he messes up on the second line he writes, you cant just square the one side. if you sub the numbers in, then 1=1 buts hes working with variables where you cant just square one side
@a1997411z the bit after the '/' explains what he's doing in the next step, it doesn't mean divide, so in fact it is you who are wrong. it's just cuz he divides both sides by zero at the end; effectively he's saying 4x0=5x0 which is obviously true.
Everyone seems to understand that dividing (B-C-A) by itself is impossible for the original values of A,B,C, but the question that needs to be asked is, if all of the algebraic steps are valid, why isn't the given solution set valid?
To those that try to find fault with my analysis, the point is that there is no fault with the algebraic steps taken. The question to be asked is why is the solution set valid for the original equation but not for A(B-C-A)=B(B-C-A). wikipedia.. One of the basic principles of algebra is that one can multiply both sides of an equation by the same expression without changing the equation's solutions...multiplication by certain expressions may introduce new solutions that were not present before
FIrst of all, you squared one side without squaring the other side. I will let you get away with that as B-A is 1. However, where the heck / - A squared come from?
Yeah it's the same as saying that by dividing both sides of 1x0=2x0 by 0 you get 1=2. No need for the long-ass variable equations, its obviously wrong...
I think u understood the whole math wrong how canu u divide 0/0 it is a indeterminate form u cAn't cancel !! u can cancel only an finite value !! u r proving that 1 /0 is 0 !! use ur iq!! dude
@topicmirsad We keep going in circles. I do realize that you can't divide by zero. For a moment, forget about the solution set (4,5,1) and look just at the algebra. What I am trying to say is that all algebraic rules used are valid, even up to the point where he writes A = B, which we both know isn't true with the given solution set. Every step of his algebra is valid. Step 2 created the anomaly. Thanks
no thats wrong. before dividing by variables e.g. (B-C-A) you have to proof that there is no combination of A,B and C for which the expression (B-C-A)=0. for sums like the one in the video THIS IS NOT POSSIBLE. you cant rule out the possibility that (B-C-A)=0 therefore you are not allowed to divide by it. one of the few expressions where you are allowed to do so is if the variable is a power, e.g. 5^x because there is no number for which 5^x=0.
You factored A out of the left side, and B out of the other... so you divided one side by 4 and the other by 5. And that was to get A*0 = B*0, which is true, but not a logical way to prove equality. It basically says: both are insignificant in the face of infinity, thus they are equal.
@topicmirsad Let (B-C-A) = X Then (A(B – C – A)) / (B – C – A) = (B(B – C – A)) / (B – C – A) becomes AX/X = BX/X which equals A=B so dividing by (B-C-A) is algebraically correct.
It is clear that A and B are two different numbers (in this case 5 and 4, but let us disregard that), then A*X=B*X, only, and only if X=0, so you can't divide by it
no its not, because A,B and C can be chosen so that (B-C-A)=0 therefore you are not allowed to divide anything by it (not even itself). before dividing by something you ALWAYS proof that its not zero.
If this equation is true than it equals 2+2 and if it equals 5 than all of that other crap is unnecessary and shows that you OBVIOUSLY did something wrong
@infitiae No. He multiplyied both sides with (B-A) . You probably mistaken that saparator in first line for a division sign, like many others here. First line is C=B-A, and not C=B-A/(B-A)
now in your algebra ...YOU decide the valour of A B and C....;then you can find the solution you want and then 2+2 =5 but only for you buddy .....;cuz you pick the number you need for A B and C to get to your response
SO DON T THINK TOO MUCH MISTER THE NUMEROLOGIST
you ll get a complete defferent result if
A =3
B=8
C=2
go on ! waste another piece of paper and a little bit of your time and see where it go ....
One of the last steps where he crosses out (B-C-A) on both sides is illegal. To do this you would need to divide both sides by (B-C-A) yet that is dividing by zero. B-C-A = 5-1-4 = 0.
@Steven01001 That all is for multiplication with zero, and other non-reversible operations. But in this case, none of that happened, because he started with all different numbers, and none of them is eqal to zero.
C=B-A which is the original equation, but the error occurs when he divides both sides of the 2nd to last step by (B-C-A) which is algebraically correct. Adding additional variables (at step 2) can cause this as I alluded to twice before. Why do you think he got the wrong answer when his math is correct?
@Steven01001 He didn't add any variables in step 2, and in any other step. First he was multiplying both with (B-A), then he subtracted A^2 from both sides, then he added AB to both side. Those are all perfectly correct algebraic and mathematical operations.
