Prove that BPRP loves math 1: "I feel like it is a logic circle, but this is what happens when you do a contradiction proof. It's a lot of fun actually."
@@Abion47 No, the multiplicative identity of 0 would state that the product = 0 means either a = 0 or b = 0, but by assumption neither a nor b are zero. It disproves the identity by a fundamental theory of numbers, and it also proves the exception of a = 0 or b = 0. Note, the or is not exclusive, which means both statements could be true.
👋👋👋 what I like best about this video is, father is letting his son understand that we should question and point out something when we know that something is wrong. 👏 Bravo It's unfortunate that the teacher and principal didn't not understand both lessons here.
Just thinking about slicing a pie also works: I can have any number of slices by cutting the pie. The more I cut, the more I have slices. But I cannot cut less than not slicing at all (which is dividing by 1). So (assuming I divide by naturals) I cannot divide by less than 1.
@Romczy -1 isnt part of the natural numbers. The op explicitly said natural numbers. Not real numbers Not complex numbers Not imaginary numbers Not irrational numbers Not transcendental numbers. Unless the OP made a mistake, using the wrong terminology.
Pascal's triangle is a great tool for binomial expansion. If you ever come across anything like (a+b)^5 you can use p's triangle to quickly find the identity of that binomial.
Il faut préciser l'ensemble de départ, sinon il faut écrire (A+B)^2 = A^2+AB+BA+B^2 (non commutativité dans l'ensemble des matrices carrées). De plus (A+B)^2 = A^2+B^2 est vrai dans Z/pZ avec p premier. Merci pour vos partage !...
Really couldn't believe that someone would actually think that was correct. it is equal to (a+b) x (a+b), which is, using FOIL, A^2 + 2ab + B^2. Guess root values and quadratics aren't being taught in elementary schools anymore.
We could also prove by contrapositive, "(A+B)^2 = A^2+B^2 implies A=0 or B=0". At the end we just need to write "2AB = 0" => "AB = 0" => "A = 0 or B = 0"
The original post has one symbol he/she didn't put resulting in the misunderstanding. They didnt put "?" On top of "=". The "?" On top of "=" denotes that the equation is asking if the left side of the equation is equal to the othe side of the equation. And because of that @bprpmathbasics thought that they are solving the equation and not proving the equation.😊
A better proof for this would be by dividing 0 by 2 in the last step leaving you with AB=0. The only way that can be true is if A or B is 0. This is not a proof by contradiction, but it is less ambiguous than dividing an undefined variable by 0.
Now I know why I didn't understand math. I was timid. They never explained this contradiction circle. It made no sense and do nothing else made sense. So I gave up. Used to be extremely good😢 at it. Only if I could have afforded to watch videos like this on the internet😢 Gosh I was soo timid when I was wrong what was wrong was that I did not appreciate the contradiction that mattered and changed the meaning of the equations and inequalities
9001x = 0. Now lets divide both sides by x, we get: 9001x/x = 0/x --> 9001 = 0, which proves once and for all that zero is over nine thousannnnnnnnnnd. What? Nine thousand? Theres no way that can be right.
We have to define why this is a math problem to begin with to get the answer... for instance 8 apples divided between 2 people equally ends up at 4 apples apiece, 2 apples divided amongst 8 people winds up in a quarter of 1 apple, and if you have 1 apple and you divide it amongst nobody, you have 1 apple, that you didnt share. And if you have 0 apples divided amongst 1 person, guess what that person has?? 0 apples. 😂 its basic math. Why are people trying to make the basics seem so deep. Lets define the "problem" (actual real world correlated issue) to resolve the math equation appropriately.
If you have one apple you aren't sharing, you still have one person to that apple. You are dividing by 1. Dividing by zero is saying "there are 4 apples and no people in that room. How many apples does each person get" that makes no sense, is undefined.
