An extreme system of equations

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[@MichaelGrantPhD]

Actually the case where y = n + 1/2 contradicts the existing discovery that y = n + b/5 so it can be eliminated super quickly.

[@volteichefe7028]

My solution: Like Michael, say floor(x)=m and floor(y)=n for simplification. Equations become: (I): m²-n=2x+y and (II): m-n=x-2y Solving for x and y, pretending m and n to be constants independent from x and y, we get: (III): x=(2m²+m-3n)/5 and (IV): y=(m²-2m+n)/5 Recall that floor(x)=m and floor(y)=n, which gives: (I): m

[@josemarsimao1211]

Great solution. It is very creative.

[@Y3shu4]

There was an error in the “2 yellow plus red” step, but magically the argument still holds despite the “typo” Good video, as always.

[@RexxSchneider]

It's not really magic. Erroneously adding 3 times the floor of y to an integer still gives an integer, just as correctly subtracting 3 times the floor of y from an integer does. You still reach the conclusion that 5x is an integer, and the incorrect equation is not made use of in any further steps.

[@Y3shu4]

Oh, I didn’t mean “magically” literally, but thanks for registering the explanation for folks that don’t follow it :)

[@vivvpprof]

@@RexxSchneider No, of course it's not really magic. It's reading boring calculations off a cheat-sheet.

[@RexxSchneider]

@@vivvpprof Well, yes. But I think Michael Penn writes his own cheat-sheet. Some of us might call it a "script".

[@gustavinho1986]

I think I found a more straightforward solution. Let x=m+r and y=n+s, with m,n integers and r,s in the interval [0,1). Plugging these, we get m^2-2m-2n=2r+s and n=r-2s . Since -2

[@yoav613]

Nice. There is one more solution,x=3/5 and y=-1/5.

[@gustavinho1986]

@@yoav613 you are right, I'm going to edit my comment. Thanks!

[@danielbranscombe6662]

the way I found the solutions is that by substituting x=m+a/5 and y=n+b/5 into each of the equations you can move all the integer values (namely the floors and m,n) to one side and between the two equations you can conclude that 2a+b=0 mod 5 and a-2b=0 mod 5 in order to make the other side (containing only r,s) also an integer from this we can conclude that the only possible combinations of (a,b) are (0,0) and (1,3) if (a,b) is (0,0) then this leads to the (0,0) and (2,0) solutions for (x,y) if (a,b) is (1,3) then this leads to the (1.2,-0.4) solution

[@jeffreycloete852]

Hi Prof Penn, just rewatched a few of your videos from 2 yrs ago. You seem so much more comfortable in front of the camera now. Thanks for generously sharing your time!

[@demenion3521]

you could have skipped a few steps. especially plugging in x=m+a/5 and y=n+b/5 into the equations tells you that 2a+b=0 (mod 5) and a-2b=0 (mod 5) - which are congruent equations. for each choice of a that gives a very simple result for b. the second equation then always gives n very easily as n=(a-2b)/5, so either n=0 or n=-1. that can then be easily used in the first equation for m. this yields the following: a=2 -> get b=1 -> no solutions. a=3 -> b=4 -> (n,m)=(-1,0) or (n,m)=(-1,2). a=4 -> b=2 -> no solutions.

[@Nusret15220]

Floor functions always create interesting ideas and problems to deal with. And this is one of them, good job man.

[@DannyPansters]

2:52 Crayon: Floor you said?

[@Willwantstobeawesome]

perfect comedy cut

[@CTJ2619]

this is interesting - however when you multiplied the (yellow dot) equation by 2 and add (red dot) equation, shouldn’t the answer be -3(y) not +3(y)?

[@jplikesmaths]

You’re right

[@marienbad2]

It is y - (-4y) so he was correct.

[@wafikiri_]

@@marienbad2 Nope, watch it again: yellow dot equation by two PLUS, not minus, red dot eq. so, (-Ly)•2 + (-Ly), that is, -3Ly.

[@jplikesmaths]

@@marienbad2 he was talking about the lhs, with the floors

[@natepolidoro4565]

3:20 that should be a minus on the front of the floor(y).

[@alejandrosaldarriaga6061]

Nice job as always Michael ! Once we know that x and y must be of the form p/5 and q/5, by writing r and s the remainders of p and q respectively by 5, we see that r=2s mod 5 which yield 5 possibilities for the couple (r,s): (0,0),(1,3),(2,1),(3,4) and (4,2). The second equation reduces then to floor(y) = r/5 - 2s/5 which completely identifies y. Then we can plug those value into the first equation to find the solutions for x if any.

[@goodplacetostop2973]

6:20 Homework 13:45 Good Place To Stop The world is kinda burning right now, so be careful. Stay hydrated and keep your house cool

[@JavierSalcedoC]

World > europe

[@goodplacetostop2973]

@@JavierSalcedoC Not only Europe, this is also the wildfire season in California for instance 🤷

[@Vladimir_Pavlov]

Substituting the definitions x= ⌊x⌋+{x}, y= ⌊y⌋+{y} into the system of given equations, using the fact that {x},{y} ∈ [0,1) and considering a small number of possible options leads to the result faster.

[@scottmiller2591]

The realization at 12:00 that this has become a land war with Asia.