The ONLY problem here is that division by zero in last step,and nothing else. Getting to that step is perfectly correct.
@Steven01001 He didn't add any variables in step 2, and in any other step. First he was multiplying both with (B-A), then he subtracted A^2 from both sides, then he added AB to both side. Those are all perfectly correct algebraic and mathematical operations.
The ONLY problem here is that division by zero in last step,and nothing else. Getting to that step is perfectly correct.
@Steven01001 He didn't add any variables in step 2, and in any other step. First he was multiplying both with (B-A), then he subtracted A^2 from both sides, then he added AB to both side. Those are all perfectly correct algebraic and mathematical operations.
The ONLY problem here is that division by zero in last step,and nothing else. Getting to that step is perfectly correct.
@topicmirsad You are correct about that step equaling zero, but in the next step he divides both sides by (B-C-A) which is algebraically correct, but results in the wrong answer. If (B-C-A) were anything but zero, the solution would be valid. In this case, it is not. The solution set he provides is valid for the original equation, but not for the final equation, which I repeat is algebraically correct.
@Steven01001 In this last step dividing by (B-C-A) is not even algebraicly correct. If we disregard given set of values and just say that A does not eqal B. Then, for this equation to be valid A(B-C-A)=B(B-C-A), (B-C-A) must be eqal to zero. In any other case eqation is not valid, and therefore you can't cross it out in any case. Only case in wich you would be able to cross them out is when A=B, but we started with assumption that A does not eqal B, just as in this video
@topicmirsad It's a proof, it doesn't matter what the actual numbers are. The last step is valid, but the proof isn't. He made an error in step two. Everything beyond that is null.
Step two should be:
C(B-A) = B-A
((B-A) / (B-A))(B-A) is not (B-A)^2. It is B-A. From then on, he is running with 1=2.
@bryanhaw1 Las step is not valid. (B-C-A) is eqal to 0 in any case, because A and B are different numbers. It would be valid if A would eqal to B, but in this case they are different numbers
Look under Fallacies based on division by zero section, It has exactly this example explained.
But Midenbinder already knew that or he wouldn't just crossed it like that, He would divide both sides by B-C-A just like he did till that very point.
This way it is less obvious that one of the golden rules in mathematics has been broken.
@odise25 Why would i need to search Wiki? I perfectly know what is divison by zero, but some other people here don't understand that "division by zero" is causing this weird result at the end, and all i'm trying to do is explain it to some people here.
i don't really understand what you're saying lol... is it math is it english i dunno... all i know is im not wrong it's been proven that you cant divide by zero. for instance (a / 0). In ordinary (real number) arithmetic, the expression has no meaning, as there is no number which, multiplied by 0, gives a (a≠0), and so we say that division by zero is undefined.
you can't say that im wrong... its math godddd lol
The error is in the factorization at the end i.e it's an illegal factorization. Because it's an equation, you have to do the same operation on both sides. If you factorize by B on one side, you need to factorize by B on the other side. If you do that in the second to last step, it'll work out perfectly.
i.e b (-ca/b -a^2/b +a) = b (b - a - c) -----------> 0 = 0. And the world is right again.
One more thing and then I'll be on my way. If you solve the last equation for "C" you should get the original equation, which you don't, therefore and even though the algebra is correct, the original equation and the last one are not similar. I again say that in some cases, multiplying by variables as done here in step 2 can create extraneous solutions. It was fun, thanks all!
If you don't believe me then solve the last equation for C and using a graphing calculator, graph out the first equation and your new one and you will see that they are totally different.
lol no /-A^2 shows that he's adding -A^2 to both sides. it only shows what his next step is gonna be. A.(B-C-A) = A(5-1-4) = A(zero) u cant divide by zero -.-.
@ZeroCool0090 wrong, 1/3 = .3 repeating, .3 repeating x 3 = .9 repeating, therefore .9 repeating = 1, therefore 1 - .9 repeating must = 0, but it actually equals 0.0 repeating with a 1 at the end of the infinite amount of 0's, therefore anything divided by 0 is infinity :3
The last equation where (B-C-A) is on both sides, he crossed them out as if he had divided both sides by (B-C-A) which is algebraically correct, except in this case, "with the given constants",... (B-C-A)=0 which makes division not mathematically correct. Everyone seems to see this, but fails to see that step 2 is where this wrong solution originated. A=4, B=5, C=1 is a solution set for C=B-A, but not for A(B-C-A)=B(B-C-A)
@0815Nolifer Agreed, the only error is at the end, but as I said... His algebra is correct except for not realizing that when you multiply by variables (as in step 2) it can create extraneous solutions as it has in this case with the constants he used. Why else would his final solution be wrong after plugging if the constants when his algebra is correct up to that point? Look it up, you'll see what I'm talking about. Multiplying by (B-A) in step 2 caused the false result in the end.
