7 ms·

This reminds me of this video about why we need to be very careful when inspecting visual proofs: https://www.youtube.com/watch?v=VYQVlVoWoPY https://www.youtub

This reminds me of this video about why we need to be very careful when inspecting visual proofs: https://www.youtube.com/watch?v=VYQVlVoWoPY https://www.youtube.com/watch?v=VYQVlVoWoPY (it includes a "proof" that pi equals exactly 4). In this case, as someone else pointed out below, this proof has unjustified assumptions in it (at least it assumes that b < a).

orangepanda 2y ago
It still holds if you’re not bothered by negative-area rectangles
- FartyMcFarter 2y ago
  But I definitely am bothered. A visual proof is supposed to appeal to our visual intuition - I don't know about you, but negative areas are not something that is visually intuitive to me.
  - Out_of_Characte 2y ago
    Negative area's can be the only practical solution. Imagine the worst case: a 10*10 room with a pillar in the middle. How would you calculate that?
    - FartyMcFarter 2y ago
      I don't see that as negative area. I see it as subtracting two areas. Both the room and the pillar have a positive area, none of them has negative lengths or areas. Kids learn subtraction before they learn negative numbers - once you learn negative numbers, you know that addition and subtraction are almost interchangeable, but this is not necessarily intuitive to begin with.
      - ses1984 2y ago
        Subtracting two areas is like adding two areas where one is negative…right?
        FartyMcFarter 2y ago
        Yes , but see my second paragraph. Just because something is true it doesn't mean it's intuitive.
        trimethylpurine 2y ago
        That's interesting. To me it seems intuitive. It's a real area that can be drawn like any other. The sign is an operator describing what function to visualize, not a property of the measured area. So thinking of it in that way eliminates any need for the term "negative area." But, intuition is subjective, so you may need to adjust the terminology to fit the visualization.
        FartyMcFarter 2y ago
        It is intuitive on some higher level, now that I know about signs, operators, negative numbers and all that. But when talking about visual information (i.e. "visual proof"), it isn't intuitive in that context. In addition to that, for all I know there could be some pitfalls involved with negative areas which I'm not aware of. Even if there aren't any pitfalls, this isn't immediately obvious to someone who isn't familiar with the concept of negative area. If I'm willing (or forced) to think in such abstract terms, I would much prefer an algebraic proof to this visual proof.
        ses1984 2y ago
        Your intuition is irrelevant if they’re mathematically equivalent.
        jacobolus 2y ago
        There are no serious pitfalls with oriented areas. Adding them to your arsenal of geometric proof tools will greatly simplify many proofs. Not having such a concept makes ancient geometry books much more complicated than they need to be, often requiring lots of detailed case analysis where the separate cases are essentially the same, just oriented opposite ways.
      - wruza 2y ago
        It’s intuitive, just in more dimensions. People have different ideas and abilities to imagine/think dimensions, but on top of that we rarely train them to do that. I think that build up to tensor fields should be in every school program. If you can’t think of a field, you’re mathematically disabled and too many basic ideas about real world are inaccessible to you. This limits the ability to vote on a set of topics and participate in non-local decisions that involve systemic understanding. Same for formal logic and statistics. Once familiarized with that, you can easily start thinking of nonlinearly signed areas, complex areas and areas simultaneously positive and negative by an attribute.
    - hansvm 2y ago
      The concept of negative area still feels like it'd get messy in a hurry. For a square pillar, the side lengths should be the same, suddenly giving you imaginary lengths just for the eventual area subtraction to work out. For a negative volume though, you need cubic roots of unity for the side lengths, throwing off your area calculations. Has anyone actually put together a system where the sort of concept you're describing is cohesive?
      - mtizim 2y ago
        You're in for a nice trip, the concept is called Geometric Algebra: https://youtu.be/60z_hpEAtD8?si=HHs_9m0IJ43nfI3S https://youtu.be/60z_hpEAtD8?si=HHs_9m0IJ43nfI3S (~50m video) TLDW: Yes, the concept is there, makes much more sense than a cross product (which is just an oriented area) and generalizes really nicely. Alternatively, read: https://en.m.wikipedia.org/wiki/Bivector https://en.m.wikipedia.org/wiki/Bivector
        hansvm 2y ago
