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proof-or-given-a-counter-example-if-n-N-n-gt-1-exist-a-number-k-N-k-0-n-such-that-n-k-is-prime-




Question Number 503 by 123456 last updated on 20/Jan/15
proof or given a counter−example:  if n∈N,n>1, exist a number k∈N  k∈(0,n] such that n+k is prime.
$${proof}\:{or}\:{given}\:{a}\:{counter}−{example}: \\ $$$${if}\:{n}\in\mathbb{N},{n}>\mathrm{1},\:{exist}\:{a}\:{number}\:{k}\in\mathbb{N} \\ $$$${k}\in\left(\mathrm{0},{n}\right]\:{such}\:{that}\:{n}+{k}\:{is}\:{prime}. \\ $$
Commented by prakash jain last updated on 20/Jan/15
Bertrand′s theorem states that   if p_i  is i^(th)  prime than p_(i+1) <2∙p_i   The statement given in question  directly follows from Bertrand theorem  for k=p_(i+1) −p_i , where p_i =n.
$$\mathrm{Bertrand}'\mathrm{s}\:\mathrm{theorem}\:\mathrm{states}\:\mathrm{that}\: \\ $$$$\mathrm{if}\:{p}_{{i}} \:\mathrm{is}\:{i}^{{th}} \:\mathrm{prime}\:\mathrm{than}\:{p}_{{i}+\mathrm{1}} <\mathrm{2}\centerdot{p}_{{i}} \\ $$$$\mathrm{The}\:\mathrm{statement}\:\mathrm{given}\:\mathrm{in}\:\mathrm{question} \\ $$$$\mathrm{directly}\:\mathrm{follows}\:\mathrm{from}\:\mathrm{Bertrand}\:\mathrm{theorem} \\ $$$$\mathrm{for}\:{k}={p}_{{i}+\mathrm{1}} −{p}_{{i}} ,\:\mathrm{where}\:{p}_{{i}} ={n}. \\ $$

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