Video

If sqrt2^sqrt2 is rational, the statement is already true; if sqrt2^sqrt2 is irrational, (sqrt2^sqrt2)^sqrt2 is rational, which shows the statement is true

@@FightingWolferine You said "if sqrt2^sqrt2 is irrational, (sqrt2^sqrt2)^sqrt2 is rational, which shows the statement is true". How?

@@bobbybannerjee5156 if (sqrt2)^(sqrt2) is irrational, and we know sqrt2 is irrational, and notice that [(sqrt^2)^(sqrt2)]^sqrt2=(sqrt2)^(sqrt2*sqrt2)=(sqrt2)^2=2 is rational. So the statement is true.

Same, still fun to try!

Sure, yet the goal of this video is not to teach calculus or college math.

Apparently yes. Thanks for bringing this up. I previously didn't know the fact that this is actually the famous two-argument Ackermann function: en.wikipedia.org/wiki/Ackermann_function

I can only think of using Taylor's series, where: sin(x)=x-x^3/3!+x^5/5!-x^7/7!+...I can make a video for that;-)

@@FightingWolferine I agree, I think it's basically how to compute fractions by hand, with this a playlist on how to navigate without a computer may be plausible.

Right, that’s why it’s not as popular/famous as the one for quadratic equations;)

See the other video linked. The definition of [x] here is not the floor of x. [-2.236]=-2 not -3 with this notation. We take the definition of integer part of x as the floor if x is positive but the ceiling if x is negative. An error in the video was that by this definition {x} is between -1 and 0 when x is negative and this was corrected in the linked video. The final solution is still correct.

It’s essentially the same but it’s easier to describe using with congruence modulo concept: basically, discuss the remainder of p, 8p-1, and 8p+1 divided by 3. In order for all three to be prime, none of the reminders should be 0.

@@FightingWolferine Yes, but why check with modulo 3? If you already know it works as a solution, it's obvious, but why would someone new to the problem check with modulo 3 instead of, say, modulo 5? While what you said does indeed work as a proof, it still leaves one question unanswered: why did you do it that specific way? My method is much less concise, but I believe it's much easier to follow *why* I'm doing what I'm doing.

You can list out the first several prime numbers of p and it is quite obvious that in many cases 8p+1 is a multiple of 3. Then you can try the idea to examine the remainder of them divided by 3.

Another way to think about this is: 8p+1, 8p, 8p-1 are three consecutive integers. One and only one of them can be divided by 3.

@@FightingWolferine Ah, writing them out as 3 consecutive numbers makes sense, though the first reason you gave is pretty much "just brute force it lol." Though, that makes the explanation a lot easier. Half of the difficulty with mine is just proving that 8p is a multiple of 3 if 8p+1 and 8p-1 are prime. I can't believe I missed that obvious explanation. I think with that in mind it's far more elegant to prove that 8p is a multiple of 3 the way you said, and then just use the logic I gave to prove that if 8p is a multiple of 3, then p is not prime, no modulo needed.

No it is not. But the other direction is true: if p and 8p-1 are prime, they cannot be multiplies of 3.

Right, you need to use the factorization of a^3+b^3+c^3-3abc, which is not obvious if you don’t already know the identity.

@GuruprakashAcademy please see my other comments. [x] is not the floor notation. It represents the integer part of x, which equals -2 for -2.56.

@@FightingWolferine I think [] denotes the greatest integer function i.e., [-2.56] =-3

@@GuruprakashAcademy Mathematical notations can be complicated. What I learned at school and referred to in this video is in alignment with the explanation in Wikipedia: en.wikipedia.org/wiki/Floor_and_ceiling_functions. But historically, the notation could be different. Anyways, hope this clarifies the misunderstanding.

@@FightingWolferine Thanks. But with this notation, I think we cannot write x = integer part + fractional part... Just check for any negative numbers.