@LeifNelandDk did you not gather the logic I was trying to convey, did you even read what I wrote? It has to apply to real world application for it to be a valid equation, is my point. And back to the 1 apple divided by 1 or more people is still just 1 apple, or a fraction thereof. What did you mean by your post? I must be missing something?
@LeifNelandDk maybe I would word the math problem as "you have 1 apple and divide it amongst 0 people, how many apples are there? Answer being 1. But if you say you have 1 apple and divide it amongst 0 people, how many apples does each person get.... would be a d***** question,.. Wouldn't it? Back to my real world application and you making up a "real world scenerio" of dividing amongst 0 people and then discussing how many of whatever of each person get out of it, when there's no people there, sounds like an unpractical, not a "REAL" Scenerio.
the actual formula for A square + B square is [a^2 + b^2 = (a + b)^2 - 2ab] or [a^2 + b^2 = (a - b)^2 + 2ab] which is true regardless of whether a or b equals to or is not equal to 0, which is basic math and is taught in primary (or in some places secondary) school. However, what is also basic math is (A+B)2=A2+B2 or to write it as a multiplier to be more obvious (A+B)x2 = Ax2 + Bx2, just basic math which some person may have mistaken it for a square instead of a multiplier "on Reddit", but I like the way you break it down for people who are not really smart (see what I did there) Subscribed, I may show your videos to my kids someday 😀
No that is not right. Also its not called a formula, its called an identity. (a+b)^2=(a+b)(a+b)=a^2+2ab+b^2 You can use pascal's triangle to quickly find the identity of any binomial expansion. Because in higher maths its easier to use p's triangle than work out (a+b)^6 by hand. You will need to know this for calc 1.
IDK about the rest of the US, but in Texas, division is not introduced with decimals. It’s introduced with the idea of remainders. You never learn the full problem. Decimals come in late in the school year for 3rd graders and is worked in 4th and 5th grade by along with fractions. But fractions is given the even greater attention. You do not even learn how to work the entire math problem when it’s first introduced because it’s believed to be too complex. I think the belief is that math at the elementary levels are suppose to have answers, not undefined answers. I think they believe it to be like saying “I don’t know.” But that’s not actually the case.
Actually a good teacher, teaching them an important lesson they will need for their entire future. You WILL be forced to believe and accept things that are obviously false. That is the way of things going forward into a meritless future.
If we solve the equation over the real numbers only, 20 is the only valid solution here. Plugging x=-4 back in the original equation gives 2log_4(-4)-log_4(1)=2. Notice that we have log_4(-4), which is undefined in the real numbers since the domain of f(x)=log_4(x) is x>0. Hence, -4 is not a valid solution. When solving equations involving functions that aren't defined on the whole real number line (such as logarithms or square roots), it is important to plug the solutions we get back into the original equation to make sure we eliminate any invalid solutions. However, if we go into the realm of complex numbers, we can accept x=-4 as a solution. Here's why: Using the change of base formula, we can rewrite the equation as 2ln(-4)/2ln(4)-ln(1)/ln(4)=2, which is equivalent to 2ln(-4)-ln(1)=2ln(4) by multiplying both sides by ln(4). We can express 1 and -4 in the form re^(i*theta), where m and n are integers: 1=e^(2n*pi*i) -4=4e^((2m+1)*pi*i) Each value of m and n gives a different complex branch for 1 and -4. Substituting this into our equation: 2ln(-4)-ln(1)=2ln(4e^((2m+1)*pi*i))-ln(e^(2n*pi*i))=2ln(4)+2(2m+1)*pi*i-2n*pi*i=2ln(4)+2*pi*i*(2m+1-n)=2ln(4) For this to be true, since the RHS is real, the imaginary part of the LHS must be 0. This means 2m+1-n=0 or n=2m+1. All this means that x=-4 is a valid solution if, for some integer m, you take the m-th branch of -4 and the (2m+1)-th branch of 1!