@Steven01001 Multiplying both sides of equation with same "things" ALWAYS gives correct solutions, and eqation we started with is the same as the one we got after we multiply eqation with "something" (number, algebraic phrase, ...)
The last eqation is same as the first one. Try to solve A(B-C-A)=B(B-C-A) for C, and you'll get C=B-A.
@ThatRubikGuy I won't argue that the last result creates the error when the given values are substituted, but as I said, step 2 is where the error is created which makes the solution set valid in the original equation, but invalid in the final equation.
There's nothing right about this. His derivations are all completely incorrect. Going to step 2, the right side should not be squared. Assuming that was even correct, going to step 3 does not introduce division by -A^2. I mean, where did that even come from? And it continues from there...
Guys, you're missing it here. Because the letters have set values, this formula will only prove true if you can plug in the values and work it out the same way. You can't without dividing by zero.
Lol at the math noobs who think he's NOT dividing by zero... to "cancel" something out you MUST divide it by itself. You don't magically make B-C-A disappear, you DIVIDE BOTH SIDES to make it happen. Sorry for the caps, just I would assume that highschool / younger kiddies need capital letters to understand what to focus on.
@mercscout Which part are you talking about near the end? I dont see him divide by zero, he has A(b-c-a)=b(b-c-a) and he divided BOTH SIDES by (b-c-a), unless your trying to state b-c-a=0, and he is diving by zero
Yes B-C-A does equal zero. And he is dividing by it. 5-4-1=0
ANY algebra trick you can see on youtube titled "OMG I BROKE MATH" is either dividing or multiplying by zero. That's why when you divide by something, you need to make condition it's not zero. When you have an equation and you want to divide it by X, you are making a statment that x=/=0, then you solve the equation for x=0.
@dabestplayer I'm not 'trying' to say it. I am saying it. In the first 30 seconds of the video he defines a,b,c and if you plug in their values we get b-c-a=0 :), 0/0 DNE
@cheftoniFTW no that part is actually correct, you dont simply add a square to both sides thats incorrect math, when given (B-A)^2 you have to F.O.I.L it, multiply the first terms then outer then inner than last, so (B-A)^2 equals (B-A)(B-A) first terms multiplied eqal B^2 outer: -AB inner: -AB and last: A^2 which equates into B^2 -2AB + A^2
@Airman32292 But even then, shouldn't the equation then be:
CB - CA = B^2 + A^2 + 2AB (As opposed to - 2AB, seeing as 2 negatives become a positive?) And in both cases, the - 2AB would make the equation as 1 = 21, and the other as +2AB would become 1 = 101?
@cheftoniFTW just to say i dont agree with this problem as a whole, but no it wouldnt be positive, once you do the F.O.I.L method you then have terms that you add together, not multiply, so you add B^2 + -AB -AB + A^2, which is the same as B^2 -AB -AB + A^2 which equals B^2 -2AB + A^2
CB-CA=B^2-2AB+A^2/-A^2 is false you would come up with 1=-14 which is wrong. to prove it. 1*5-1*4=1 so 1=5^2-2*4*5+4^2/-4^2 so reduced it is 25-40+16/16 so reduced even more it is 25-40+1 so -15+1 which = -14 and last time i check -14 doesn't = 1
@ajerrils That's not what he meant rofl, the /-A^2 part refers to the step being taken, not a division. The notation isn't ideal but I'd have thought at least that was obvious.
Love how so many people have gone through and said "Oh you've done done all teh steps wrong herp derp". The original statement says C=B-A, so naturally 0=B-A-C or for here 0=B-C-A. Of course if you have that then it's no different to multiplying that part with say A, or B or C; you're just multiplying 0 by something. The statement's perfectly valid, it's just that if you put any values in you'll get 0=0. So many have given thumbs down for illegitimate reasons it's painful.
@MakiRole Lets set aside the many errors in math throughout the problem... in the end in order to cancel out (B-C-A) on each side of the equation you must divide by zero, which you cannot do.
