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.
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.
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 !...
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.
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.
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.
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 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.
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.
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.
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.
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.
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 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.
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 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.
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
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?
