The Freshman's Dream (a classic mistake) - Numberphile

It's a classic mistake - but The Freshman's Dream can come true. See the extra footage at • The Freshman's Dream (... - featuring Kevin Tucker. More links & stuff in full description below ↓↓↓
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[@numberphile]

See the extra footage at kzbin.info/www/bejne/npjVhWarhNWcitk

[@guepardiez]

2ab or not 2ab, that is the question.

[@hancocki]

Still better in the original Klingon.

[@goatgamer001]

Indeed

[@Sbence92]

Well, matrix multiplication says not 2ab. If you know what I mean.

[@oliviervancantfort5327]

2ab or not 2ab, that is the equation...

[@RC_Engineering]

I read it as Italian. "to uh-be or not to uh-be"

[@Mikee512]

Even a stopped mod 2 clock is wrong once a day.

[@Rdac0]

A running mod 2 clock could be wrong twice a day

[@santiagoorozco1886]

Well, twice a day is no times a day mod 2

[@Arnikaaa]

Someone please explain!!

[@Rdac0]

@@Arnikaaa a running mod 2 clock that's wrong twice a day mod 2 is wrong 0 times a day because you just read the opposite number

[@derphysiker1062]

The most common example of this mistake i see in students homework is the case for n = -1. So: 1/(a+b) = 1/a + 1/b

[@MK-13337]

Well "everything is linear" is so common it is memeworthy. Sin(x+y) = sin(x) + sin(y) because why not...

[@enpeacemusic192]

@@MK-13337oh but we all know sin(x) = x so that's actually true /s

[@MK-13337]

@@enpeacemusic192 True! One of the things that deducted the most points from a calculus mini exam was when people tried to simplify arcsin(2sin(x)) = 2x.

[@enpeacemusic192]

@@MK-13337lmaoo first order Taylor approximations go hard

[@willthecat3861]

IMO... its lazy or incompetent teaching.... because straight away... when at least the universe is real numbers... then when (a,b) = (1,2), 1 > 1/(a+b). By the time you are 12, you know what "greater than" is, what "equals"... or at least "identical to" means... and what a no-exception 'rule' means. If this stuff hasn't been incorporated into a students 'mathematical' understanding... by the time they are 'qualified' freshmen... then that's the result of a sh*t educational system, at least for mathematics. Sorry if that is too opinionated; but, that is my opinion.

[@Johnjkngbaup]

Hate to admit it but the square root of a sum being equal to the sum of both square roots separately is definitely something I've done like 15 steps deep into solving a differential equation on an exam before.

[@Adamreir]

This was super exiting! «Here is a pattern. Does it always work: no. But it always works for primes, and here is the proof.» Great story with a great punchline, and some tension in between!

[@barutjeh]

One reason the mistake is so common is that a similar rule does hold for multiplication. (ab)² = a²•b². It can make the same step feel more plausible for addition.

[@doomtho42]

Thank you for making me feel less insane. I was watching the video thinking I was an absolute moron who’d somehow gotten this wrong for years without ever noticing it. Then I read your comment and realized I was thinking of the multiplication “version” of the equation. Phew.

[@adambellcomp]

Thank you numberphile! As a composer, your video on superpermutations gave me a new way to structure elements of a composition. And now, after seeing the Pascal triangle, even though I've seen it numerous times before, has given me the answer to deriving the total number of each sub-ensemble for a given group of players. I assumed there was an underlying mathematical answer to that problem, and I had never found the time to sit and work out an answer, so thank you!

[@Steve_Stowers]

Yes, I can confirm that freshman (and even non-freshmen) make this mistake all too often. My theory is that they do this because they've learned to think of the Distributive Law as a rule for removing parentheses, rather than a description of the relationship between addition and multiplication.