@@GuruprakashAcademy That's a good question. The fractional part (or decimal part) of x can actually have different definitions: en.wikipedia.org/wiki/Fractional_part. The odd function definition actually persists the relationship of x=[x] + {x}, however, {x} is between (-1,0] when x<0. So the solution process in the video was wrong to claim 0 <= {x} < 1 universally. The solution was right for x>0. When x<0, -1 < {x} <= 0, thus -1 < x^2 + x - 3 < 0, this leads to -2.3 < x < -2 (only take the negative range), which still leads to [x] = -2. So the final solution is still the same. But thanks for pointing out the issue.

Let me clarify the notation: Floor(-2.56)=⌊-2.56⌋=-3, ceil(-2.56)=⌈-2.56⌉=-2, integer part of -2.56 = [-2.56]=-2.

Right, but this solution is not complete. Need to capture all solutions.

Ah, ok...

I think those are two only solutions, though I didn't check to see...x^2 = 3 - floor[x], which means x^2 is an integer, so x if of the form +-sqrt(n) for some integer n (I mean, lol, I assume these are supposed to be real numbers)...so, for positive x, x^2 = 3 - floor [x], just check all the squares for x^2, there are finitely many...very few, actually, lol...

As for negative x, then it's the equivalent of an equation using y=-x, which would give us y^2 = 3 + floor [y], y positive...now, y^2 > 2y>3+y>=3+floor[y], if y>3, which is obvious...I mean, I could prove it too, lol...remembering that y>=floor[y], since y is positive...so that means we need only check y up to 3, no need to go further...which means that we need to descend only to -3 for x negative...so, lol, since x^2 is an integer, it's a short and finite search, I suspect sqrt(2) and -sqrt(6) are the only solutions, but maybe I'm wrong...

Basically, just try all x^2 from 0 to 9 or so (3^2 = (-3)^2 = 9), etc...

The downside of your approach is if the right hand side is a much greater number (e.g. A=1000], then iterating A-[x] will be tedious.

The integer part of x is defined as the floor of x when x>0, the ceiling of x when x<0. The integer part of -2.56 is -2, not -3.

@@FightingWolferine that’s true the integer part is defined that way but the notation in your video looks like x^2 + the floor of x.

​@@moeberry8226 your comment is wrong. The floor notation is square bracket with top part open.

Yes you're actually right, the floor of that interval is -3

The other solution is -√6 not -√5

Right, for every solution obtained under y>x, you can swap x and y and get the counterpart solution under y<x.

Oh, misunderstood the comment. Yes, it is not very hard to guess ONE solution but it takes some interesting thoughts to prove that’s the ONLY solution to the problem.

The last part is wrong: you are not supposed to use the conclusion that the upper bound is e<3. The convergence is based on the fact that this function is monotonously increasing and has an upper bound and we denote lim(1+1/x)^x=e. So the upper bound needs to be proved separately.

@@FightingWolferine Are you saying that stating e<3 is sort of a circular reasoning for the boundedness of the sequence (?)

Yes, since that’s how e is defined

My bad. For any natural number n > 1.

Right. Should have stated that here n is a natural number

quick mafs

Consider f(x,y)=x^k+y^k, because f(x,-x)=0, thus y=-x is a root to f(x,y)=0. Thus x+y is a factor of f(x,y). This is similar to e.g.: f(x)=x^3-x, f(1)=0, thus x-1 is a factor of f(x)

This substitution simplifies the condition: from a relationship between x and y to only on the radius r (essentially from Cartesian to polar). It’s useful in solving range problems.

q is prime and not 2, therefore it's odd

@@hyperpsych6483And it has to satisfy the modulo condition

@@hyperpsych6483 Oh, I see. That was obvious.

Also the fact that p|a and p|b does not imply that gcd(a,b)=p, but merely that p|gcd(a,b).

Right, mentioned in the narration but didn’t add that in the slide;)

It implies gcd(a,b) is at least p

@@FightingWolferine Yes.

Thanks 🔥