noname900
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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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RDKGames
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(Original post by 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.
A question I wrote a while back.

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

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_gcx
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(Original post by RDKGames)
A question I wrote a while back.

Don't think the notation a|b is in a-level
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(Original post by _gcx)
Don't think the notation a|b is in a-level
You're right, I forgot to mention that in my post :s
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Sir Cumference2
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(Original post by _gcx)
Don't think the notation a|b is in a-level
Only in number theory in further maths.
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(Original post by RDKGames)
You're right, I forgot to mention that in my post :s
Don't worry, I understand the notation. I'm just trying my best to do the infinite prime number proof, I have seen it before but I've forgotten it.
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(Original post by RDKGames)
A question I wrote a while back.

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

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.
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RDKGames
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(Original post by 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.
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.

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
This is a good counter-example.

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.
Pretty much. As you say, b^2 = p_1^2 p_2^2 \ldots p_n^2. Since a|b^2 it means a can be any product combination of these p_i factors and their squares, however we know a is prime, therefore it is EXACTLY equal to one of the p_i. Hence it definitely divides b since p_i is a factor.

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.
By contradiction.

Suppose \sqrt{p} is rational. Hence \sqrt{p} = \dfrac{a}{b} where a,b \in \mathbb{Z} and b \neq 0. The fraction is of course irreducible, as usual.

So pb^2 = a^2. Notice that p|a^2. Can you proceed from here?
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(Original post by 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.
A) The smallest prime being 2 is irrelevant to the proof, so is the definition that "pn is the biggest prime number" (how is this useful in your proof?). The claim "P has a remainder of one when divided by its factors" doesn't make much sense (in general the wording here is really confusing), and very importantly P is not necessarily a prime itself: take 2 \times 3 \times 5 \times 7 \times 11 \times 13+1=30031=59 \times 509, which is not prime. There are two cases: P is prime, and P is not prime, but considering both cases will lead to the contradiction.

C) Very serious mistake here. What about b=2^2 \times 3 \times 5 ? Then b^2 cannot be written in the form b^2=p_1^2 \times p_2^2 \times ... \times p_n^2 where p_i are prime for all i. The real prime factorisation of b is p_1^{\alpha_{1}} \times p_2^{\alpha_{2}} \times ... \times p_n^{\alpha_{n}} where p_i are distinct primes and \alpha_{i} 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 a|b^2, and b^2= p_1^{2\alpha_{1}} \times p_2^{2\alpha_{2}} \times ... \times p_n^{2\alpha_{n}}, then a=p_i for some 1 \leqslant i \leqslant n. Note that b=p_1^{\alpha_{1}} \times p_2^{\alpha_{2}} \times ... \times p_n^{\alpha_{n}}, which has a factor of p_i, so a | b.

D) Start with the idea that \sqrt{a}=m/n 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).
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(Original post by RDKGames)
Pretty much. As you say, b^2 = p_1^2 p_2^2 \ldots p_n^2. Since a|b^2 it means a can be any product combination of these p_i factors and their squares, however we know a is prime, therefore it is EXACTLY equal to one of the p_i. Hence it definitely divides b since p_i is a factor.
Ah, you beat me to the reply, but I don't think this part is entirely correct. That isn't the general factorisation of b^2, I think.
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(Original post by I hate maths!)
C) Very serious mistake here. What about b=2^2 \times 3 \times 5 ? Then b^2 cannot be written in the form b^2=p_1^2 \times p_2^2 \times ... \times p_n^2 where p_i are prime for all i.
I would just write that as b = 2 \times 2 \times 3 \times 5 then it's OK to say b^2 = p_1^2 p_2^2 \ldots p_n^2.
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(Original post by RDKGames)
I would just write that as b = 2 \times 2 \times 3 \times 5 then it's OK to say b^2 = p_1^2 p_2^2 \ldots p_n^2.
Ah, that is true! Though I was lead to believe that p_i were distinct from the phrase "If b isn't prime then b will have prime factors p1...pn" but I suppose that list of primes could contain repeats (although that would be redundant). Would be strange, for example, to say that 60 has prime factors 2, 2, 3, 5, but I guess not technically incorrect.
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noname900

Here's also more contradiction questions that may require more thinking. Feel free to ask for help if you're stuck.



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(Original post by 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, b^2 = p_1^2 p_2^2 \ldots p_n^2. Since a|b^2 it means a can be any product combination of these p_i factors and their squares, however we know a is prime, therefore it is EXACTLY equal to one of the p_i. Hence it definitely divides b since p_i is a factor.



By contradiction.

Suppose \sqrt{p} is rational. Hence \sqrt{p} = \dfrac{a}{b} where a,b \in \mathbb{Z} and b \neq 0. The fraction is of course irreducible, as usual.

So pb^2 = a^2. Notice that p|a^2. Can you proceed from here?
As a^2 has a factor of p, a will also have a factor of p as a^2 will have prime factors (P1)^2...(Pn)^2 and a will have prime factors P1...Pn. a can be written as (p*m), where m is some factor of a. pb^2=(pm)^2 => pb^2=p^2m^2 => b^2=pm^2. There is a contradiction as we assumed a/b is a fraction in its simplest form, however there is a common factor of p in (a/b). I had thought to use that contradiction proof just as I would have done with sqrt2, but the fact that I didn't have a number put me off, however I saw where part c comes into play.

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.
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