[@jessehammer123]

I tend to concur. The “everything is linear idea” seems to me, best I can tell, to derive from a mental shortcut of wanting to simplify things by making more complicated, harder-to-understand stuff apply to smaller chunks of information. Like with sqrt(a+b)=sqrt(a)+sqrt(b), what can you possibly do with sqrt(a+b) otherwise? It looks complicated, but students know how to deal with sqrt(a) and sqrt(b), so boom- sqrt(a+b)=sqrt(a)+sqrt(b). Certainly distributivity is a nice property, but it seems to me a lot of students consider it applicable in all cases as a result of that niceness.

[@rociopaoloni5080]

​​@@jessehammer123At least where I'm from it's nos just that students apply it because it's nice, it's because that's the extent of what we are taught. I just knew that some parenthesis can be removed that way and that's it. All this modules and relationships between valor that it's being talk about it's alien to me, and so many others.

[@davidwright5719]

As a PhD physicist, I don’t make the freshman version of this mistake, but I do occasionally make more advanced versions of it. For example, when computing processes to which multiple Feynman diagrams contribute, I have occasionally forgot to include the interference terms.

[@md-sl1io]

assuming that (a^b)^c is the same as a^(b^c) is one that a lot of people make the mistake of

[@krystofdayne]

Yeah powers without parentheses is one that sometimes still catches me out. I was recently confused by an exercise in probability theory where there was a sequences of real random variables X_n that, for some λ > 0, had probability 1/2 of either being n^λ or -n^λ. And at first I thought, hang on, X_n is supposed to be real, but wouldnt -n^λ give you complex numbers if λ isn't an integer? But of course it doesnt because -n^λ = -(n^λ) ≠ (-n)^λ

[@pinkraven4402]

Yeee, dat one's sneaky fox

[@djdrack4681]

I think many of these (similar) mistakes students make are prob because of the order of progression of math in school: algebra-->geometry-->trig/pre-calc-->calc-->linear algebra I think the issue is between class tiers 1/2...at least in US they stress the theorems (memorizing the format of them etc), rather than the underlying logic behind the theorem. So kids will remember various transitive/associative theorems in algebra, similar but diff ones in geometry, then (being overloaded w/ equations cuz they're not mathematicians) they start to think these sort of 'simple' associative/transitive theorems can be 'universally applied' to other similar (but diff) formulas/scenarios.

[@EebstertheGreat]

2:55 "One o'clock, three o'clock, five o'clock" *ROCK* "Two o'clock, four o'clock, six o'clock" *ROCK* *Nine, ten, eleven o'clock, twelve o'clock, rock* *We're gonna rock around the clock tonight*

[@Zacks.C-land]

Glad I wasn’t the only one who thought of that song as he said the numbers out loud. 😂

[@coreC..]

@@Zacks.C-land Here's another one..

[@yanntal954]

Fermats little theorem gives you this. (a+b)^(p-1) = 1 (mod p) And so we have that: (a+b)^p = a+b (mod p) = a^p + b^p (mod p)

[@zswu31416]

Nice. Although the proof in the video works for all rings of characteristic p

[@proloycodes]

did yall not watch the video? he shows this too

[@minamagdy4126]

Well, the form with an exponent of p-1 given is only true for the base not being a multiple of p. A more general form of Fermat's Little Theorem is the form with an exponent of p, which works for all integers. Otherwise, that's a different valid proof.

[@yanntal954]

@@minamagdy4126 You can even get the Freshman's dream using the more general Euler's totient function. For example, you can get it for exponent 24 using mod 35

[@Kaelygon]

I remember a teacher saying reducing addition from fractions was a common mistake, for example (x+2)/(3x+2) != x/3x . So each time you did this in test, he wrote the 5 "S" reminder Senkin Suuri Sössö Supistit Summasta which from Finnish roughly translates to "you big sissy reduced from sum"

[@jimmyh2137]

Little trick: if you don't remember if a rule is valid or not, try with small numbers! Not sure if you can reduce a sum? X=1 ---> (1+2)/(3+2) = 3/5 != 1/3

[@CodeNascher_]

in german it's similar: "Differenzen und Summen kürzen nur die Dummen", which translates to "only the dumbs are reducing the differences and sums"

