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    (Original post by Renzhi10122)
    Solution 553

    We construct a sequence that works. Start with  a_1=1 then let  a_2 be some large number such that we may have  a_3=2 , i.e  a_2 \equiv 3 \pmod{6} . Now let  a_4 be some even larger number compared to  a_2 such that we may have  a_5=3 , and such a number exists by CRT. Hence, we keep repeating this, alternating the sequence between large number and the sequence  \{ 1,2,3... \} , missing out a part of the sequence if that number has already been used as one of our large numbers, and hence, we have constructed a sequence that works.
    An interesting solution! I was hoping for something more elementary though (one where people who don't have experience in number theory could potentially solve through brute force).
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    Blazy's problem reminded me of this problem

    Problem 554 *

    Let  a_1,a_2... be a sequence of integers with infinitely many positive and infinitely many negative terms. Suppose that for every positive integer  n , the numbers  a_1,a_2...a_n leave  n distinct remainders upon division by  n .

    Prove that every integer occurs in the sequence exactly once.
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    Problem 555

    If n^q\in\mathbb{N}\;\forall n then q\in\mathbb{N}.
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    (Original post by Lord of the Flies)
    Problem 555

    If n^q\in\mathbb{N}\;\forall n then q\in\mathbb{N}.
    What is the domain of n?

    Are you stating
    ( \mathbb{P}_q \Leftrightarrow \exists q \in \mathbb{R}  \mathrm{ s.t. }  \forall n \in \mathbb{N} , n^q \in \mathbb{N} \implies q \in \mathbb{N} \Leftrightarrow \mathbb{Q}_q ) \Leftrightarrow \mathbb{T}

    Else if we can take any real n, then

    Assume \mathbb{P}_q

    Let n=\sqrt[q]{ \sqrt{2}} \implies n^q = \sqrt{2} \not \in \mathbb{N} \implies \neg \mathbb{Q}_q \implies \neg \mathbb{T}
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    (Original post by Johann von Gauss)
    What is the domain of n?

    Are you stating
    ( \mathbb{P} \Leftrightarrow \exists q \in \mathbb{R}  \mathrm{ s.t. }  \forall n \in \mathbb{N} , n^q \in \mathbb{N} \implies q \in \mathbb{N} \Leftrightarrow \mathbb{Q} ) \Leftrightarrow \mathbb{T}

    Else if we can take any real n, then

    Assume \mathbb{T} holds for some integer q

    Let n=\sqrt[q]{ \sqrt{2}} \implies n^q = \sqrt{2} \not \in \mathbb{N} \implies ( \mathbb{P} \implies \neg \mathbb{Q} )
    n is integer, seeing as it doesnt work in the reals nor the rationals. Also, what on earth does your first line mean?
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    (Original post by Renzhi10122)
    n is integer, seeing as it doesnt work in the reals nor the rationals. Also, what on earth does your first line mean?
    (There exists a real q such that for all natural n, we have n^q is a natural number) implies q is a natural number

    i.e. there exist no non-natural real q for which his statement is true

    Statements within statements is a key feature of my work

    I find it more logically intuitive
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    (Original post by Johann von Gauss)
    ...
    Oh come on obviously \mathbb{N}. And you could've said all that in a single sentence without the silly syntax.
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    (Original post by Johann von Gauss)
    (There exists a real q such that for all natural n, we have n^q is a natural number) implies q is a natural number

    i.e. there exist no non-natural real q for which his statement is true

    Statements within statements is a key feature of my work

    I find it more logically intuitive
    Ah ok, im just not a fan of the big P, Q and T business.
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    (Original post by Lord of the Flies)
    Oh come on obviously \mathbb{N}. And you could've said all that in a single sentence without the silly syntax.
    This isn't a strictly number theory thread, and I don't like making obvious assumptions without stating them - like "clearly I can assume P, Q,..." because then you just reduce problems to triviality. E.g. if we can make the 'obvious' assumption that, for distinct primes, an Nth root of prime P multiplied by the Mth root of prime Q is irrational if the Nth root of prime P is irrational, and the Mth root of prime Q is irrational, then your problem becomes trivial, since we can reduce it to proving that for all prime factors N of n, the qth root of N must be natural for it to be rational, and this is easy to show.

    And the syntax isn't silly, its unambiguous, and quick to read

    (Original post by Renzhi10122)
    Ah ok, im just not a fan of the big P, Q and T business.
    It saves inkBtw what level is your problem? I'm a little afraid to give it a serious attempt, if its some IMO level problem
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    (Original post by Johann von Gauss)
    It saves ink

    Btw what level is your problem? I'm a little afraid to give it a serious attempt, if its some IMO level problem
    Easy IMO, give it a try. Doesn't require any special knowledge.

    Posted from TSR Mobile
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    (Original post by Johann von Gauss)
    What is the domain of n?

