alfadog
Final Approach
I can see how it might. I had no idea that you could "tune" a square like that. I like that a lot. I do know the analogous way to check a level.
Pythagoras' Theorem
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I can see how it might. I had no idea that you could "tune" a square like that. I like that a lot. I do know the analogous way to check a level.
Cap'n Jack
Final Approach
Chicken, or egg?
The problem won't be resolved right away, since most STEM occupations need a BS degree, minimum, and the interest needs to be developed in high school. A lot of students won't be bothered enough, or interested in the classes, to do STEM. I have the impression parents don't show STEM as much of an option to their children. And finally, speaking about chemistry, a lot of those jobs went overseas. That's why, this year, I went to Germany, and will go to Poland, Japan, and possibly China.
alfadog
Final Approach
But yours is not a proof. It is a representation of the formula. You go into it knowing the lengths of all sides.
SoonerAviator
Final Approach
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THAT IS MY POINT! The image I provided was an application (or representation) of the theorem which shows how the area of a square attached to each side of a right triangle would have areas consistent with the theorem formula. What I don't understand, is how the teacher's example is a "proof" of that theorem as opposed to another example of an application of the theorem. Hence why I said I feel like I'm missing the connection.
Matthew
Touchdown! Greaser!
I think every US President should find a proof for that theorem.
https://www.maa.org/press/periodica...-a-garfields-proof-of-the-pythagorean-theorem
https://www.maa.org/press/periodica...-a-garfields-proof-of-the-pythagorean-theorem
I have no problem with the way the information was presented, nor the impact that a great teacher has on maintaining the interest of his students (especially in STEM fields). My point was that I feel like I was missing some "connection" that was to be made with the example. I was hoping someone might expound on why the example of relating the area of the empty space and the area of the right triangles was critical to the theorem itself, or if it was just an illustration on how to use a2xb2=c2 to solve a slightly more difficult scenario.
This was the image which I was more familiar with: which shows that a square with 4-equilateral sides of "a", and a square with 4-equilateral sides of "b", then the sum of those equals a square with 4-equilateral sides of "c".
As I noted above, the proof that a^2+b^2 = c^2 was skipped in the original video. He got real close, but it was skipped. The missing part the proof from the video in the original post is illustrated by the link in this post:
This latter graphic shows how a^2 + b^2 must = c^2. The white area remains the same after the triangles are slid over as it was before you moved the triangles. You know that the area is the same because the white area, both before and after you slide the triangles, is simply the difference between the area of the big outer square minus the area of the four triangles. Because the white area is the same before and after you slide the triangles, you can now prove the theorem.
Before you slide the triangles, the white area is a square that is has an area of c^2. Then, after you slide the triangles over, the white area is now two squares, one with an area of a^2, and the other is b^2. Because all you have done is displaced the white area, but not changed the total amount of the white area, the area of the two white squares after sliding the triangles over must equal the area of the first white square before you slid the triangles over. Therefore a^2 plus b^2 must equal c^2.
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alfadog
Final Approach
1. A square is constructed using four triangles of sides a, b, c (c being the hypotenuse) with length of each side of the square being a+b.
2. This "missing area" inside is clearly a square of side length c and area c^2.
3. Rearrange the four triangle to make another square of the same side being a+b.
4. This time the missing areas are clearly two squares, one of side length a and the other of side length b.
5. The missing area then is a^2 + b^2.
6. Omitting a couple obvious steps then clearly a^2 + b^2 = c^2
7. Profit!!?!!
Matthew
Touchdown! Greaser!
Kinda fun reviewing all this after so many years.
I have trouble viewing the video for some reason, but I can make out the proof he's showing.
Another way to look at it is with the 4 triangles surrounding the square in the center.
Remember that the area of a triangle is 1/2 base * height, or 1/2 AB. The area of the center square is C^2.
So the area of a large square with those 4 triangles is both:
1) (A + B)^2, where the outer edges of the 4 triangles are used, or
2) 4x ( 1/2 AB) + C^2, where the areas is the center square plus the area of all 4 triangles
Since they are both the same, then (A + B)^2 = 4(1/2 AB) + C^2
and simplifying -> (A + B)^2 = A^2 + 2 AB + B^2
and -> 4(1/2 AB) + C^2 = 2AB + C^2
so A^2 + 2AB + B^2 = 2AB + C^2
after subtracting 2AB from both sides, you end up with A^2 + B^2 = C^2
I have trouble viewing the video for some reason, but I can make out the proof he's showing.
Another way to look at it is with the 4 triangles surrounding the square in the center.
Remember that the area of a triangle is 1/2 base * height, or 1/2 AB. The area of the center square is C^2.
So the area of a large square with those 4 triangles is both:
1) (A + B)^2, where the outer edges of the 4 triangles are used, or
2) 4x ( 1/2 AB) + C^2, where the areas is the center square plus the area of all 4 triangles
Since they are both the same, then (A + B)^2 = 4(1/2 AB) + C^2
and simplifying -> (A + B)^2 = A^2 + 2 AB + B^2
and -> 4(1/2 AB) + C^2 = 2AB + C^2
so A^2 + 2AB + B^2 = 2AB + C^2
after subtracting 2AB from both sides, you end up with A^2 + B^2 = C^2
Daniel L
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I can relate with the teacher because i majored in math. I can relate with the enthusiasm because there's something exciting about the axioms of mathematical expression.
