When Two Numbers Can Make the Rest

[@WrathofMath]

Happy New Year, gang! Thanks for all of your support in 2024, now - 2025 is going to be tubular. also, sorry for my camera's GARBAGE autofocus in this video Join Wrath of Math to get exclusive videos, lecture notes, and more: kzbin.info/door/yEKvaxi8mt9FMc62MHcliwjoin More math chats: kzbin.info/aero/PLztBpqftvzxXQDmPmSOwXSU9vOHgty1RO

[@Inky_149]

this year is going to be mathmagical!!

[@Happy_Abe]

Why tubular? 😅

[@GhostyOcean]

Bezout's identify. For any integers a,b (not both zero) there exist integers x,y such that ax+by=gcd(a,b), and this is the mininal positive solution. Any other combination of ax+by must be a multiple of gcd(a,b).

[@ccbgaming6994]

💯

[@gregorymorse8423]

Actually gcd(0, 0) is usually defined as 0 which preserves the relation in Bezouts identity.

[@tylerpullen8556]

This video is basically the beginning of linear algebra without all of the fancy terminology

[@liamturman]

Your videos are always so high quality and so informative! I love the music and your content. Keep it up!

[@xysrs]

i love when i know what a video is talking about before watching it

[@graf_paper]

I've always found the lack of an intos on this channel a little clunky, like I am 2 minutes in and not sure where the video is going.

[@wyattstevens8574]

Fun fact: if you iterate the algorithm starting with a and b (6:00) the values q you get are exactly the terms of a/b as a continued fraction!

[@Caramelldanson]

Since this is non-constructive, do we know a method for finding x and y for any a and b besides guess-and-check?

[@jamestan9422]

extended euclid’s algorithm

[@stormswindy3013]

i love linear combinations of scalars

[@davidbailis8415]

My guess is that for any two integers an and b, where a>b and a/b is not an integer, there exists an x and y value such that xa-yb=1. For all values of an and b where a/b is an integer, there does not exist integers such that xa-yb=1

[@StefanReich]

That's not correct... Consider a=10 and b=6

[@robel9tail]

I think it is where a - c = 1 and b is all factors of c.

[@m3morizes]

Your second statement is true for b>1.

[@redcleon]

4:50 I don't understand why r isn't multiplied by b, as well. a / b = q + r. What am I missing?

[@WrathofMath]

This is division with a remainder. It mimics how we begin talking about division in school. For example, what is 20 divided by 8? Well, 8 goes into 20 two times with a remainder of 4, that is: 20 = 8*2 + 4. What is a divided by b? Well, b goes into a q times with a remainder of r, that is: a = bq + r.

[@thevikifalcon7670]

Numberphile put out a similar video recently with David Eisenbud. Great minds think alike!

[@cs9ib]

Great video

[@WrathofMath]

Thank you!

[@ufailowell]

my guess is 3 and 2 but I am ready to be shown why thats not right

[@SeNate7776]

I'd theorize it would work for any relatively prime number pairing

[@StefanReich]

@@SeNate7776 Yes

[@TannerJ07]

Any coprime pairing

[@phongthaiquoc833]

hi

[@odemura]

why express an integer through a pair of possibly bigger integers used as coefficients?

[@mikemian]

If you can manufacture 2 known weights or volumes, what unknown weights or volumes can you measure. Consider the problem: if you need to measure 7 gallons of water and you have 8, 12, 16, 20, and 25 gallon containers, can you do it and what's the least number of containers you need to bring.

[@andrewkarsten5268]

@@mikemiandid you mean 24 instead of 25? As written it’s quite easy, fill 20 all the way and fill 12 all the way, pour 20 into 25, then pour the remaining 5 from 12 to fill 25 and you’ll have 7 left. There may be a way to do it with 2 containers and if there is I could maybe find it by thinking longer, but as written you had the gallons increasing by 4 gallons until the last jug, hence why I ask.

[@mikemian]

@@andrewkarsten5268 if the last container was 24 and not 25, you wouldn't be able to measure 7 gallons.

[@andrewkarsten5268]

@@mikemian right, but as stated the first part of your question was “can you do it?” Obviously, part of the problem is figuring out if it’s possible to begin with. If you asked with 24, the answer is no, just like how in this video you cannot always find integers x and y given a and b such that a linear combination of them gives some integer. If gcd(a,b)=4, then you can’t make 3 for example. I agree if it’s 24 you can’t do it because the best you can do is measure multiples of 4.

[@mikemian]

@@andrewkarsten5268- you can do it simply using the 25 and 8 gallon containers (fill the 25 from the 8 4 times and you'll have 7 left in the 8), but the point of asking if it's possible, is knowing you will need the 25 gallon container in conjuction with any of the others apart from the 20, if you use 2 containers.

[@neuralwarp]

Processes is an English word, not a Greek or American one. So it's [ PROE•sess•iz ] not [ PRAH•sess•eez ].

[@kyay10]

"American" is an accent, so this is valid. This type of pedantary is useless

[@gregorymorse8423]

Wrong. If a=0, everything said here is wrong. a=0, b=1, gcd(0,1)=1 per definition of gcd. Therefore neither a nor b can be 0 as an extra condition for this to be true.

[@WrathofMath]

so in other words, everything i said is right since I explicitly said a and b are positive. thanks for clarifying

[@gregorymorse8423]

@WrathofMath ah didnt catch it. But this isn't necessary either? Negative numbers also work fine.