    Are you stating
    ( \mathbb{P}_q \Leftrightarrow \exists q \in \mathbb{R}  \mathrm{ s.t. }  \forall n \in \mathbb{N} , n^q \in \mathbb{N} \implies q \in \mathbb{N} \Leftrightarrow \mathbb{Q}_q ) \Leftrightarrow \mathbb{T}

    Else if we can take any real n, then

    Assume \mathbb{P}_q

    Let n=\sqrt[q]{ \sqrt{2}} \implies n^q = \sqrt{2} \not \in \mathbb{N} \implies \neg \mathbb{Q}_q \implies \neg \mathbb{T}
    I don't think you can have a logical statement which is a function of q with q in the quantifier.
    You can drop the s.t. also, to make it more compact and avoid those big bold letters for propositions as they're used for sets.
    I like it though.
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    (Original post by Johann von Gauss)
    ...
    I realise the name of the thread is "the proof is trivial" but with a little common sense you'll see that the idea is actually not to post problems whose proof is trivial.

    In the case of 555 it is crystal clear what the domain should be, since otherwise the question wouldn't be worth asking, as you've needlessly pointed out.

    And terribly sorry, but unless you're writing a paper on predicate logic it most definitely is silly, unnecessary, and unimpressive. Open any article by any mathematician ever and you can be sure that any statements made won't be written like that - for good reason.

    Not going to elaborate if you bother responding because I have better things to do than convince a sixth-former who probably discovered quantifiers a week ago that their syntax would make any undergrad chuckle.
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    ^

    Fair enough, but would someone else be kind enough to explain why this kind of syntax isn't used outside logic?

    I had thought that University level maths and above were very rigorous in their logical connectives

    Doesn't using words (which are ambiguous) go against this, especially when not all mathematicians speak English well?


    From a quick search, I have come across https://en.wikibooks.org/wiki/Mathem...cal_Connectors which seems to encourage this sort of syntax:

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    (Original post by Johann von Gauss)
    ^

    Fair enough, but would someone else be kind enough to explain why this kind of syntax isn't used outside logic?

    I had thought that University level maths and above were very rigorous in their logical connectives

    Doesn't using words (which are ambiguous) go against this, especially when not all mathematicians speak English well?


    From a quick search, I have come across https://en.wikibooks.org/wiki/Mathem...cal_Connectors which seems to encourage this sort of syntax:

    Words aren't ambigous all the time. And even if they are, it's written in a way that is easily dis-ambiguated correctly by the person reading them.

    Mathematics is inherently an art and constricting yourself to expressing it through a particular sybtax destroys that inherent beauty, imo.

    Sentences and English do a much better job in coveying mathematical ideas clearly, plus a nice diagram or two if required.

    Do excuse my spelling, I'm typing on my phone and that's never good.
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    (Original post by Johann von Gauss)
    ^

    Fair enough, but would someone else be kind enough to explain why this kind of syntax isn't used outside logic?
    (a) Because it very rapidly becomes unreadable. In reading a piece of mathematics, one is interested in what the ideas are; if the ideas are obscured behind a dense mat of notation, the motivation to read on becomes ever more diluted!

    (b) It is just as easy to make mistakes in using complex logical symbolism as it is in expressing the idea in English. In fact, easier to make such mistakes.

    I had thought that University level maths and above were very rigorous in their logical connectives
    We are; but one must not mistake rigour for its means of expression.

    Doesn't using words (which are ambiguous) go against this, especially when not all mathematicians speak English well?
    Words are not ambiguous if they are used well. After all, the symbols that one uses in mathematics tend to get translated into natural language: "for all epsilon greater than zero, there is a delta greater than zero...". The use of quantifier notation to express this is certainly compact; the question is whether it is clearer or more obscure to get right and to understand.
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    (Original post by Gregorius)
    ...
    Thank you for your helpful response

    Personally, I usually find translating problems into logical syntax lets me see what is required easier - but this is clearly not a view shared by most people - so I shall cease and desist EDIT: in using it for anything other than organising my own thoughts

    (Original post by Zacken)
    Sentences and English do a much better job in coveying mathematical ideas clearly, plus a nice diagram or two if required.

    I guess this depends on how you visualise the problems, and your's seems to be the majority view. But I have to say I find sentences ugly.
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    (Original post by Johann von Gauss)
    Thank you for your helpful response

    Personally, I usually find translating problems into logical syntax lets me see what is required easier - but this is clearly not a view shared by most people - so I shall cease and desist.
    Ah now, I wouldn't want you to stop doing something if you found it helpful for your own private thoughts! All I would ask is that when you come to write for other people, express that clarity you've gained in a way that is friendly to the reader.
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    Problem 556*

    Solve the equation

     \displaystyle \left(\sqrt{2-\sqrt{3}}\ \right)^{x} + \left(\sqrt{2+\sqrt{3}}\ \right)^{x} = 2^{x}
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    (Original post by ThatPerson)
    Problem 556*

    Solve the equation

     \displaystyle \left(\sqrt{2-\sqrt{3}}\ \right)^{x} + \left(\sqrt{2+\sqrt{3}}\ \right)^{x} = 2^{x}
    x=2 as squaring the LHS gives + sqrt(3) - sqrt(3) [=0], and 2+2 which equals 4 which = RHS. Simples.
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    (Original post by qwertzuiop)
    x=2 as squaring the LHS gives + sqrt(3) - sqrt(3) [=0], and 2+2 which equals 4 which = RHS. Simples.
    Justify that that is the only solution, or otherwise, pls.
 
 
 
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