@@youngmathematician9154 thanks for your reply, I told him some of that and actually solved it for him twice but he said because there's "2" we can move it and continue solving, + in wolformalpha the -4 sol. Is not valid for some reason 🙂
@@batdisker8917 The reason WolframAlpha does not cite -4 as a valid solution might be because WolframAlpha takes the principal branch of both logarithms, which gives an invalid answer.
These videos are great refreshers just in case you're stupid like me and forgot how to do simple math but want to try online college several years later after graduating high school.
(1+2)^2 = 9 This is because 1+2 becomes 3, and 3 squared is 9. 1^2 + 2^2 = 5 This is because 1 squared is 1, and 2 squared is 4. Therefore, 1+4 = 5. 9 =/= 5
The equation can be true in specific circumstances. To find such circumstances we need to look at the last correct line from the notebook photo. Namely it is 2AB=0. The equation is true for A=0 OR B=0. As a result, in the following step we get an attempt to divide by 0, a bad no-no.
I feel like things are too complicated. If A or B equals 0 then why would it be involved in the equation? What benefit would it serve? If (A+B)^2 equals 36 because A=4 and B=2, I get that. But if you told me (A+B)^2 equals 36 because A=6 and B=0, I would ask why are we involving B in the first place? In my earlier example of (A+B)^2 equals 36, I could also say (A+B+C+D+E+F)^2 equals 36 so long as C,D,E,F each equal zero but it seems unnecessary to do that. Therefore, it should be a given that since a variable represents a mathematical object, which is something, a some thing, a variable should, as a result, represent something that holds a value and that can never be zero because zero, by definition, holds no value. But maybe I'm wrong. I just watched the video to relax my brain and think about something else for a minute and thought I'd share my thought. Happy to receive responses.
by your logic, the equation "x-1 = -1" is literally imposdible to solve. there is NOTHING in the definition of variables that says they must hold value, only that they be a number of any kind that correctly solves their function.
In many places there are examples where a variable HAS to be zero. Like physics there is many equations where you put 0 into a variable to find another variable. This video is discussing basic high school algebra where you learn what to do with variables holding no value. It is not until later high school/college physics where you start to apply that and use it. If you didn't learn what to do with those variables, physics and many other applied algebra classes would be drastically harder bc you'd have no idea what to do
I am responding, but I have only watched this one video. Who is this video for? What level? What age? For many (dare I say most) students, a simple example like (3+4)^2 does not equal 3^2 + 4^2 is sufficient. So many individuals end up hating math because math teachers "force" them to think via proof and only proof. BS and BS in math education & statistics, and MS pure mathematics.
Thod must be a joke. Dont make me go buy a maths book on purpose. (A+b)^2. As much as i eant to fo this quickly. Ehy doninknow it is a polynomial and needs to be multiplied Yeah it is (A^2 +2AB+ B^2)
The statement only holds true if 2AB = 0, which means either A or B (can be both) = 0. 2 x 0 x 0 is also equals to 0. edit : And if either A or B (can be both) = 0 you cant divide with AB, because its dividing by zero which is undefined
I'm afraid you missed the "plus" in the formula and the multiplier by the "2". The answer is a squared plus 2ab plus b squared equally, if cubing instead the answer given is: a cubed, plus a squared x b, plus a x b squared, plus b cubed I thought binominal theory was fully understood. Cross reference any calculations with Pascal's Triangle
I think you can also prove this rooting, or even the pythagorean theorem, Because if (a+b)^2 = a^2 + b^2 Then (a+b) must equal to "c" If you let c = (a+b) , then you a rightside triangle with the sides a, b and (a+b) if you then let b = a, and you have c^2 = a^2 + b^2 then you get (2a)^2 = a^2 + a^2 -> 4a^2 = 2a^2 or you can do (2a)(2a) = a^2 + a^2 Either which way, you end up with a triangle that is invalid. because it breaks c^2 = a^2 + b^2. But if a and b have to be different I think you can show it by square root.
Dont. Change. The equation. You dont need to make it harder and longer. That is the opposite reason of why equations can be changed in the first place.