[@Kaelygon]

​@@topherthe11th23In programming != means "not equal" which is same as crossed equal sign. There's space between the parenthesis and '!=' , it's a bit hard writing equations in text. For example if we substitute x with 1, (x+2)/(3x+2) = (1+2)/(3*1+2) 3/(3+2)=3/5 which is not same as x/3x = 1/3 hence (x+2)/(3x+2) '!=' x/3x I didn't mean factorial

[@xchurricane]

Hearing someone say "back when i was in school we called it foil," completely agreeing with him, then realizing we're probably around the same age 😂 Don't make early career sound so old haha!

[@tomholroyd7519]

People still say foil it out (in English anyway)

[@henryhill7991]

​@@tomholroyd7519 some schools are trying to move away from FOIL, since it only works for 2 binomials. Distribute or sometimes double distribute is more clear and accurate.

[@frydaze13]

I always hated the mnemonic "FOIL". I know it's so clever trick that sort of works, but I felt like it wasn't really helping my kids learn mathematics. By the time binomial multiplication was being introduced, the kids already knew and understood distribution of multiplication over addition. I.E., 4(x+3)=4x+12 They knew to distribute the 4 over the (x + 3) So when it came time to multiply binomials, I would just call it double distribution. Tell them to distribute the first term over the other two, and then the second term over the other two. They got it just fine, and perform just as well as their classmates who learned to "FOIL". And I felt like my kids understood just a little bit better what was going on during the process without the cutesy little name

[@henryhill7991]

@@frydaze13 thats awesome! In my class I try very hard not say FOIL for exactly that reason. Many of the kids can follow that process, but make one of them a trinomial and now they are lost. But the ones one understand it was distribution are able to do those higher level problems.

[@Ryan_Thompson]

The clock thing might be nice as an example to introduce modulo arithmetic, but as a programmer, the constant switching between 1-based and 0-based systems gave me conniptions.

[@orterves]

I'm glad I'm not the only zero

[@thecommexokid]

Well 0 = n (mod n), so whether you count from 1 to n or from 0 to n-1 doesn’t matter, since 0 and n are equivalent.

[@Ryan_Thompson]

@@thecommexokid But 0 % n = 0, and 1 % n = 1. Actually the whole example is off by one if you don’t catch any of the many points where he switches back and forth between the two systems. It didn’t really trip me up but it was a unnecessarily a bit confusing in the same way mixing 0-based and 1-based arrays or loops all over the same code might be. That’s it. Not the end of the world.

[@TheCalabiYauManifolder]

​@@Ryan_ThompsonIn mathematics, the integers mod n aren't actually integers at all. They are a separate structure, in which n is actually literally equal to 0, so it doesn't matter how specifically you're referring to them because they aren't actually the integers that they're being referred to as. What is described as "2" here, for example, isn't actually the integer 2 but its projection into this separate structure.

[@skylark.kraken]

In year 7 maths my teacher showed Pascal's triangle and without explaining how it was constructed asked my class what we noticed about it, I was thinking about something unrelated, nobody offered to answer, I assume my teacher looked for who was the most distracted and chose me, within a second of being asked I said that the sum of each row is equal to 2 to the power of the second number which was not an answer my teacher was expected, he asked me what else I noticed and I said that the rows with prime numbers in the second position are all divisible by that number; his reaction both times was kinda funny, he was a very animated guy. It took me until my third try to mention how it was constructed It was beginning of the first year of secondary school and that moment was something he brought up at parents' evening, my parents bring it up whenever they can - he stood up while talking to my parents because he couldn't contain his excitement. He ended up offering me tutoring 1 or 2 lunchtimes a week, was able to get way ahead, and it all started with Pascal's triangle and prime numbers being special, he did always enjoy showing me something I had no clue about and see what I made of it. I'm a human calculator (as in a non-non-human calculator, going back to the beginning) btw

[@faithgrins]

To answer Brady's question about "OK, but where do we use this?": modular arithmetic is _super_ important to a lot of algorithms that are used in common computing scenarios. E.g.: the most common cryptographic scheme on the internet is RSA, which only works because raising a number to the correct power in the right mod system yields the original number.

