is incomplete hence the contradiction.

This is a good counter-example.

Pretty much. As you say, . Since it means can be any product combination of these factors and their squares, however we know is prime, therefore it is EXACTLY equal to one of the . Hence it definitely divides since is a factor.

By contradiction.

Suppose is rational. Hence where and . The fraction is of course irreducible, as usual.

So . Notice that . Can you proceed from here?

**RDKGames**)Pretty much, altough you've fell into the common misconception trap of saying P must be prime. It's not.

If our list of primes is {2,3,5,7,11,13} then P = 2*3*5*7*11*13 + 1 = 30,031 which can be decomposed as 59*509. Clearly, P is not prime.

What the statement instead tells us is that since P leaves us with a remainder of 1 upon division by any prime we know, its EITHER a prime itself, or it has a prime factor that is not already in our list (59 and even 509 are both new primes from the example above).

Either way, our list of finitely many primes is incomplete hence the contradiction.

This is a good counter-example.

Pretty much. As you say, . Since it means can be any product combination of these factors and their squares, however we know is prime, therefore it is EXACTLY equal to one of the . Hence it definitely divides since is a factor.

By contradiction.

Suppose is rational. Hence where and . The fraction is of course irreducible, as usual.

So . Notice that . Can you proceed from here?

For the proof that there were infinitely many primes. I did notice that the new number P isn't always a prime since I had tested out a few list of finite primes I made. I realised that these numbers weren't prime and that they had a bigger prime factors. However all these youtubers said that p1p2p3p4...pn+1 was a prime, so I accepted it. Then I remedied it in my mind by telling myself that if we did have all the primes as the assumption(since all non prime P values have prime factors>pn, so theoretically if there were none left, P would be prime) said then there would have been a new prime when 1 was added since all the primes we knew would no longer divide into the new number P.

Last edited by noname900; 4 weeks ago

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