@@authorttaelias4483 well the first line i was referring to in my original comment is the same. A^2 + B^2 =/= (AB)^2 Unless A and B are both equal to zero. One of the first things you learn in algebra is that (AB)^2 = A^2 + 2AB + B^2. Or is this whole proof dependant on dividing out zero from the 0,0 position?
![[LIVE] : ONE ลุมพินี 88 | คู่เอก "ป้อมเพชร vs อัสลามจอน"](http://i.ytimg.com/vi/1ZqTVVbMgbo/mqdefault.jpg)
"2=0 proof" from Reddit r/askmath
th-cam.com/video/U-sBnlqPahg/w-d-xo.html
A division by zero has occurred, since your final equation is a contradiction.
1 ÷ 0 = can't divide 0
0 ÷ 1 = 0
🤍🌈🥰
@@charlesmbennett0000 such an equation is a contradiction, and so there is no solution.
@@byronrobbins8834 This is simply what the calculator not my words but it's.🤍🌈🥰
It’s a bit circular - you have to assume that A commutes with B and A times B is not zero
Prove that BPRP loves math
1: "I feel like it is a logic circle, but this is what happens when you do a contradiction proof. It's a lot of fun actually."
The majority of humans do not understand linear algebra.
I feel like it makes more sense to take 2AB=0 AND divide both sides by 2, that tells you A×B=0 so its only true if at least one equals 0.
That would be the approach if you are actually trying to solve the equation rather than disprove the identity.
This is my preferred demonstration too. Divide 2B from both sides and A=0, contradicting your initial premise.
@@Abion47 No, the multiplicative identity of 0 would state that the product = 0 means either a = 0 or b = 0, but by assumption neither a nor b are zero. It disproves the identity by a fundamental theory of numbers, and it also proves the exception of a = 0 or b = 0. Note, the or is not exclusive, which means both statements could be true.
👋👋👋 what I like best about this video is, father is letting his son understand that we should question and point out something when we know that something is wrong. 👏 Bravo
It's unfortunate that the teacher and principal didn't not understand both lessons here.
Also fun fact, any number you use will make the left side equal double the value of the right side
If you do 2, you get 16 = 8
3 is 36 = 18
Darn I like your teaching style! How I wish you had been my teacher. I could see myself enjoying your class.
Just thinking about slicing a pie also works: I can have any number of slices by cutting the pie. The more I cut, the more I have slices. But I cannot cut less than not slicing at all (which is dividing by 1). So (assuming I divide by naturals) I cannot divide by less than 1.
Wrong. -1 is an integer, smaller than 1 and there are no issues with dividing by it. Natural numbers are fitting to your rule
@@Romczythanks! I corrected 😅
@Romczy
-1 isnt part of the natural numbers.
The op explicitly said natural numbers.
Not real numbers
Not complex numbers
Not imaginary numbers
Not irrational numbers
Not transcendental numbers.
Unless the OP made a mistake, using the wrong terminology.
One glance at the equation in very first line and the error is painfully obvious.
Im pretty bad at math, but you are so good at explaining that i actually slightly understood all of the numbers and letters on yohr board
Same for me, but I must admit, the word "slightly" is still putting in a _lot_ of work in my case.
Even my middle school algebra teacher drilled that into me.
When you factor a trinomial you get a binomial squared and many others.
Pascal's triangle is a great tool for binomial expansion.
If you ever come across anything like (a+b)^5 you can use p's triangle to quickly find the identity of that binomial.
I’m 26 and no longer studying or working in a field that requires maths. But why can’t I stop watching all your videos 😂
a^2 + b^2 = (a + b)^2, when 2ab = 0, which is the case if either a or b (or both) = 0.
Eons of wind erosion & the way the light hits it bro..its just a typical rock formation.
(a + b)^2 = a^2 + 2ab + b^2
(a - b)^2 = a^2 - 2ab + b^2
(a + b) (a - b) = a^2 - b^2
Students learn the binomial formulas in year 7 at latest.