[@SomeNerdOutThere]

Modular arithmetic shows up in Bresenham's line algorithm. It's a beautiful function that traces the integer values of the xy coordinates along an arbitrary line segment. Instead of using division to handle the slope, it uses modular arithmetic to tell when it should increment along the shorter axis. Because integer math is far faster on a computer than floating point, this is a very rapid way to trace a line.

[@lenskihe]

Pascal's triangle always coming back to surprise you

[@krystofdayne]

Tbf, this was probably the least surprising appearance of Pascal's triangle ever ^^' the fact that it expands into the binomial coefficients is probably the most well-known thing about it

[@lenskihe]

@@krystofdayne Very true. But the fact, that it obeys this pattern connected to prime numbers, did surprise me. I should have written: Every time you think you know everything there is to know about Pascal's triangle, some pattern shows up that you've never heard about.

[@the1exnay]

I think he undersold how useful modular arithmetic is. Anytime you just care about the remainder, it’s modular arithmetic. If you’re giving someone $4.25 in change and you want to know how many quarters to give them, it doesn’t matter that 4.25/0.25=17. All that matters is that 4.25 mod 1 = 0.25 and 0.25/0.25=1 Similarly when calculating what day of the week it is or what time of day it is, that’s modular arithmetic Additionally modular arithmetic is great for double checking something. You can run the calculation faster in modular arithmetic and double check that matches the answer you got in normal math. There are also a ton of ways it is used in computers and for fancy mathematical proofs. But you don’t need to know it in order to benefit from your computer using it.

[@TealJosh]

For me it's just inherently ingrained from the start. Never had issue with freshman's dream. Just memorized it. a^2{+/-}2ab+b^2 = (a{+/-}b)^2 And the true dream: a^2-b^2 = (a-b)(a+b)

[@AnotherRoof]

Ohhh this came up in my latest video on polygon constructions in showing the irreducibility of certain cyclotomic polynomials (which I promise is related to constructing regular polygons).

[@kingoreo7050]

My two favourite maths youtbers in one place!

[@TomRocksMaths]

Even better: 1 + 1 = 0 (on the two clock)

[@keatonconnell1694]

Don't tell Terrence Howard!

[@maynardtrendle820]

​@@keatonconnell1694My thoughts exactly!😂

[@darkpulcinella9690]

which is bascially 1 XOR 1

[@CommanderdMtllca]

A wild Tom rocks appears!

[@TomRocksMaths]

@@CommanderdMtllca used a master ball

[@CraigGidney]

Fast multiplication algorithms are based on making this dream come true. The Schonnage Strassen algorithm transforms the problem into a basis where the a*b term is zero because a and b are perpendicular vectors. The specific transform they use is called the number theoretic transform; very similar to the Fourier transform. More abstractly it's an application of the convolution theorem.

[@Khannesjo]

you can just ditch the square entirely in mod 2, as an odd squared stays odd, and an even squared stays even.

[@chrisdaley2852]

This extends to the prime case with Fermat's Little Theorem: (a+b)^p≡a+b≡a^p+b^p (mod p)

[@hansolo6831]

@@Khannesjo what about 1? 1 is odd but it's square is even

[@maighstir3003]

@@hansolo6831 1 to any power is still 1.

[@antosha4812]

@@hansolo6831 terrence howard?

The maximum modulus/"clock size" that works is the greatest common divisor of the corresponding Pascal's triangle row: 1,1,2,3,2,5,1,7,2,3,1,11,1,13,… Surprisingly, this is *not* in OEIS! There are multiple with the first entry removed, but I could not figure out whether any of those always has to be the same and why.

[@AidanRatnage]

I learned this in GCSE Maths, I wonder what else freshmen forget from when they were 16.