        Gotcha; it's less that my complaint doesn't apply, but more that it isn't relevant (i.e., "squares" and "cubes" aren't especially interesting constructs which need to coexist nicely, and if you relax that constraint then directional geometry can be very interesting). Does that sound right?
  - mithametacs 2y ago
    It seems like an elegant addition to the understanding. Maybe if you viewed an animation where `a` starts larger than `b`, and steps till it's smaller. Then you could see where the negativity happens. And seeing is nice.
  - deleted 2y ago
    [deleted]
  - thaumasiotes 2y ago
    If you're bothered by the idea of negative-area rectangles, there's no need to justify the assumption that b < a, because that's the only way you can assign any meaning to either side of the equation.
    - FartyMcFarter 2y ago
      The equation isn't about rectangles at all, so I disagree. The rectangles only exist for the purposes of this proof. Any resulting complications are on the proof, not the reader's burden.
  - seanhunter 2y ago
    One easy place to get some intuition for signed areas is in the context of integrals. You know position is the integral of velocity right? So say you walk in a straight line from your starting point, then you keep slowing down until you come to a stop and walk backwards past your starting point. If you were to graph your velocity vs time at some point it would dip below the t axis because your velocity would be negative. Ok cool. If you integrate from the point you came to a stop and started walking backwards you’re calculating the area above the negative velocity curve(between it and the time axis). You’ll find it is a negative area. You know it has to be negative because you walked backwards past your starting point so it gets so negative that it cancels out all the positive area from when you were walking forwards.
- foxglacier 2y ago
  This sort of thing is more useful for teaching than being an actual proof. You teach the same concept in different ways and the students can form a more solid understanding of the underlying concept independent of symbols or shapes, or at least they may understand it one way if they don't get the other ways. You can do all sorts of factorizations the same way and handle negative areas by drawing them in a different color.
kleiba 2y ago
> at least it assumes that b < a But you can fairly assume that wlog.
- im3w1l 2y ago
  No
  - School-Cotton 2y ago
    Why not? Either b < a, b = a or b > a. If b = a both sides are zero and the result is trivial. If b > a just swap the names of a and b and multiply both sides by -1.
    - im3w1l 2y ago
      With this additional justification I accept it, yes. But (imo) now you are doing algebra. And then you might as well prove it just by distributing the RHS of the original expression.
      - dagss 2y ago
        This additional justification is so trivial it's just left out for brevity. In general a lot of statements are accepted even if not each and every detail is spelled out.
        davrosthedalek 2y ago
        Sorry, but no. First, "just switch a and b" is clearly wrong, and we had multiple people propose that (Without the adjustment of sign). So not that trivial, I would argue. At least not more trivial than the underlying equality one wants to prove in the first place.
        deleted 2y ago
        [deleted]
        im3w1l 2y ago
colejohnson66 2y ago
Yes, but that’s because it relies on infinite repetitions of the perimeter adjustment. Rearranging some boxes here does not involve infinity.
amelius 2y ago
Here's a similar one: https://en.wikipedia.org/wiki/Missing_square_puzzle https://en.wikipedia.org/wiki/Missing_square_puzzle
blablabla123 2y ago
I vaguely remember from an Analysis lecture when Mathematical rigor wasn't such a thing, the professor mentioned a famous Mathematician (maybe Euler?) who had a notebook with hundreds of proofs where many of the visual ones didn't turn out right. Definitely nice for intuition though since this is oftentimes lacking
TheRealPomax 2y ago
No, it assumes "one side is smaller than the other": the labels are arbitrary, so there iss no case where b is larger than a as we can just swap the labels and get a>b again. Even though you might think there are two cases that we need to look at because we're using two letters, there's actually only on case to demonstrate because we're demonstrating a property of the inequality. The only "but what if..." would be if a=b, which has no geometric proof, but also doesn't need one because "zero = zero times anything" is (by definition) true for fields.
- davrosthedalek 2y ago
  The problem is not symmetric between a and b (because a-b isn't). You cannot swap them. (And even if, this swapping is not part of the "proof", so in any case the proof is incomplete.)
  - xlii 2y ago