@OonslaughtoO There is no division by zero, he is simply removing from both sides a similar equation seeing a both (B -C - A)s are equal, they cancel themselves out.
berybite its ok and can fesics law or some one some how tell me why 0/0 its not ok it is undeterminate u can not reduce every thing in 1 equation becose 0 i dont untherstand how fisics can reduce every thing in one ecuation.
C=B-A/(B-A) is ok :L then you tried to multiply both sides by B-A? so you should get C(B-A)=B-A not C(B-A)=(B-A)2 you cant just say 'oh ill add a little squared sign on it because my balls hurt..'
I was expecting something funny... without commentary all I see is a mistake with steps 2-5... (B-A) gets squared for no reason, divide the second equation by -A^2 for no reason.. dvide by -AB, IDK what your doing at all here... the whole thing is a giant mistake.
Why use letters instead of the numbers to write it? It would be 10 times easier to spot your mistake, although I already know you must have made one. Haven't you even been to elementary school? I think they taught the two times table there already!
3rd line is false. cb-ca =1 so 1= bb-2ab+aa/-aa, so 1=25-40+16/16, so 1=1/16 is the 3rd line result which is a false statement. I am sure later in the video they devide by zero like all the other silly tricks, but they usually come after a false statement.
Theres something wrong at the beginning.
C= (B-A) / (B-A) (Correct up to here)
Then you multiply everything (B-A) and it becomes:
C(B-A)=(B-A) not C(B-A)=(B-A)^2!! [It is very well known that (B-A)/(B-A) * (B-A) = (B-A)]
Hence it is:
CB-CA=B-A
Now replace letters with numbers:
1*5 - 1*4 = 5 - 4
It is obvious that it is true..
dennis0mus 35 seconds ago
he messes up on the second line he writes, you cant just square the one side. if you sub the numbers in, then 1=1 buts hes working with variables where you cant just square one side
nicky48473 8 minutes ago
i guess you could ignore it all and say the answer is wrong
ryanroberts1234 1 hour ago
It´s because in his last step he divided by (B-C-A) which is (5-1-4) which is 0
you cannot divide by zero thats all. If A B and C were other numbers you could not get to the
last line.
Meomur 2 hours ago 3
third equation is wrong. Didn't bother to watch anymore
Photonface 3 hours ago
@Photonface What do you mean the third equation is wrong?
garylu1999 8 minutes ago
CB-CA=(B^2-2AB+A^2)/-A^2 ????
if sub the real numbers into the equations
1(5)-1(4)=[5^2-2(5)(4)+4^2]/-(4^2)
which gives me
1=-1/16
already wrong there asshole
a1997411z 3 hours ago
@a1997411z the bit after the '/' explains what he's doing in the next step, it doesn't mean divide, so in fact it is you who are wrong. it's just cuz he divides both sides by zero at the end; effectively he's saying 4x0=5x0 which is obviously true.
bluepansylove 2 hours ago
Everyone seems to understand that dividing (B-C-A) by itself is impossible for the original values of A,B,C, but the question that needs to be asked is, if all of the algebraic steps are valid, why isn't the given solution set valid?
Steven01001 4 hours ago
To those that try to find fault with my analysis, the point is that there is no fault with the algebraic steps taken. The question to be asked is why is the solution set valid for the original equation but not for A(B-C-A)=B(B-C-A). wikipedia.. One of the basic principles of algebra is that one can multiply both sides of an equation by the same expression without changing the equation's solutions...multiplication by certain expressions may introduce new solutions that were not present before
Steven01001 4 hours ago
@Creepyseven Checking the solution in reverse...
A(B-C-A)=B(B-C-A)
AB-AC-A^2=B^2-BC-AB
AB-A^2-B^2+AB=AC-BC
AB-A^2-B^2+AB=C(A-B)
C=(2AB-A^2-B^2)/(A-B)
C=(2AB-A^2-B^2)/(A-B) =(A-B)(B-A)/(A-B)
C=B-A
...which is the original equation
As I said before, his algebra is valid. but step 2 created anomalies.
Steven01001 5 hours ago
what a cheap trick...
In the first line C=(B-A)/(B-A) the B-A gets cancelled out.