[@michaelbauers8800]

I can understand how someone might confuse the distribution that allows a(b + c) with the square. But I also thought that early on in algebra, the square of the binomial is taught.

[@Sfx654]

i love how he makes a duck quack sound every time he says "right?"

[@Mason-lr5dz]

I really hope he never reads this, but I'm glad I did 😂

[@rtpoe]

Mod 2 is basically "Odd or Even?"

[@kelli217]

I note that while (a + b)^4 = a^4 + b^4 does not work for mod 4... it _does_ work for mod 2. So it's possible that other modulos might apply to the freshman's dream, but not the actual degree of the polynomial.

[@mongmanmarkyt2897]

The relationships are prime :)

[@MK-13337]

@@kelli217 (a+b)^(2k) = a^(2k) + b^(2k) in mod 2 for any k (even non natural numbers make some sense although then you run into the problem that you can't choose any representative from an equivalence class).

[@Moinsdeuxcat]

Pythagoras theorem is kind of an instance of Freshman's dream being true (||u+v||²=||u||²+||v||² if u and v are orthogonal), not because 2=0, but because the (scalar) product of both vectors ("cross-term") vanishes. Same thing happens in any algebra for commuting zero-divisors a, b such that ab=0 as then (a+b)²=a²+b². And, more generally, for any pair of anticommuting elements (in any characteristic, but of course mod 2 this just means commuting!)

[@iabervon]

You can extend modular arithmetic to powers of primes in a way where the calculations come out strangely (but usefully for error correction and cryptography) and the freshman's dream still works the same as it does for those primes. In the arithmetic used in AES encryption, (a+b)²=a²+b² for all 256 possible values of each of a and b.

[@G.Aaron.Fisher]

It's odd that he said it's "not clear" whether Fermat's Little Theorem (FLT) or Fermat's Last Theorem is more useful. Maybe he just didn't want to go off topic, but I'm sure he knows that it's not particularly close. There's a reason that the initialism FLT is understood to refer to Fermat's Little Theorem, despite Fermat's Last Theorem being so widely known. FLT is a strong candidate for most useful theorem in all of number theory, and regularly come up in all sorts of problems and applications. Fermat's Last Theorem is more of a novelty item, similar to Goldbach's Conjecture. I've seen the n=3 case come up a few times in other problems. But I'm unaware of any applications for Fermat's Last Theorem in its entirety.

[@strifen117]

This made me interested in when there exist solutions for certain pairs of an and b for N non prime. I even took out a notebook and tried a few examples to see if I could find a pattern. I found that for N=4 you could get one, and for N=9, so I thought maybe squares. Then I tried 16 and it didn’t work, so I thought maybe squares of primes. I tried 25 and that didn’t work either. I reached the end of my math knowledge and hung it up there, but it made for a fun half hour of math exploration :) I’m really appreciative of all the videos you put out, they always make me think, thanks

[@malavoy1]

The z-transform is a use for this Fermat in discrete mathematics (it's a discrete version of the Laplace transform). If you sample a signal, you have to sample at at least twice the highest frequency. However when you reconvert back to continuous math you can only represent the original frequency range. So, for example, if you sample audio and there is noise above 20 Khz, it will wrap around to a low frequency, hence the sampling device needs to first filter out anything over 20 Khz. Or alternatively you can increase the sample rate to include enough of the noise as well, so it appears in it's proper place and be filtered out while in 'sample' form (i.e. digital in this case).

[@ni5439]

I never did that mistake. But after watching the explanation at the beginning, I'm sure it will get stuck in my subconscious mind and it will pop out one day

[@garrysekelli6776]

A freshman in highschool could easily make this mistake. A freshman in college would have to be a football scholarship.

[@onethreetheshaman]

Well, that's elitist of you.

[@Omnifarious0]

On the (a+b)^6 case, you can scrap three terms if you use the 3 clock (you still have to keep 20a^3b^3). It's the greatest common factor that you want.

[@EPMTUNES]

This video was great! Would love to see more from professor tucker

[@MK-13337]

Also (a+b)^2 = a^2 + b^2 when multiplication is anticommutative.