    IMO GP is about the labelling: Instead of a2-b2 use longer2-shorter2 (and on visual representation one side will always be longer, as per GPs explanation). Diagram doesn't show proof for "shorter2-longer2". I believe showing negative area would spark more controversy (imagine that negative area is painted orange, visual area would still be positive). Shorter2-longer2 gives you same absolute value with reverse signed, so it feels symmetric to me (can't remember what the formal definition is), i.e: 1. a2 - b2 = ( a-b)(a+b) # * -1 2. -(-a2 + b2) = -(-a+b)(a+b) 3. -(b2 - a2) = -( b-a)(b+a) # / -1 4. b2 - a2 = ( b-a)(b+a) Edit: https://imgur.com/a/olETWfr https://imgur.com/a/olETWfr - crude image with labels swap as it seems it's bringing some controversy.
  - TheRealPomax 2y ago
    It's fully symmetric. Basic arithmetic covers the following identities: 1. (a² - b²) = -(b² - a²), because of even powers 2. (a - b) = -(b - a) So the following two statements are the same statement: 3. (a² - b²) = (a - b)(a + b) 4. -(b² - a²) = -(b - a)(b + a) Let's assume this only holds for a>b (because we're content that the geometric proof shows that): 3a. (a² - b²) = (a - b)(a + b), a > b But we don't know if it also holds for b>a... after all, how would you show a negative areas? What does that even mean Turns out: it doesn't matter, the b>a relation reduces to the same formulae as the a>b relation, so the geometric proof covers both. To see why, some more elementary algebra: we can invert both sides of (4), provided we also invert the relation between a and b, so this: 4a. -(b² - a²) = -(b - a)(b + a), b > a Is the same as this: 4b. (b² - a²) = (b - a)(b + a), a > b, by inversion Of course, algebra doesn't care about which labels you use, as long as the identities and relations between them are preserved, so we can swap "a" for "b" and "b" for "a" in both the identity and relation in (4b) to get: 4c. (a² - b²) = (a - b)(a + b), b > a And we found a symmetry that we (maybe) didn't realize was there: 3a. (a² - b²) = (a - b)(a + b), a > b 4c. (a² - b²) = (a - b)(a + b), b > a Same formula, inverse relation. It turns out that it doesn't matter whether we start with a>b or b>a, they reduce to the same expression, thanks to those even powers, and a geometric proof for one is by definition a proof for the other.
petermcneeley 2y ago
Blasphemy! The value of pi is 3 https://religions.wiki/index.php/Biblical_value_of_pi https://religions.wiki/index.php/Biblical_value_of_pi
ComplexSystems 2y ago
There is no problem with the proof except the assumption that the value in the limit is the same as the value at infinity. If we simply define pi(n) as a function from N U {inf}, which gives the value that "pi" takes at the nth step of the process, and pi(inf) as the value that it actually takes for the circle, then we simply have a function where lim n->inf pi(n) ≠ pi(lim n->inf). For all finite n, it equals 4, and then at infinity it equals 3.1415... . There are ways to reformulate the above so that "infinity" isn't involved but this is the clearest way to think of it. It isn't much different than the Kronecker delta function delta(t), which is 1 at t=0 and 0 elsewhere. We have lim t->0 delta(t) ≠ delta(lim t->0 t).
saghm 2y ago
My geometry teacher in 9th grade was adamant that we couldn't assume lengths and angles in diagrams except where explicitly labeled. In particular, he said that we should never assume that a diagram is drawn to scale; if the diagram happened to depict a quadrilateral as a square, unless something stated that it was a square (or we had sufficient information to determine that it was), we should treat it as an unidentified quadrilateral and proceed only with the actual information provided or else he would "take more points off than the question was worth" on tests. On one of our quizzes, he did provide a diagram that looked like a kite (two pairs of adjacent sides each with the same length but different length than the other pair) but listed the angles in a way that could only work for a parallelogram that was not a kite, and he made good on his word to take off extra points for people who misidentified it as a kite.
- brookst 2y ago
  I like this teacher.
  - saghm 2y ago
    He was pretty great! He was super smart, super passionate, and had a great sense of humor; everything I could have asked for in a teacher.
mmooss 2y ago
> we need to be very careful when inspecting visual proofs This is not a visual proof but a nice visualization, like a written explanation that is not a proof. For an actual novel proof, nobody would imagine that they could eyeball it for a few minutes and conclude it was complete, correct, and consistent - maybe with the exception of professional mathematicians examining for simple proofs. You might eyeball it and follow its logic and not see any immediate flaws, but that's different.

