squaring the circle is easy. the cartesian product of two circles is a clifford torus.
@kiti_cat52410 сағат бұрын
5:15 I thought he would say a googol (10¹⁰⁰)
@kiti_cat52411 сағат бұрын
4:08 i get it
@user-iy6dt4xp5o3 ай бұрын
5:20 A STRANGE GAME. THE ONLY WINNING MOVE IS NOT TO PLAY.
@01juniorpen3 ай бұрын
7:27 i dont even qualify as an american
@captaincy73 ай бұрын
as a geometry failure, the ASS theorem is my favorite.
@xavier93
3 ай бұрын
Also as a geometry failure, I got my ASS beaten after failing the geometry test for using the ASS theorem to prove two triangles similar 😂😂
@o_s-243 ай бұрын
9:20 ah yes, the great mathematician, Dess Carts
@NerdWithLaptop3 ай бұрын
5:28 YO
@tianyuema47973 ай бұрын
5:29 If A is correct (25%), then the correct answer is not 25%, contradicting the statement. If B is correct (0%), then the correct answer is not 0%, contradicting the statement. If C is correct (50%), then the correct answer is not 50%, contradicting the statement. If D is correct (25%), then the correct answer is not 25%, contradicting the statement.
@catloverplayz3268
3 ай бұрын
The Who Wants To Be a Millionaire paradox a classic
@rodrigoqteixeira3 ай бұрын
6:36 We know you're secret, we use you're secret (no, actually I uninstalled it)
@Frymeme13 ай бұрын
2:47 bro 😅
@newwaveinfantry83623 ай бұрын
11:45 - Okay, I will change your mind, because you are blazenly wrong. There are 3 real algebras up to ring isomorphism: 1) R[x]/(x^2) i.e. R^2, the ring of the real plane 2) R[x]/(x^2-1) i.e. R[j] where j^2=1 and j is different from 1,-1, the ring of the split real numbers. 3) R[x]/(x^2+1) i.e. C, the field of complex numbers. You can replace (x^2) with the ideal generated by any perfect square of a linear function ((mx+c)^2) as long as m=/=0 and what you get is isomorphic. You can replace (x^2-1) with the ideal generated by any product of linears with different roots and you get an isomorphism. You can replace (x^2+1) by the ideal generated by any irreducible quadratic in R[x] and you still get something isomorphic. (x^2) = (x)^2 and (x^2-1) = (x+1)(x-1), so the ideals are not prime, thus R^2,R[j] are not even integral domains, while (x^2+1) is not only prime, but maximal, so C is a field, and an algebraically closed one at that. Thus C and R^2 are not even close to being isomorphic as rings. R^2 has uncountably many zero divisors, C has zero. They have different multiplicative algebras. R^2 has inner products (a,b)*(c,d)=(ac,bd) which is never invertible if a,b,c or d is 0 and C has inner product (a,b)*(c,d)=(ac-bd,bc+ad) which is always invertible unless one is (0,0). Every polynomials with coefficients in C factors into linears and can be turned into a monic polynomial, but not every polynomials with coefficients in R^2 factors into linears and you can't always make it monic by multiplication. In fact, since R^2 is not a UFD, neither is R^2[x], so factorization of polynomials is not unique. R^2 isn't even an integral domain, as stated before, so neither is R^2[x], so you can get polynomials that literally factor into higher order polynomials and in different ways at that, which is pain. Eg. [(a,0)x^6+(b,0)]*[(0,c)x^17-(d,0)] = (-ad,0)x^6+(-bd,0). So, in conclusion, to a topologist, (maybe) measure theorist or group theorist that's just looking additive groups, they are the same, but to a ring theorist or someone studying vector spaces, they are very different.
@trendygaming7953 ай бұрын
0:11 I just realized
@kexcz82763 ай бұрын
Lol. That example on the title screen is wrong yet in this specific case it is true 😅😂
@JaiveerSingh-zh2rc3 ай бұрын
Oh wtf percentages are reversible lol true
@c.jishnu3783 ай бұрын
I do the same things with my school notebook page numbers as you do with your titles, if It's 65 i write 2⁶+1. If it's 9 i wrote (2+1)(2²-2x1+1⅖)
@WalacaVencano3 ай бұрын
2:15 Hum, actually, there is no solution: x+2=x-2 is equivalent to x+2-x = x-2-x equivalent to 2=-2 which is wrong. 10:53 0/0 = X with X = the number you wants. Asks me why do I think that if you want to know.