[@Moinsdeuxcat]

This includes the case of (characteristic 2 + commutative) as commutative and anticommutative are synonymous when 1=-1

[@good_deeds_always_get_punished]

A freshman's dream is graduating without any student loans.

[@dibenp]

17:15 How did he say that (2 x 4^5) is mod-7-equivalent to (2^3 x 4)? Oh. I think I got it… (2 x 4^5) = (2 x 4^2 x 4^2 x 4) {= mod 7} (2 x 2 x 2 x 4) = (2^3 x 4)

[@INT41O]

Set a = {{1,0},{0,-1}} and b = {{0,1}, {-1,0}} and it is true as well. This is related to the way Dirac derived his equation.

[@Chalisque]

if a and b are zero divisors such that ab=0, then a^n+b^n=(a+b)^n as all the other terms in the expansion vanish. (e.g. a=2, b=5 working mod 10)

[@roberttalada5196]

I love the transition to 3D for mod(3) with cubes

[@coolworship6704]

16:11 "Give me your favourite prime" Me: (thinks 7) Says 7 "Give me a number less than 7" Me: (thinks 4) Says 4

[@tomholroyd7519]

I like these types of equations, which I say are false if you are an undergraduate, but they become true again in graduate school

[@ffc1a28c7]

I kind of wished that the proof of the binomial theorem was went into a little. It's really easy to do using Pascal's triangle and is quite illuminating (for (x+1)^n, you have that (x+1)^(n-1) is some polynomial, say p. Then (x+1)^n=xp+p, and the adjacent terms end up with the same exponent of x and can be added).

[@Rubrickety]

I’ll bet this video generated a lot of excited “I spotted you in the background of a Numberphile video!” messages. 😊

[@TSotP]

We weren't taught Foil. I always used a grid ......a +b a.. | a² | ab +b |ba | b² a²+ab+ba+b² = a²+2ab+b²

[@linuxp00]

That's true for orthogonal vectors, think about it c² = (a + b)² = a² + b² - 2ab cos(90°) = a² + b²

[@MrClivesinger]

Degrees or radians work exactly like clock arithmetic, and circular coordinates systems are extremely useful

[@AdrianBoyko]

I don’t know the name of the disembodied voice but it’s funny that he thinks that there is one “real world math” and that the infinite number of maths modulo N are “fake”.

[@louisreinitz5642]

I had a professor who named this "The Law of Universal Linearity"

[@djsyntic]

Perhaps I'm projecting a touch, but the way the guy in the video keeps stepping forward a little bit and then stepping backward a little bit makes me think he has to go to the bathroom. But as I say perhaps I'm projecting because I got to go to the bathroom.

[@chonchjohnch]

Funnily enough, when doing partial fraction decomposition for laplace transforms, you often can use this as a shortcut in my experience

[@markhagerman3072]

I've seen a variation of this error "in the wild". A friend mistakenly assumed that sin(x+y) was the same as sin(x) + sin(y).

[@toohip]

This is kind of how parity works in data storage RAID. Just mathing out to an odd or even

[@Sfx654]

i think at this point we can generalize the freshman problem to any function f: f(x+y)=f(x)+f(y)

[@elvest9]

But it's much easier and more fun to calculate if (a+b)^2 = a^2 + b^2

[@johnny_eth]

Most computer integer arithmetic is mod 2**32 or mod 2**64, plus 2-complement trickery to encode negative numbers. Computers do a lot of quirky stuff with arithmetic which developers need to know and handle in code.

[@pavfrang]

When it's not identity it's an equation with possible solutions -> a²+b² = (a+b)² is true if a or b is zero, which means that this has infinite solutions..

[@usnbostx2]

One of my Navy instructors called this the “fuzzy bunny”, as in Monty Python.

[@PDex-r6g]

"two kinds of people argument" has a new meaning to me after this.