I learned start of year 8
Yes!
Exactly
Was really hoping to hear more about the resolution with the school but thank you
Il faut préciser l'ensemble de départ, sinon il faut écrire (A+B)^2 = A^2+AB+BA+B^2 (non commutativité dans l'ensemble des matrices carrées). De plus (A+B)^2 = A^2+B^2 est vrai dans Z/pZ avec p premier. Merci pour vos partage !...
Thank you
I don't understand French, but :"Oui". Your math is correct.
Really couldn't believe that someone would actually think that was correct. it is equal to (a+b) x (a+b), which is, using FOIL, A^2 + 2ab + B^2. Guess root values and quadratics aren't being taught in elementary schools anymore.
In czech republic they taught us these equations in elementary school and even in first grade of secondary school we had it
1 apple less nothing results in 1 apple.
We could also prove by contrapositive, "(A+B)^2 = A^2+B^2 implies A=0 or B=0". At the end we just need to write "2AB = 0" => "AB = 0" => "A = 0 or B = 0"
The original post has one symbol he/she didn't put resulting in the misunderstanding.
They didnt put "?" On top of "=". The "?" On top of "=" denotes that the equation is asking if the left side of the equation is equal to the othe side of the equation.
And because of that @bprpmathbasics thought that they are solving the equation and not proving the equation.😊
A better proof for this would be by dividing 0 by 2 in the last step leaving you with AB=0. The only way that can be true is if A or B is 0. This is not a proof by contradiction, but it is less ambiguous than dividing an undefined variable by 0.
Now I know why I didn't understand math. I was timid. They never explained this contradiction circle. It made no sense and do nothing else made sense. So I gave up. Used to be extremely good😢 at it.
Only if I could have afforded to watch videos like this on the internet😢
Gosh I was soo timid when I was wrong what was wrong was that I did not appreciate the contradiction that mattered and changed the meaning of the equations and inequalities
what's the catch? This is basic linear algebra, it is taught to students in the first year. Here it is even checked logically in 10 seconds.
linear?
@@churros3289 wikipedia.org/wiki/Linear_algebra
It's just algebra not linear algebra lol
@@FenShen-us9tv perhaps for me both were more than 20 years ago :)
My theory is, some people are born without math genes. I know order of operation and probability could solve for Y. But that’s as far as it goes.
9001x = 0. Now lets divide both sides by x, we get: 9001x/x = 0/x --> 9001 = 0, which proves once and for all that zero is over nine thousannnnnnnnnnd. What? Nine thousand? Theres no way that can be right.
What he was saiyan ^
9001x=0 is true ONLY for x=0, so when you try to divide both sides by x, you actually divide by 0. We DO NOT divide by 0!!!!
Doesn't L'hopitals rule say that the quotient of 1/0 equals positive infinity?
You can, too, have in both sides AB=-AB, which is an even more notorious contradiction
I’d love to have a conversation with the teacher about limits and L’Hopitals.
If 1÷0=0 is true, then 0×0=1 must also be true. As it is not, then 1÷0=0 is false.
We have to define why this is a math problem to begin with to get the answer... for instance 8 apples divided between 2 people equally ends up at 4 apples apiece, 2 apples divided amongst 8 people winds up in a quarter of 1 apple, and if you have 1 apple and you divide it amongst nobody, you have 1 apple, that you didnt share. And if you have 0 apples divided amongst 1 person, guess what that person has?? 0 apples. 😂 its basic math. Why are people trying to make the basics seem so deep. Lets define the "problem" (actual real world correlated issue) to resolve the math equation appropriately.
If you have one apple you aren't sharing, you still have one person to that apple. You are dividing by 1.
Dividing by zero is saying "there are 4 apples and no people in that room. How many apples does each person get" that makes no sense, is undefined.
