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Question Number 12487 by geovane10math last updated on 23/Apr/17

Someone has been solved Goldbach′s  Conjecture??

$$\mathrm{Someone}\:\mathrm{has}\:\mathrm{been}\:\mathrm{solved}\:\mathrm{Goldbach}'\mathrm{s} \\ $$$$\mathrm{Conjecture}?? \\ $$

Commented by FilupS last updated on 24/Apr/17

Is this not a proof??     2n=p_1 +p_2      n∈Z>2  p_i ∈P     p_i =2k_i +1        k∈Z≥1     2n=2k_1 +1+2k_2 +1  2n=2(k_1 +k_2 +1)  RHS∈E  ∴true

$$\mathrm{Is}\:\mathrm{this}\:\mathrm{not}\:\mathrm{a}\:\mathrm{proof}?? \\ $$$$\: \\ $$$$\mathrm{2}{n}={p}_{\mathrm{1}} +{p}_{\mathrm{2}} \:\:\:\:\:{n}\in\mathbb{Z}>\mathrm{2} \\ $$$${p}_{{i}} \in\mathbb{P}\:\:\:\:\:{p}_{{i}} =\mathrm{2}{k}_{{i}} +\mathrm{1}\:\:\:\:\:\:\:\:{k}\in\mathbb{Z}\geqslant\mathrm{1} \\ $$$$\: \\ $$$$\mathrm{2}{n}=\mathrm{2}{k}_{\mathrm{1}} +\mathrm{1}+\mathrm{2}{k}_{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{2}{n}=\mathrm{2}\left({k}_{\mathrm{1}} +{k}_{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\mathrm{RHS}\in\mathbb{E} \\ $$$$\therefore\mathrm{true} \\ $$

Commented by prakash jain last updated on 24/Apr/17

Hi Filup, your argument proves  that sum of 2 odd primes is even.  What is required is that every  even number>4 can be written as  sum of 2 primes.

$$\mathrm{Hi}\:\mathrm{Filup},\:\mathrm{your}\:\mathrm{argument}\:\mathrm{proves} \\ $$$$\mathrm{that}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{2}\:\mathrm{odd}\:\mathrm{primes}\:\mathrm{is}\:\mathrm{even}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{required}\:\mathrm{is}\:\mathrm{that}\:\mathrm{every} \\ $$$$\mathrm{even}\:\mathrm{number}>\mathrm{4}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{2}\:\mathrm{primes}. \\ $$

Commented by geovane10math last updated on 24/Apr/17

No, the problem isn′t so easy.

$$\mathrm{No},\:\mathrm{the}\:\mathrm{problem}\:\mathrm{isn}'\mathrm{t}\:\mathrm{so}\:\mathrm{easy}. \\ $$$$ \\ $$

Commented by FilupS last updated on 25/Apr/17

thanks. I now undwrstand the problem

$$\mathrm{thanks}.\:\mathrm{I}\:\mathrm{now}\:\mathrm{undwrstand}\:\mathrm{the}\:\mathrm{problem} \\ $$

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