[@AdmiralJota]

Doesn't the Pascal's Triangle thing still work fine for rows that are *powers* of primes? E.g., row 4 works if you use mod 2; row 9 works if you use mod 3; etc.

[@timseguine2]

Interestingly Pascal's triangle mod 2 looks like Sierpinski's triangle.

[@maksampece947]

Wow, that is beautiful. Never realized it

[@veschyoleg]

“I have basically the slickest proof to convince someone that this really works” Did he say something concerning margins? 🤔

[@JamesVestal-dz5qm]

In my teaching experience I didn't have a lot of students make this mistake. My favorite example is (a+b)(a-b)=a^2-b^2, so the freshman's dream is sort of true in this instance!

[@weare2iq376]

We're gonna rock around the clock tonight🎶 We're gonna rock, rock, rock, 'til broad daylight 🎶

[@danedickerson]

log(1) + log(2) + log(3) = log(6)!

[@megachonker4173]

You just have to think about the definition of the objects you're working with: What is 𝐚²? 𝐚•𝐚. What is (𝐚 + 𝐛)²? (𝐚 + 𝐛)•(𝐚 + 𝐛). If the students spent more time thinking instead of solving problems like machines, such mistakes would not happen.

[@hansolo6831]

Yes exactly

[@philw6056]

Mistakes always happen. Sometimes you make an error, maybe a sign error and have trouble finding where it went wrong. And as soon as you notice where to look, you also notice the error. That's relatable. It's more of a problem when you can't find the error after someone made you aware of the operation that went wrong.

[@megachonker4173]

​@@philw6056yeah, mistakes can happen. When it happens, you have to think about the definition of the objetcs and do it again.

[@megachonker4173]

​@@philw6056anyway... let's analyze this situation: "Oh! What is (𝐚𝐛)²? I have no idea! But my teacher said that it is equal to 𝐚²𝐛², so... it must be true! Hmm... What is (𝐚 + 𝐛)²?! How can I know if my teacher did not say a thing about it!? Well.. If I repeat what he did, then: (𝐚 + 𝐛)² mist be equal to 𝐚² + 𝐛²! Obviously! I did not I think about it?"

[@megachonker4173]

​@@philw6056anyway... let's analyze this situation: "Oh! What is (𝐚𝐛)²? I have no idea! But my teacher said that it is equal to 𝐚²𝐛², so... it must be true! Hmm... What is (𝐚 + 𝐛)²?! How can I know if my teacher did not say a thing about it!? Well.. If I repeat what he did, then: (𝐚 + 𝐛)² mist be equal to 𝐚² + 𝐛²! Obviously! Why did not I think about it?"

[@bertblankenstein3738]

All works fine if a or b or both are 0. No clock required.

[@bethhentges]

Or just one of them is zero.

[@zombi1034]

I am a programmer. For me O( (a+b)^2 ) == O( a^2 + b^2 ) 😂

[@Ryan_Thompson]

This guy big-Os.

[@hedlund]

​@@Ryan_ThompsonA proper-endian lad if ever I saw one.

[@moogzoliver]

Logically, the square is not composed of a sum of smaller squares. A square of area 16 cannot be constructed by adding two squares with side lengths of 2 together.

[@mrzorak4532]

In linear algebra, if a and b are orthogonal, then (a+b)^2 = a^2 + b^2

[@MinMax-kc8uj]

I don't care about modular arithmetic unless it's in a trig function in some way. I think the multi angle tangent function has it, if I remember correctly.

[@wobaguk]

People naturally do modular arithmentic without realising it when they are thinking about days. Its Tuesday, I have to do something in 8 days. OK next Wednesday then. How many more than a multiple of 7.

[@samueldeandrade8535]

You are overestimating people a LOT, but ok.

[@samueldeandrade8535]

You are overestimating people a LOT, but ok.