@LeifNelandDk did you not gather the logic I was trying to convey, did you even read what I wrote? It has to apply to real world application for it to be a valid equation, is my point. And back to the 1 apple divided by 1 or more people is still just 1 apple, or a fraction thereof. What did you mean by your post? I must be missing something?
@LeifNelandDk maybe I would word the math problem as "you have 1 apple and divide it amongst 0 people, how many apples are there? Answer being 1. But if you say you have 1 apple and divide it amongst 0 people, how many apples does each person get.... would be a d***** question,.. Wouldn't it? Back to my real world application and you making up a "real world scenerio" of dividing amongst 0 people and then discussing how many of whatever of each person get out of it, when there's no people there, sounds like an unpractical, not a "REAL" Scenerio.
the actual formula for A square + B square is [a^2 + b^2 = (a + b)^2 - 2ab] or [a^2 + b^2 = (a - b)^2 + 2ab] which is true regardless of whether a or b equals to or is not equal to 0, which is basic math and is taught in primary (or in some places secondary) school. However, what is also basic math is (A+B)2=A2+B2 or to write it as a multiplier to be more obvious (A+B)x2 = Ax2 + Bx2, just basic math which some person may have mistaken it for a square instead of a multiplier "on Reddit", but I like the way you break it down for people who are not really smart (see what I did there)
Subscribed, I may show your videos to my kids someday 😀
No that is not right.
Also its not called a formula, its called an identity.
(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2
You can use pascal's triangle to quickly find the identity of any binomial expansion.
Because in higher maths its easier to use p's triangle than work out (a+b)^6 by hand.
You will need to know this for calc 1.
All of that is true, but none of it is relevant to this video.
@duckyoutube6318 A formula is simply an equation with 2 or more variables. Every formula is an identity, but not every identity is a formula.
@trwent Thank you for the correction. Its important to be honest and precise in math.
IDK about the rest of the US, but in Texas, division is not introduced with decimals. It’s introduced with the idea of remainders. You never learn the full problem. Decimals come in late in the school year for 3rd graders and is worked in 4th and 5th grade by along with fractions. But fractions is given the even greater attention.
You do not even learn how to work the entire math problem when it’s first introduced because it’s believed to be too complex. I think the belief is that math at the elementary levels are suppose to have answers, not undefined answers. I think they believe it to be like saying “I don’t know.” But that’s not actually the case.
Actually a good teacher, teaching them an important lesson they will need for their entire future. You WILL be forced to believe and accept things that are obviously false.
That is the way of things going forward into a meritless future.
I like proof by contradiction, but sometimes I get lost in the sauce in my own supposed statement lol
I remember my algebra teacher in high school proving that 2 + 2 = 5. Hidden in the proof was a division by zero.
No it is not equivalent. It is a perfect square trinomial. Famous formula right here in the liner of all algebra and slightly higher textbooks.
It’s basically reductio ad absurdum in mathematical form, and I’m all for it.
Thank you
This was covered extensively in Algebra 1. If you do not remember this one, you never understood Algebra.
My teacher says that the solutions for (2log4(x) - log4(x+5)=2) are 20 and "-4"!!!!, what do you think? And thanks
If we solve the equation over the real numbers only, 20 is the only valid solution here. Plugging x=-4 back in the original equation gives 2log_4(-4)-log_4(1)=2. Notice that we have log_4(-4), which is undefined in the real numbers since the domain of f(x)=log_4(x) is x>0. Hence, -4 is not a valid solution.
When solving equations involving functions that aren't defined on the whole real number line (such as logarithms or square roots), it is important to plug the solutions we get back into the original equation to make sure we eliminate any invalid solutions.
However, if we go into the realm of complex numbers, we can accept x=-4 as a solution. Here's why:
Using the change of base formula, we can rewrite the equation as 2ln(-4)/2ln(4)-ln(1)/ln(4)=2, which is equivalent to 2ln(-4)-ln(1)=2ln(4) by multiplying both sides by ln(4).