orangepanda 2y ago

It still holds if you’re not bothered by negative-area rectangles

FartyMcFarter 2y ago
But I definitely am bothered. A visual proof is supposed to appeal to our visual intuition - I don't know about you, but negative areas are not something that is visually intuitive to me.
- Out_of_Characte 2y ago
  Negative area's can be the only practical solution. Imagine the worst case: a 10*10 room with a pillar in the middle. How would you calculate that?
  - FartyMcFarter 2y ago
    I don't see that as negative area. I see it as subtracting two areas. Both the room and the pillar have a positive area, none of them has negative lengths or areas. Kids learn subtraction before they learn negative numbers - once you learn negative numbers, you know that addition and subtraction are almost interchangeable, but this is not necessarily intuitive to begin with.
    - ses1984 2y ago
      Subtracting two areas is like adding two areas where one is negative…right?
      - FartyMcFarter 2y ago
        Yes , but see my second paragraph. Just because something is true it doesn't mean it's intuitive.
        trimethylpurine 2y ago
        That's interesting. To me it seems intuitive. It's a real area that can be drawn like any other. The sign is an operator describing what function to visualize, not a property of the measured area. So thinking of it in that way eliminates any need for the term "negative area." But, intuition is subjective, so you may need to adjust the terminology to fit the visualization.
        FartyMcFarter 2y ago
        It is intuitive on some higher level, now that I know about signs, operators, negative numbers and all that. But when talking about visual information (i.e. "visual proof"), it isn't intuitive in that context. In addition to that, for all I know there could be some pitfalls involved with negative areas which I'm not aware of. Even if there aren't any pitfalls, this isn't immediately obvious to someone who isn't familiar with the concept of negative area. If I'm willing (or forced) to think in such abstract terms, I would much prefer an algebraic proof to this visual proof.
        ses1984 2y ago
        Your intuition is irrelevant if they’re mathematically equivalent.
        jacobolus 2y ago
        There are no serious pitfalls with oriented areas. Adding them to your arsenal of geometric proof tools will greatly simplify many proofs. Not having such a concept makes ancient geometry books much more complicated than they need to be, often requiring lots of detailed case analysis where the separate cases are essentially the same, just oriented opposite ways.
    - wruza 2y ago
      It’s intuitive, just in more dimensions. People have different ideas and abilities to imagine/think dimensions, but on top of that we rarely train them to do that. I think that build up to tensor fields should be in every school program. If you can’t think of a field, you’re mathematically disabled and too many basic ideas about real world are inaccessible to you. This limits the ability to vote on a set of topics and participate in non-local decisions that involve systemic understanding. Same for formal logic and statistics. Once familiarized with that, you can easily start thinking of nonlinearly signed areas, complex areas and areas simultaneously positive and negative by an attribute.
  - hansvm 2y ago
    The concept of negative area still feels like it'd get messy in a hurry. For a square pillar, the side lengths should be the same, suddenly giving you imaginary lengths just for the eventual area subtraction to work out. For a negative volume though, you need cubic roots of unity for the side lengths, throwing off your area calculations. Has anyone actually put together a system where the sort of concept you're describing is cohesive?
    - mtizim 2y ago
      You're in for a nice trip, the concept is called Geometric Algebra: https://youtu.be/60z_hpEAtD8?si=HHs_9m0IJ43nfI3S https://youtu.be/60z_hpEAtD8?si=HHs_9m0IJ43nfI3S (~50m video) TLDW: Yes, the concept is there, makes much more sense than a cross product (which is just an oriented area) and generalizes really nicely. Alternatively, read: https://en.m.wikipedia.org/wiki/Bivector https://en.m.wikipedia.org/wiki/Bivector
      - hansvm 2y ago
        Gotcha; it's less that my complaint doesn't apply, but more that it isn't relevant (i.e., "squares" and "cubes" aren't especially interesting constructs which need to coexist nicely, and if you relax that constraint then directional geometry can be very interesting). Does that sound right?
- mithametacs 2y ago
  It seems like an elegant addition to the understanding. Maybe if you viewed an animation where `a` starts larger than `b`, and steps till it's smaller. Then you could see where the negativity happens. And seeing is nice.
- deleted 2y ago
  [deleted]
- thaumasiotes 2y ago
  If you're bothered by the idea of negative-area rectangles, there's no need to justify the assumption that b < a, because that's the only way you can assign any meaning to either side of the equation.
  - FartyMcFarter 2y ago
    The equation isn't about rectangles at all, so I disagree. The rectangles only exist for the purposes of this proof. Any resulting complications are on the proof, not the reader's burden.
- seanhunter 2y ago
  One easy place to get some intuition for signed areas is in the context of integrals. You know position is the integral of velocity right? So say you walk in a straight line from your starting point, then you keep slowing down until you come to a stop and walk backwards past your starting point. If you were to graph your velocity vs time at some point it would dip below the t axis because your velocity would be negative. Ok cool. If you integrate from the point you came to a stop and started walking backwards you’re calculating the area above the negative velocity curve(between it and the time axis). You’ll find it is a negative area. You know it has to be negative because you walked backwards past your starting point so it gets so negative that it cancels out all the positive area from when you were walking forwards.
foxglacier 2y ago
This sort of thing is more useful for teaching than being an actual proof. You teach the same concept in different ways and the students can form a more solid understanding of the underlying concept independent of symbols or shapes, or at least they may understand it one way if they don't get the other ways. You can do all sorts of factorizations the same way and handle negative areas by drawing them in a different color.