@that_guy12113 ай бұрын
oh man, i'm too dumb for this, lol
@SnowghostFilms3 ай бұрын
Don’t worry, the thumbnail is still correct. Just used a method that wouldn’t work over 99% of the time
@RobWithfold3 ай бұрын
day 1 waiting for e^i math memes
@victoriamacarthur89063 ай бұрын
2:14 isn’t correct, x doesn’t exist, it’s a false equation
@Kryptonite0993 ай бұрын
3:14 💀💀
@gachanimestudios8348
3 ай бұрын
3:14... 3.14 ≈ π... Holy shit.
@Kryptonite099
3 ай бұрын
🧐🧐@@gachanimestudios8348
@ianweckhorst32003 ай бұрын
2:37 it’s actually much easier to prove they’re similar, just use x, y and the right angle, no need to use Pythagorean theorem Edit: ...3? Three triangles? Similar to the original one? You mean the pink one doesn’t count? What?!?
@rudramukherjee1663
3 ай бұрын
The third triangle is the other two triangles combined together (x and y must equal 90 degrees due to the pink triangle), you have an x, y, and a right angle
@Tekdruid3 ай бұрын
11:58 Yeah, _escape_ velocity.
@ianweckhorst32003 ай бұрын
1:27 htf did you get that answer
@sivaprasath3638
3 ай бұрын
just take ln on both sides and substitute 1 = e^(2πn) , there you got it.
@brycenbg93723 ай бұрын
lol 2:11 😂 😂
@catloverplayz3268
3 ай бұрын
Every math problem has a solution if you break a couple rules
@brycenbg9372
3 ай бұрын
bro ....@@catloverplayz3268 is not about breaking any math rule but suppose this equation has no solution 😂 😂 😂 😂
@ethannguyen27543 ай бұрын
Idk man finding the roots of (x-1)^100 seems pretty easy to me
Пікірлер: 60
Can't wait for Math Memes 2(2³)
@catloverplayz3268
3 ай бұрын
I can't wait for Math memes 3x+1 X=5
@Impatient1v3r50n
3 ай бұрын
I can't wait for Math Memes sqrt256
@kylezdancewicz7346
3 ай бұрын
@@catloverplayz3268you mean 7/4x-10=18
@catloverplayz3268
3 ай бұрын
@kylezdancewicz7346 3/4x+88=100
@catloverplayz3268
3 ай бұрын
@kylezdancewicz7346 3/4x+88=100
Can't wait for math memes 5²-3²
squaring the circle is easy. the cartesian product of two circles is a clifford torus.
5:15 I thought he would say a googol (10¹⁰⁰)
4:08 i get it
5:20 A STRANGE GAME. THE ONLY WINNING MOVE IS NOT TO PLAY.
7:27 i dont even qualify as an american
as a geometry failure, the ASS theorem is my favorite.
@xavier93
3 ай бұрын
Also as a geometry failure, I got my ASS beaten after failing the geometry test for using the ASS theorem to prove two triangles similar 😂😂
9:20 ah yes, the great mathematician, Dess Carts
5:28 YO
5:29 If A is correct (25%), then the correct answer is not 25%, contradicting the statement. If B is correct (0%), then the correct answer is not 0%, contradicting the statement. If C is correct (50%), then the correct answer is not 50%, contradicting the statement. If D is correct (25%), then the correct answer is not 25%, contradicting the statement.