We can express 1 and -4 in the form re^(i*theta), where m and n are integers:
1=e^(2n*pi*i)
-4=4e^((2m+1)*pi*i)
Each value of m and n gives a different complex branch for 1 and -4.
Substituting this into our equation:
2ln(-4)-ln(1)=2ln(4e^((2m+1)*pi*i))-ln(e^(2n*pi*i))=2ln(4)+2(2m+1)*pi*i-2n*pi*i=2ln(4)+2*pi*i*(2m+1-n)=2ln(4)
For this to be true, since the RHS is real, the imaginary part of the LHS must be 0. This means 2m+1-n=0 or n=2m+1.
All this means that x=-4 is a valid solution if, for some integer m, you take the m-th branch of -4 and the (2m+1)-th branch of 1!
@@youngmathematician9154 thanks for your reply, I told him some of that and actually solved it for him twice but he said because there's "2" we can move it and continue solving, + in wolformalpha the -4 sol. Is not valid for some reason 🙂
@@batdisker8917 The reason WolframAlpha does not cite -4 as a valid solution might be because WolframAlpha takes the principal branch of both logarithms, which gives an invalid answer.
@@youngmathematician9154 OK, again thanks for answering ❤️
These videos are great refreshers just in case you're stupid like me and forgot how to do simple math but want to try online college several years later after graduating high school.
Mathematics is the art of simplifying things. (a+b)^2 = a^2 + 2ab + b^2
Why trying to solve problems that does not mean to us?
correct, he used the term undefined. i was always taught, not allowed
Either i am a genius or there are way too many dummies out there. How can any adult not know this?
(1+2)^2 = 9
This is because 1+2 becomes 3, and 3 squared is 9.
1^2 + 2^2 = 5
This is because 1 squared is 1, and 2 squared is 4. Therefore, 1+4 = 5.
9 =/= 5
The equation can be true in specific circumstances. To find such circumstances we need to look at the last correct line from the notebook photo. Namely it is 2AB=0. The equation is true for A=0 OR B=0. As a result, in the following step we get an attempt to divide by 0, a bad no-no.
I feel like things are too complicated. If A or B equals 0 then why would it be involved in the equation? What benefit would it serve? If (A+B)^2 equals 36 because A=4 and B=2, I get that. But if you told me (A+B)^2 equals 36 because A=6 and B=0, I would ask why are we involving B in the first place? In my earlier example of (A+B)^2 equals 36, I could also say (A+B+C+D+E+F)^2 equals 36 so long as C,D,E,F each equal zero but it seems unnecessary to do that. Therefore, it should be a given that since a variable represents a mathematical object, which is something, a some thing, a variable should, as a result, represent something that holds a value and that can never be zero because zero, by definition, holds no value. But maybe I'm wrong. I just watched the video to relax my brain and think about something else for a minute and thought I'd share my thought. Happy to receive responses.
by your logic, the equation "x-1 = -1"
is literally imposdible to solve.
there is NOTHING in the definition of variables that says they must hold value, only that they be a number of any kind that correctly solves their function.
@@DarthZ01 by my logic, it is not impossible. It is unnecessary.
In many places there are examples where a variable HAS to be zero. Like physics there is many equations where you put 0 into a variable to find another variable. This video is discussing basic high school algebra where you learn what to do with variables holding no value. It is not until later high school/college physics where you start to apply that and use it. If you didn't learn what to do with those variables, physics and many other applied algebra classes would be drastically harder bc you'd have no idea what to do
Hopefully that helps ! :)
You should have a conversation with Terrance Howard. Maybe you can crack the mathematical errors.....
Great video!!
I am responding, but I have only watched this one video. Who is this video for? What level? What age? For many (dare I say most) students, a simple example like (3+4)^2 does not equal 3^2 + 4^2 is sufficient. So many individuals end up hating math because math teachers "force" them to think via proof and only proof. BS and BS in math education & statistics, and MS pure mathematics.