kleiba 2y ago

> at least it assumes that b < a But you can fairly assume that wlog.

im3w1l 2y ago
No
- School-Cotton 2y ago
  Why not? Either b < a, b = a or b > a. If b = a both sides are zero and the result is trivial. If b > a just swap the names of a and b and multiply both sides by -1.
  - im3w1l 2y ago
    With this additional justification I accept it, yes. But (imo) now you are doing algebra. And then you might as well prove it just by distributing the RHS of the original expression.
    - dagss 2y ago
      This additional justification is so trivial it's just left out for brevity. In general a lot of statements are accepted even if not each and every detail is spelled out.
      - davrosthedalek 2y ago
        Sorry, but no. First, "just switch a and b" is clearly wrong, and we had multiple people propose that (Without the adjustment of sign). So not that trivial, I would argue. At least not more trivial than the underlying equality one wants to prove in the first place.
      - deleted 2y ago
        [deleted]
      - im3w1l 2y ago

colejohnson66 2y ago

Yes, but that’s because it relies on infinite repetitions of the perimeter adjustment. Rearranging some boxes here does not involve infinity.

amelius 2y ago

Here's a similar one: https://en.wikipedia.org/wiki/Missing_square_puzzle https://en.wikipedia.org/wiki/Missing_square_puzzle

blablabla123 2y ago

I vaguely remember from an Analysis lecture when Mathematical rigor wasn't such a thing, the professor mentioned a famous Mathematician (maybe Euler?) who had a notebook with hundreds of proofs where many of the visual ones didn't turn out right. Definitely nice for intuition though since this is oftentimes lacking

TheRealPomax 2y ago

No, it assumes "one side is smaller than the other": the labels are arbitrary, so there iss no case where b is larger than a as we can just swap the labels and get a>b again. Even though you might think there are two cases that we need to look at because we're using two letters, there's actually only on case to demonstrate because we're demonstrating a property of the inequality. The only "but what if..." would be if a=b, which has no geometric proof, but also doesn't need one because "zero = zero times anything" is (by definition) true for fields.

