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There is, interestingly, a counter argument to this strategy of de-platforming others online. This has to do with movements that use this method to attack the powerful anonymously, like with the #Metoo movement on Twitter. So, there is room for debate on this issue. I think it should be discussed more in a moderate tone in the public. There should be frank discussions like this about how to utilize information in ways that can benefit or harm society because as it stands now the form is simply a brainwashing technique, or like you said: propaganda
⢠Other Questions
What are your thoughts on Donald Trump's ABC News town hall in Philadelphia?
HE WON! HE WON!TRUMP WINS NOBEL PRIZE FOR MEDICINE!Or, rather, he will have won once it's announced.Recall that during the ABC Town Hall, the sly little George Stephanopoulos tried his best to ambush our President by tricking him into saying "Herd Mentality".Trump had George right where he wanted him.
George thought that Trump was going for "Herd Immunity". The whole world thought the same thing, especially the libs, because their pundits had been talking about the newly acquired scientist on Trump's team, Dr. Scott Atlas, newly arrived from Fox News. Because this Atlas fellow, who carries the globe's Covid concerns on his shoulders, promotes the idea of Herd Immunity.But that's not it. That's not where Trump was going at all.Every news pundit on CNN and MSNBC were now making fun of Trump. Again. Everyone thought that Trump misspoke and confused the word "mentality" with the word "immunity". Wrong!As always, Trump knew exactly what he was saying.Trump had been going for herd mentality in his administration from the get-go. And now he is closer than ever. Trump had ditched or fired or coerced the bad apples out of the administration; cleared 'em all out; took three years.Now, everybody had a mind-set just like his, thought the same way, viewed the world similarly. Shared the same mentality as Trump.He had achieved his herd mentality alright, and announced it to the world right there in front of ABC's most prized possession, Stephanopoulos. Trump knocked that little bird right off his perch. But Trump got more than just the one bird with that genius stone of his; he got all of 'em.For two days now, we hear their bleating, their crowing about how Trump can't even get his words right. But Trump always gets his words right.
And today, during her Presser, little Kayleigh will announce that the Nobel Committee will have bestowed upon her boss the prestigious Prize for Medicine. Specializing in the field of Herd Mentality, specifically the study of Herd Mendacity, Trump wins his Nobel.Trump is the first President to achieve Herd Mendacity throughout his Administration. A great achievement. Only Trump could have done. HERD MENDACITY!Every other administration may have lied a little bit, about small things. And maybe big things, like during Nixon or Johnson or W. And maybe even the POTUS himself told a couple whoppers. But this! This achievement makes them pale in comparison. A full 80% or so of the Trump team lies almost daily, and always when you push a microphone in their face. The whole team is under qualified, so even as they suspect that they are telling the truth, they don't have enough knowledge in their fields or departments to even understand the questions asked. They lie even unwittingly.
But Trump! He is their leader. He lies all the time. He leads by example. Follow me! Watch what I say, not what I do. (Rachel has it backward.)So today, finally, Trump proclaims on TV that he has achieved in his administration "Herd Mentality". More specifically in the sub-genre of Herd Mendacity. Wherein everyone lies. Just watch 'em. Pay attention.
And then bend the knee.
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Assume a, b, c are integers. What is the proof that abc(a³-b³)(b³-c³)(c³-a³) is a multiple of 7?
Nice question :). Just to say, simpler way of proving it in this case, from Himanshu Chaudhary answer (which is sufficient by itself but leaves out some steps) combined with Mathew Crawford and Anunay Kulshrestha, answer:Idea is just to use inspection to prove the result about all of the cubes congruent to 1 or -1.In all this I use for modular arithmetic, e.g. to mean mod 7. First if a b or c are divisible by 7, are done. So the only remaining case is if they are all non zero.Any non zero integer cubed is either 1 or -1 mod 7 - we can just see this by inspection without any number theory:n1³ 1n2³ 8 1n3³ 27 -1nThen for the remaining ones use 4 -3, 5 -2, 6 -1n4³ -3³ 1netc.So na³, b³, and c³ - are either 1 or -1, and so it's impossible to have three different values for the cubes. So of(a³-b³)(b³-c³)(c³-a³)one of those factors must be 0 mod 7, so divisible by 7, proving the result.So you can see that without any number theory, just arithmetic and basic algebra.But the proof Mathew Crawford gave lets you see why it works,nbased on the simple result from number theory that n^k pm1:mod:p where p :(prime) 2k1.
n nAnd that then lets you generalize to e.g.abc (a^5-b^5)(b^5-c^5)(c^5-a^5) is a multiple of 11Follows because n^5 pm1:mod:11 (for any non zero integer n) so there are only two possible (non zero) values for the fifth powers, so we have the same situation as before.Either one of a, b or c is divisible by 11, or else, one of (a^5-b^5), (b^5-c^5), (c^5-a^5) is divisible by 11(Because of the theorem, we don't need to test all the numbers to check that n^5 pm1:mod:11)abc (a^6-b^6)(b^6-c^6)(c^6-a^6) is a multiple of 13Because n^6 pm1:mod 13 (for any non zero integer n)and so on, we have infinitely many results following the same pattern.
The general result is:abc (a^k-b^k)(b^k-c^k)(c^k-a^k) is a multiple of pwhere p is prime (greater than 2) and p 2k1Testing, e.g.n2*3*4 (2^6-3^6)(3^6-4^6)(4^6-2^6)n 216668874240n 16666836480*13:) - maths is fun isn't it :).(In this case the term which is divisible by 13 is (3^6-4^6) -259*13 )Oh, and proof of the theorem Mathew stated - not sure the name, but it's a consequence of Fermat's little theoremnn^p n:(mod:p)n(which has many Proofs of Fermat's little theorem) So starting from that, then if n is non zero we can divide both sides by nn(in all this, means mod p)nn^(p-1) 1ni.
e.
nn^2k 1nson(n^k)^2 - 1 0 (n^k- 1 ) (n^k 1 ) 0 mod p(i.e. is divisible by p)And because p is prime, it has to divide one or other of the factors there,nso we get the result that for any n non zero,n(n^k- 1 ) 0 mod pnorn (n^k 1 ) 0 mod pnso n^k pm1(see comment to Anunay Kulshrestha's proof for special case for k 3)