@catloverplayz3268
3 ай бұрын
The Who Wants To Be a Millionaire paradox a classic
6:36 We know you're secret, we use you're secret (no, actually I uninstalled it)
2:47 bro 😅
11:45 - Okay, I will change your mind, because you are blazenly wrong. There are 3 real algebras up to ring isomorphism: 1) R[x]/(x^2) i.e. R^2, the ring of the real plane 2) R[x]/(x^2-1) i.e. R[j] where j^2=1 and j is different from 1,-1, the ring of the split real numbers. 3) R[x]/(x^2+1) i.e. C, the field of complex numbers. You can replace (x^2) with the ideal generated by any perfect square of a linear function ((mx+c)^2) as long as m=/=0 and what you get is isomorphic. You can replace (x^2-1) with the ideal generated by any product of linears with different roots and you get an isomorphism. You can replace (x^2+1) by the ideal generated by any irreducible quadratic in R[x] and you still get something isomorphic. (x^2) = (x)^2 and (x^2-1) = (x+1)(x-1), so the ideals are not prime, thus R^2,R[j] are not even integral domains, while (x^2+1) is not only prime, but maximal, so C is a field, and an algebraically closed one at that. Thus C and R^2 are not even close to being isomorphic as rings. R^2 has uncountably many zero divisors, C has zero. They have different multiplicative algebras. R^2 has inner products (a,b)*(c,d)=(ac,bd) which is never invertible if a,b,c or d is 0 and C has inner product (a,b)*(c,d)=(ac-bd,bc+ad) which is always invertible unless one is (0,0). Every polynomials with coefficients in C factors into linears and can be turned into a monic polynomial, but not every polynomials with coefficients in R^2 factors into linears and you can't always make it monic by multiplication. In fact, since R^2 is not a UFD, neither is R^2[x], so factorization of polynomials is not unique. R^2 isn't even an integral domain, as stated before, so neither is R^2[x], so you can get polynomials that literally factor into higher order polynomials and in different ways at that, which is pain. Eg. [(a,0)x^6+(b,0)]*[(0,c)x^17-(d,0)] = (-ad,0)x^6+(-bd,0). So, in conclusion, to a topologist, (maybe) measure theorist or group theorist that's just looking additive groups, they are the same, but to a ring theorist or someone studying vector spaces, they are very different.
0:11 I just realized
Lol. That example on the title screen is wrong yet in this specific case it is true 😅😂
Oh wtf percentages are reversible lol true
I do the same things with my school notebook page numbers as you do with your titles, if It's 65 i write 2⁶+1. If it's 9 i wrote (2+1)(2²-2x1+1⅖)
2:15 Hum, actually, there is no solution: x+2=x-2 is equivalent to x+2-x = x-2-x equivalent to 2=-2 which is wrong. 10:53 0/0 = X with X = the number you wants. Asks me why do I think that if you want to know.
oh man, i'm too dumb for this, lol
Don’t worry, the thumbnail is still correct. Just used a method that wouldn’t work over 99% of the time
day 1 waiting for e^i math memes
2:14 isn’t correct, x doesn’t exist, it’s a false equation
3:14 💀💀
@gachanimestudios8348
3 ай бұрын
3:14... 3.14 ≈ π... Holy shit.
@Kryptonite099
3 ай бұрын
🧐🧐@@gachanimestudios8348
2:37 it’s actually much easier to prove they’re similar, just use x, y and the right angle, no need to use Pythagorean theorem Edit: ...3? Three triangles? Similar to the original one? You mean the pink one doesn’t count? What?!?
@rudramukherjee1663
3 ай бұрын
The third triangle is the other two triangles combined together (x and y must equal 90 degrees due to the pink triangle), you have an x, y, and a right angle
11:58 Yeah, _escape_ velocity.
1:27 htf did you get that answer
@sivaprasath3638
3 ай бұрын
just take ln on both sides and substitute 1 = e^(2πn) , there you got it.
lol 2:11 😂 😂
@catloverplayz3268
3 ай бұрын
Every math problem has a solution if you break a couple rules
@brycenbg9372
3 ай бұрын
bro ....@@catloverplayz3268 is not about breaking any math rule but suppose this equation has no solution 😂 😂 😂 😂
Idk man finding the roots of (x-1)^100 seems pretty easy to me
@NamanNahata-zx1xz
3 ай бұрын
So does infinite mclauren series of sin(x)
🤯=3^1/2
@fiprandom3783
3 күн бұрын
±
i dont understand