In the second line C(B-A) = (B-A)^2 is wrong as well, it is C(B-A) = (B-A)
karanjoe5 5 hours ago
C = B - A / ( B - A ) ...aren`t u suppose to look for c not for A&B and why do the solving when u already have the answer given??idiot
dontkizzme03 7 hours ago
where does that quad came form?
jimmytaker 8 hours ago
the last step is 0 = 0 not a = b
rabaduej 10 hours ago
FIrst of all, you squared one side without squaring the other side. I will let you get away with that as B-A is 1. However, where the heck / - A squared come from?
nsakic 10 hours ago
Yeah it's the same as saying that by dividing both sides of 1x0=2x0 by 0 you get 1=2. No need for the long-ass variable equations, its obviously wrong...
decheater64 11 hours ago
Could you pick a more annoying song?
GraingerSays 12 hours ago
I think u understood the whole math wrong how canu u divide 0/0 it is a indeterminate form u cAn't cancel !! u can cancel only an finite value !! u r proving that 1 /0 is 0 !! use ur iq!! dude
thyhuh 13 hours ago
*brain deactivates*
fsjal117 15 hours ago
(B-C-A) WRONG ! to divide by 0 is not allowed
MrSchoko99 17 hours ago
random shit
Minecraftmonkeys 17 hours ago
im 11
AnimeFreak623623 18 hours ago
@AnimeFreak623623 would you like a chocolate medal?
fsjal117 15 hours ago
how did he get (B-A)^2, seems wrong
truehate101 18 hours ago
division by zero, we meet again. It's always the same with those "amazing math tricks".
saemj 19 hours ago
Wtf? The letters A, B, and C are only used in the Alphabet, what the hell is wrong with u?!!
DaNO324 19 hours ago
@DaNO324 You sir, are a retard. Don't say such things if you are only in 4th grade or something.
bannahispanicpower 13 hours ago
What is wrong:
3d line. Dividing by -a^2 for no reason.
Also, the 2x2=5 part.
53prime 19 hours ago
@53prime oh wait.. that's what you're dooooing, not the actual equation. Gotcha.. nvm
53prime 19 hours ago
im in the nerdy side of youtube again
Awsomeace123 19 hours ago
all the math is correct untill he multiplys A and B by 0 and gets anything other than 0=0, in this case, getting 4=5
MrFloresent 20 hours ago
wait, fuck this, im going back to dr. michio kaku
GarrettDavisHustle 20 hours ago
Most every algebraic fallacy is founded either on division by 0 or incorrect extraction of a root.
appa609 20 hours ago
4 cannot = 5 so its false.
55GamePlayer 1 day ago 7
@topicmirsad We keep going in circles. I do realize that you can't divide by zero. For a moment, forget about the solution set (4,5,1) and look just at the algebra. What I am trying to say is that all algebraic rules used are valid, even up to the point where he writes A = B, which we both know isn't true with the given solution set. Every step of his algebra is valid. Step 2 created the anomaly. Thanks
Steven01001 1 day ago
@Steven01001
no thats wrong. before dividing by variables e.g. (B-C-A) you have to proof that there is no combination of A,B and C for which the expression (B-C-A)=0. for sums like the one in the video THIS IS NOT POSSIBLE. you cant rule out the possibility that (B-C-A)=0 therefore you are not allowed to divide by it. one of the few expressions where you are allowed to do so is if the variable is a power, e.g. 5^x because there is no number for which 5^x=0.
Creepyseven 1 day ago
is 3:08 wrong?
MoRe2PyroBass 1 day ago
i dare you click F13
OhHowTo 1 day ago 7
@OhHowTo >:(
lol
WOLFGANGFIGHTER52 13 hours ago
@OhHowTo if you have a mac, that's possible
combatReady1538 3 hours ago
omg what a noob! how old is he? 8 maybe :D
fanff3000 1 day ago
lol he made a youtube account for 1 video
thebiggestidiot1 1 day ago
with my magic fairy 2+2= fish jajajajajjajajajaja
firebeetlese 1 day ago
translate to english
FPSmini 1 day ago
Why the hell can you not just write the numbers dammit.
dalynthesk8r 1 day ago
This has been flagged as spam show
You factored A out of the left side, and B out of the other... so you divided one side by 4 and the other by 5. And that was to get A*0 = B*0, which is true, but not a logical way to prove equality. It basically says: both are insignificant in the face of infinity, thus they are equal.
Dragon20942 1 day ago
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Dragon20942 1 day ago
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HappyFolkPinoy 1 day ago
ur stupid as hell
iluvsasukeful 1 day ago
...you lost me at A=4
duckpondproductions 1 day ago
fail, division by zero -.-
exagi 1 day ago
(B-C-A)=0 Ax0=Bx0 即使A與B不相等 相乘的結果還是=0
a29938198 1 day ago
@topicmirsad Let (B-C-A) = X Then (A(B – C – A)) / (B – C – A) = (B(B – C – A)) / (B – C – A) becomes AX/X = BX/X which equals A=B so dividing by (B-C-A) is algebraically correct.