(1+2)^2 = 9
1^2 + 2^2 = 5
its just simple math why some do it wrong
This was quite obvious to me, and I took algebra back in 1969.
1 apple with nothing taken from it results in 1 apple.
How simple huh?
Do something about your audio. Otherwise, you're great.
Redditor tried to divide by zero, that is all I need to know
Except when a eq 0 or b eq 0
We,, obviously ...
Instant thought you can’t divide AB because it could and is 0
But otherwise the equation could be true for 0
Pretty much the only scenario where you can't simplify 2AB=0 is the scenario where (A+B)²=A²+B² is correct. There are no accidents :p
Thod must be a joke.
Dont make me go buy a maths book on purpose.
(A+b)^2. As much as i eant to fo this quickly. Ehy doninknow it is a polynomial and needs to be multiplied
Yeah it is (A^2 +2AB+ B^2)
Now what if both a and b are zero
then you're dividing by 0
The statement only holds true if 2AB = 0, which means either A or B (can be both) = 0.
2 x 0 x 0 is also equals to 0.
edit : And if either A or B (can be both) = 0 you cant divide with AB, because its dividing by zero which is undefined
Then you have a trivial case where both (a+b)^2 = 0 and a^2 + b^2 = 0.
Memories of Sister Mary Angelique
Nah. Everything = One.
(1+1)² = 2² = 4
1²+1² = 1+1 = 2
@MaximeB-hc4oz 1 = 0, 0 = 1.
1 ÷ 0 = can't divide 0
0 ÷ 1 = 0
🤍🌈🥰
I should have paid attention in algebra. 😕 ¯\_(ツ)_/¯
all this to say you cannot divide by zero
I'm afraid you missed the "plus" in the formula and the multiplier by the "2". The answer is a squared plus 2ab plus b squared
equally, if cubing instead the answer given is: a cubed, plus a squared x b, plus a x b squared, plus b cubed
I thought binominal theory was fully understood. Cross reference any calculations with Pascal's Triangle
i got to the end, apologies!
I think you can also prove this rooting, or even the pythagorean theorem, Because if (a+b)^2 = a^2 + b^2 Then (a+b) must equal to "c"
If you let c = (a+b) , then you a rightside triangle with the sides a, b and (a+b)
if you then let b = a, and you have c^2 = a^2 + b^2
then you get (2a)^2 = a^2 + a^2 -> 4a^2 = 2a^2
or you can do (2a)(2a) = a^2 + a^2
Either which way, you end up with a triangle that is invalid. because it breaks c^2 = a^2 + b^2.
But if a and b have to be different I think you can show it by square root.
Dont. Change. The equation.
You dont need to make it harder and longer.
That is the opposite reason of why equations can be changed in the first place.
🫶🏾
The first step is already wrong, as (a+b)^2 = a^2+b^2+2ab
My guy said 2AB = 0 ==> 2 = 0 😂
2AB=0 is correct when A or B =0
These redditors will make me to punch the screen on my phone real hard.
Basically:
"Let's solve the equation 3x=0, cancel "x", bingo, now we have 3=0"
Sad result of lack of knowledge what function domain and the "cancelation" is. Quite unfortunate
Lmao, the first line is such a low level error. This must be written by a Jr. High kid.
Its an equation not an identity, it is not a mistake
Plus it was actually a proof by contradiction so it was meant to show that the first thing wasn’t true
@@authorttaelias4483 So you're saying that the proof is set up as an "IF" statement? Then you're saying i could make a proof like "2A = A"
@@tyty8484 if you wanted to prove that not true then yes but also if you wanted to find WHEN it was true it’d be true for A=0
@@authorttaelias4483 well the first line i was referring to in my original comment is the same. A^2 + B^2 =/= (AB)^2 Unless A and B are both equal to zero. One of the first things you learn in algebra is that (AB)^2 = A^2 + 2AB + B^2.
Or is this whole proof dependant on dividing out zero from the 0,0 position?
Foil!