[@tommyhuffman7499]

In your 2 clock, you can't have a^2 + b^2. There's only two numbers, 1 and 0. Without trying to write it all out on my phone, for a and b to be different, one most be 0. So, you only have two variables that equal 1, so a^2 + a^2, which is 0 in the end; two variables equal to 0, which is a^2 + a^2 and results in 0, or a and b are different, meaning 1 of them is 0, so it should probably cancelled with the middle term, and whose final result is 1.

[@davidmilhouscarter8198]

10:55 2, 3, and 5 are primes. That’s what they have in common.

[@rickseiden1]

I was just working something out last night and involved FOIL. I still say "First, outer, inner, last" as I do the multiplication.

[@FaerieDragonZook]

(a+b)^2 is actually (a*a+a*b+b*a+b*b). If you live in a 'space' where a and b 'anticommute' with each other, a*b+b*a = 0 so (a+b)^2 = (a^2+b^2). For example, in the quaternions, if a = i and b = j, where i,j, and k are the quaternion nonreal basis elements, then a*b = k and b*a = -k. so (a+b) = (i+j)^2 = -2 and a^2 +b^2 = -1+-1 = -2

[@rainrunya]

I clicked on this thinking it said "The Frenchman's dream" and I didn't realise until the Wikipedia snippet that I was wrong

[@edmundwoolliams1240]

Michael Penn did a video about this, and this explains it much better

[@DouglasZwick]

1, 2, 1 o'clock, 2 o'clock rock 1, 2, 1 o'clock, 2 o'clock rock 1, 2, 1 o'clock, 2 o'clock rock We're gonna rock around the clock tonight

[@jimboslice4468]

Maybe I'm crazy but I can't say I've ever seen anyone write an 8 like that 4:08

[@marksteers3424]

Why do Americans call the combinations P choose K? We were taught P - permutations - C- Combinations. Just a quirk I suppose. We also say minus 5 and not negative 5.

[@EastBurningRed]

great introduction to modular arithmetic

[@pierreabbat6157]

4^7=16384, and 1638 is made by overlapping 14, 21, and 28.

[@andrewclifton9772]

I learned that when I was 12. WTF kind of freshman - I assume they are talking college students - thinks that (a+b)^2 = a^2 + b^2 ??

[@bertblankenstein3738]

One, two, tree o'clock, four o'clock rock Five, six, seven o'clock, eight o'clock rock Nine, ten, eleven o'clock, twelve o'clock rock We're gonna rock around the clock tonight.

[@jasper265]

On the 2-clock a^2 = a and sqrt(a) = a. With that in mind, it seems rather silly to be talking about the Freshman's dream in that context...

[@boopsnoop5469]

Complex variables - second year, did this mistake on some HW near the end of the semester, got it back and both the professor and I were shocked that I just did not FOIL it lol

[@andrewbesso4257]

I have heard this error called a "wishful thinking theorem".

[@d18c7db]

I'm confused, why for example with the 2 o'clock we evaluate as if 2=0 for coefficients only, but we evaluate a² and b² to the power of 2 instead of to the power of 0 as for the coefficients?

[@ZEPHYRZHANG-mg8zi]

I mean you dont even need to expand it to see why it's bad. You can break up any number into just ones and then you'd get any number to any power is just itself

[@ILikeEpicurus]

This Mr Tucker explains brilliant (not all Tuckers are the same), but moves around like the seconds-hand on the clock. How does that compute?

[@GNARGNARHEAD]

the pattern at the heart of mathematics 😵‍💫

[@Matthew.Morycinski]

How can anyone forget the mixed terms if the equation is utilized over and over throughout high school math?

[@umi3017]

Isn't that makes (a+b)^2=(a+b)^0=1?, then it's different to a^0+b^0=2

[@shadamethyst1258]

The exponent is actually a natural number (or an integer, if you have a group structure), not a modulo number. It's a subtle but important detail

[@maksampece947]

The exponent is just a representation of how many times the multiplication is done, so it stays the same. The only reason we simplify ab into a (mod p) * b (mod p) (e.g. 3 * 1 = 1 * 1) is because of the unique property that ab (mod p) = a (mod p) b (mod p), the way I understand it.