His passion is what made me work as a tutor for the better part of a decade. Though, I didn't teach kids because I learned quickly that it takes actual training to do that. I focused on college students in calculus.
I can bet you a dollar bill, crisp fresh out of the mint, part of his passion is seeing the glimmer in a student's eyes when they finally get it. There's a smile in a person's face that you only see when they get math. Priceless.
I can't describe the satisfaction felt after helping a student get it. Like, they truly understand it. It kicks in and... wow, it's my fondest memories as a tutor.
Sent from my SM-T350 using Tapatalk
His passion is what made me work as a tutor for the better part of a decade. Though, I didn't teach kids because I learned quickly that it takes actual training to do that. I focused on college students in calculus.
I can bet you a dollar bill, crisp fresh out of the mint, part of his passion is seeing the glimmer in a student's eyes when they finally get it. There's a smile in a person's face that you only see when they get math. Priceless.
I can't describe the satisfaction felt after helping a student get it. Like, they truly understand it. It kicks in and... wow, it's my fondest memories as a tutor.
Sent from my SM-T350 using Tapatalk
Daniel L
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Daniel
If anyone has time, look up Eratosthenes and how he used basic geometry to discover that Earth is spherical and how he derived the circumference of earth within an error of a few percent.
All done in the Great library of Alexandria (where he worked).
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All done in the Great library of Alexandria (where he worked).
Sent from my SM-T350 using Tapatalk
EdFred
Taxi to Parking
I have no problem with the way the information was presented, nor the impact that a great teacher has on maintaining the interest of his students (especially in STEM fields). My point was that I feel like I was missing some "connection" that was to be made with the example. I was hoping someone might expound on why the example of relating the area of the empty space and the area of the right triangles was critical to the theorem itself, or if it was just an illustration on how to use a2xb2=c2 to solve a slightly more difficult scenario.
This was the image which I was more familiar with: which shows that a square with 4-equilateral sides of "a", and a square with 4-equilateral sides of "b", then the sum of those equals a square with 4-equilateral sides of "c".
Fun fact: It doesn't have to be the side of a square.
Try it with equilateral hexagons where a side of the triangle is a side of a hexagon, or half circles, or...
EdFred
Taxi to Parking
It is absolutely a proof. Graphical representations can be proofs. Do it with unknown sides, then.
alfadog
Final Approach
Your graphic has no meaning with unknown sides. That is why it is not a proof. The fellow in the video does a graphical representation as a proof.
EdFred
Taxi to Parking
It absolutely does. I can't believe I even have to explain this, but I will anyway, because, damn, people be dumb. You can even do this with a compass and straight edge like the Greeks did - oh, and a get balance scale. Take a piece of sheet metal (or any consistent and uniform material) and on it construct a right triangle with arbitrary length legs. Does not matter the length of the legs. And since you are only using a straight edge and compass you don't know the lengths anyway. After constructing the triangle, construct 3 squares using each of the sides of the triangle like I show in my drawing using only the compass and straight edge. You do know how to do this, because you're so smart. Cut the squares from the material. Take the two smaller squares and place them on one side of the balance scale. Take the larger square and place it on the other side. Holy. ****ing. ****. The G.D. Scale balances! How is that since I don't know what the **** I'm talking about?
Just because I labeled the sides in the example to help people out doesn't negate the proof.
bflynn
Final Approach
I vaguely remember doing a proof of the PT somewhere. The one I was taught was to bisect the triangle to make two new smaller right triangles, then use ratios to construct a formula which reduced to the PT formula. I can barely remember doing 40 years or so ago, but the googles tell me now it’s called the Bhaskara proof.
The visual proof is nice, but it wouldn’t have been acceptable in any of the proof courses I took.
The visual proof is nice, but it wouldn’t have been acceptable in any of the proof courses I took.
Matthew
Touchdown! Greaser!
Matthew
Touchdown! Greaser!
I finally was able to view the video. I also think there’s a step missing at the end. At least for me there was a jump to the final a^2+b^2=c^2
alfadog
Final Approach
You continue to confuse a demonstration with a proof. I don't think that's going to change by virtue of anything I say here.
bflynn
Final Approach
He did jump at the end and got a little fuzzy.
Square 2ab = green triangles + blue square
Square 2ab = green triangles + yellow + red
Green triangles + yellow + red = green triangles + blue
Take away the green triangles on both sides.
Therefore yellow + red = blue.
There is a mathematical proof in the method, it just requires calculating areas.
Square 2ab = green triangles + blue square
Square 2ab = green triangles + yellow + red
Green triangles + yellow + red = green triangles + blue
Take away the green triangles on both sides.
Therefore yellow + red = blue.
There is a mathematical proof in the method, it just requires calculating areas.
EdFred
Taxi to Parking
There are over 300 different proofs of the theorem and the one I showed is one of the accepted ones.