[@carstenmeyer7786]

At 3:56 there may be a sign error: It should be *... - 3 ⌊y⌋ = 5x* --------------------------------------------- At that point we can introduce the fractional part *{x}* to get to the cases for *⌊y⌋* directly: *x =: ⌊x⌋ + {x} with {x} in (0; 1), x in R* Use this definition to eliminate both floors in the given "red" equation and solve for *y* . Both fractional parts determine the domain of *y* : *y = {x} - {y} in (-1; 1) => ⌊y⌋ in { -1; 0 }* We plug the two distinct cases for *⌊y⌋* into *5x* at 3:56 to get: *⌊y⌋ = -1: 2 ⌊x⌋^2 + ⌊x⌋ + 3 = 5x => x in { 3/5; 6/5; 13/5 }* *⌊y⌋ = 0: 2 ⌊x⌋^2 + ⌊x⌋ = 5x => x in { 0; 2 }* We sketch the graphs of both sides to find all intersection points. Using *5y* at 3:56 we finally obtain: *(x; y) in { (0; 0), (3/5; -1/5), (6/5; -2/5), (2; 0), (13/5; -1/5) }*

[@TechSY730]

Surely there has to be something a bit more elegant than splitting this proof into 5*5 = 25 cases. Yeah, in this case there are some easy to spot cases to eliminate by equality or contradictions, but that isn't assured in the general case for other such floor/ceil function problems.

[@jursamaj]

See other response here (for example, daniel branscombe): you don't have to check 25 cases. Indeed, even following prof's Penn's example in the video, you'd only have 5 cases. Sure, not *every* possible system of equations can be narrowed down so, but we aren't solving every possible system.

[@thomashoffmann8857]

Around 9:30: how can you conclude that y=n + b/5? There is still 3y on the left side 🤔 Y could be also 1.0666... Multiplied by 3 gives a natural number plus 1/5

[@becomepostal]

Nope.

[@RexxSchneider]

It was postulated much earlier that y = n + b/5 where n ∈ ℤ and b ∈ {0 ... 4 }. That's how n and b were defined because 5y must be an integer. At 9:10 Michael Penn has found that 3y is an integer + 4/5 by manipulation of the original two equations. However, given that y is of the form n + b/5, it must be true that 3y is also equal to 3n + (3b/5). Clearly the fractional parts must be 4/5 in either representation of 3y. That means that when you divide 3b by 5 it must leave a remainder of 4, so 3b ≡ 4 (mod 5), and that makes b ≡ 3 (mod 5), so b = 3 since b ∈ {0 ... 4 }.

[@thomashoffmann8857]

@@RexxSchneider thanks! Only the x condition was shown later on but you are right. Also y has conditions 👍

[@josephmartos]

Yes! I love homeworks!

[@photinodecay]

But it is integral due to the floor function always being an integer!

[@coc235]

The proof that n=3 mod 5 in the second case canbe done much quicker, using just the first equation!

[@RexxSchneider]

Spoiler alert! For those who want to compare homeworks: For the case a=2 following the same logic, I get 3b ≡ 3 (mod 5), hence b = 1. So x = m+2/5 and y = n+1/5. Then m - n = x - 2y = m+2/5 - 2n-2/5 which gives n = 0. Hence y = 1/5 That leads to m² - 2m - 1 = 0 which has no integer solutions, so there are no solutions of the form x = m+2/5. For the case a=3 following the same logic, I get 3b ≡ 2 (mod 5), hence b = 4. So x = m+3/5 and y = n+4/5. Then m - n = x - 2y = m+3/5 - 2n-8/5 which gives n = -1. Hence y = -1/5 That leads to m² - 2m = 0 which has integer solutions m=0 or m=2. Hence x = 3/5 or x = 13/5. For the case a=4 following the same logic, I get 3b ≡ 1 (mod 5), hence b = 2. So x = m+4/5 and y = n+2/5. Then m - n = x - 2y = m+4/5 - 2n-4/5 which gives n = 0. Hence y = 2/5 That leads to m² - 2m - 2 = 0 which has no integer solutions, so there are no solutions of the form x = m+4/5. I'm happy to be corrected if I've slipped up in the calculations.

[@SuperYoonHo]

Subscribed or have i?!! oh well i have i subbed

[@vivvpprof]

Definite Vsauce vibes there but this channel just doesn't have the potential.

[@SuperYoonHo]

@@vivvpprof wow vsauce is famouse lol everyone knows it

[@Pradowpradow]

Didn't watch the whole video. I only found (y=x=0), (x=2, y=0), and (x=1.2, y=-0.4)

[@sadeghsadeghi7783]

You make a mistake. please correct the solution to this problem . our mistake is in 2 yellow plus red.

[@becomepostal]

And once again there is a mistake that makes the video harder to follow. That’s becoming cumbersome.

[@RexxSchneider]

As the equation that contains the mistake is only used to show that 5x is an integer, and is not used again in the solution, the error has no consequences (since adding 3 times the floor of y results in an integer in just the same way that subtracting 3 times the floor of y does). It was a bit disconcerting, though.

[@becomepostal]

@@RexxSchneider sorry but I can't follow a double discourse: the one that is written and the one that is true. It's not the first time it happens here, far from it.

[@RexxSchneider]

@@becomepostal There's no double discourse to follow. The set of equations that contained an error were only used to show that 5x is an integer and the error didn't affect that outcome. From that point on, the working went back to the original equations and the erroneous equations were never used again. Absolutely nothing in what was written subsequently was anything but true. You have only one line of argument to follow.

[@becomepostal]

@@RexxSchneider OK, but true doesn't follow from false, so even if here it didn't change a thing in the end, it was purely by chance, so he has not proved anything because there was a mistake, as usual.

[@RexxSchneider]

@@becomepostal Actually, truth can very often follow from false. You only have to look at an OR gate, for example. Anyway, I used to mark maths exams, and always had to "follow though" any mistake and give credit for correct methods, with very little deducted for trivial arithmetic errors. So I'm quite happy to give credit to Michael Penn for an accurate method and a correct result, which in no way depended on the inconsequential mistake of swapping a plus for a minus.