davrosthedalek 2y ago
The problem is not symmetric between a and b (because a-b isn't). You cannot swap them. (And even if, this swapping is not part of the "proof", so in any case the proof is incomplete.)
- xlii 2y ago
  IMO GP is about the labelling: Instead of a2-b2 use longer2-shorter2 (and on visual representation one side will always be longer, as per GPs explanation). Diagram doesn't show proof for "shorter2-longer2". I believe showing negative area would spark more controversy (imagine that negative area is painted orange, visual area would still be positive). Shorter2-longer2 gives you same absolute value with reverse signed, so it feels symmetric to me (can't remember what the formal definition is), i.e: 1. a2 - b2 = ( a-b)(a+b) # * -1 2. -(-a2 + b2) = -(-a+b)(a+b) 3. -(b2 - a2) = -( b-a)(b+a) # / -1 4. b2 - a2 = ( b-a)(b+a) Edit: https://imgur.com/a/olETWfr https://imgur.com/a/olETWfr - crude image with labels swap as it seems it's bringing some controversy.
- TheRealPomax 2y ago
  It's fully symmetric. Basic arithmetic covers the following identities: 1. (a² - b²) = -(b² - a²), because of even powers 2. (a - b) = -(b - a) So the following two statements are the same statement: 3. (a² - b²) = (a - b)(a + b) 4. -(b² - a²) = -(b - a)(b + a) Let's assume this only holds for a>b (because we're content that the geometric proof shows that): 3a. (a² - b²) = (a - b)(a + b), a > b But we don't know if it also holds for b>a... after all, how would you show a negative areas? What does that even mean Turns out: it doesn't matter, the b>a relation reduces to the same formulae as the a>b relation, so the geometric proof covers both. To see why, some more elementary algebra: we can invert both sides of (4), provided we also invert the relation between a and b, so this: 4a. -(b² - a²) = -(b - a)(b + a), b > a Is the same as this: 4b. (b² - a²) = (b - a)(b + a), a > b, by inversion Of course, algebra doesn't care about which labels you use, as long as the identities and relations between them are preserved, so we can swap "a" for "b" and "b" for "a" in both the identity and relation in (4b) to get: 4c. (a² - b²) = (a - b)(a + b), b > a And we found a symmetry that we (maybe) didn't realize was there: 3a. (a² - b²) = (a - b)(a + b), a > b 4c. (a² - b²) = (a - b)(a + b), b > a Same formula, inverse relation. It turns out that it doesn't matter whether we start with a>b or b>a, they reduce to the same expression, thanks to those even powers, and a geometric proof for one is by definition a proof for the other.

petermcneeley 2y ago

Blasphemy! The value of pi is 3 https://religions.wiki/index.php/Biblical_value_of_pi https://religions.wiki/index.php/Biblical_value_of_pi

ComplexSystems 2y ago

There is no problem with the proof except the assumption that the value in the limit is the same as the value at infinity. If we simply define pi(n) as a function from N U {inf}, which gives the value that "pi" takes at the nth step of the process, and pi(inf) as the value that it actually takes for the circle, then we simply have a function where lim n->inf pi(n) ≠ pi(lim n->inf). For all finite n, it equals 4, and then at infinity it equals 3.1415... . There are ways to reformulate the above so that "infinity" isn't involved but this is the clearest way to think of it. It isn't much different than the Kronecker delta function delta(t), which is 1 at t=0 and 0 elsewhere. We have lim t->0 delta(t) ≠ delta(lim t->0 t).

saghm 2y ago

My geometry teacher in 9th grade was adamant that we couldn't assume lengths and angles in diagrams except where explicitly labeled. In particular, he said that we should never assume that a diagram is drawn to scale; if the diagram happened to depict a quadrilateral as a square, unless something stated that it was a square (or we had sufficient information to determine that it was), we should treat it as an unidentified quadrilateral and proceed only with the actual information provided or else he would "take more points off than the question was worth" on tests. On one of our quizzes, he did provide a diagram that looked like a kite (two pairs of adjacent sides each with the same length but different length than the other pair) but listed the angles in a way that could only work for a parallelogram that was not a kite, and he made good on his word to take off extra points for people who misidentified it as a kite.

brookst 2y ago
I like this teacher.
- saghm 2y ago
  He was pretty great! He was super smart, super passionate, and had a great sense of humor; everything I could have asked for in a teacher.

mmooss 2y ago

> we need to be very careful when inspecting visual proofs This is not a visual proof but a nice visualization, like a written explanation that is not a proof. For an actual novel proof, nobody would imagine that they could eyeball it for a few minutes and conclude it was complete, correct, and consistent - maybe with the exception of professional mathematicians examining for simple proofs. You might eyeball it and follow its logic and not see any immediate flaws, but that's different.