Steven01001 1 day ago
@Steven01001 You are dividing by zero again!
It is clear that A and B are two different numbers (in this case 5 and 4, but let us disregard that), then A*X=B*X, only, and only if X=0, so you can't divide by it
topicmirsad 1 day ago
@Steven01001
no its not, because A,B and C can be chosen so that (B-C-A)=0 therefore you are not allowed to divide anything by it (not even itself). before dividing by something you ALWAYS proof that its not zero.
Creepyseven 1 day ago
B-C-A=0 sO simple!
hope8414 1 day ago
your face is wrong
ProdigyofHappiness 1 day ago
If this equation is true than it equals 2+2 and if it equals 5 than all of that other crap is unnecessary and shows that you OBVIOUSLY did something wrong
kidn00b1 1 day ago
Comment removed
kidn00b1 1 day ago
I have 2 apples in one hand 2 apples in the other, Put them in hand I have 4, next question?
kidn00b1 1 day ago
5-1-4 = 0
SoulAcid1 2 days ago
in step two he multiplies the left side by (B-A) and the right side by (B-A)^2
infitiae 2 days ago
@infitiae No. He multiplyied both sides with (B-A) . You probably mistaken that saparator in first line for a division sign, like many others here. First line is C=B-A, and not C=B-A/(B-A)
topicmirsad 1 day ago
I would like to prove you wrong but then I took an arrow to the knee
Dhanushenr 2 days ago
How do Line 3,4 work?
5up3rp3rs0n 2 days ago
ok .....2+2=4 it s elementary
now in your algebra ...YOU decide the valour of A B and C....;then you can find the solution you want and then 2+2 =5 but only for you buddy .....;cuz you pick the number you need for A B and C to get to your response
SO DON T THINK TOO MUCH MISTER THE NUMEROLOGIST
you ll get a complete defferent result if
A =3
B=8
C=2
go on ! waste another piece of paper and a little bit of your time and see where it go ....
TheLio666 2 days ago
@TheLio666 He's perfectly aware of that ... In case you didn't notice this video is titled "Good math TRICK"
topicmirsad 1 day ago
Why am I watching this ????
seanyoutub2010 2 days ago
U stupid? U cant devide by zero duh
ionjo2000 2 days ago
One of the last steps where he crosses out (B-C-A) on both sides is illegal. To do this you would need to divide both sides by (B-C-A) yet that is dividing by zero. B-C-A = 5-1-4 = 0.
WarmWeatherGuy 2 days ago 26
I SUCK AT MATH AND HATE IT, WHY AM I HERE
MrPrzyballa 2 days ago 2
i feel like an idiot
peruvianamerican101 2 days ago
This has been flagged as spam show
google "( algebraic Extraneous solutions: multiplication wikipedia"
Steven01001 2 days ago
@Steven01001 That all is for multiplication with zero, and other non-reversible operations. But in this case, none of that happened, because he started with all different numbers, and none of them is eqal to zero.
topicmirsad 1 day ago
Dang, I need to read before I post. Multiplying the 2nd step by (B-A) caused this error is what I meant to say.
Steven01001 2 days ago
@Steven01001 :) ... no bad ... I understand what you wanted to say ...
topicmirsad 1 day ago
Opps I meant to say the second step, not the 2nd to last
Steven01001 2 days ago
A(B-C-A)=B(B-C-A)
AB-AC-A^2=B^2-BC-AB
AB-A^2-B^2+AB=AC-BC
AB-A^2-B^2+AB=C(A-B)
C=(2AB-A^2-B^2)/(A-B)
C=(2AB-A^2-B^2)/(A-B) =(A-B)(B-A)/(A-B)
C=B-A which is the original equation, but the error occurs when he divides both sides of the 2nd to last step by (B-C-A) which is algebraically correct. Adding additional variables (at step 2) can cause this as I alluded to twice before. Why do you think he got the wrong answer when his math is correct?
Steven01001 2 days ago
@Steven01001 He didn't add any variables in step 2, and in any other step. First he was multiplying both with (B-A), then he subtracted A^2 from both sides, then he added AB to both side. Those are all perfectly correct algebraic and mathematical operations.
The ONLY problem here is that division by zero in last step,and nothing else. Getting to that step is perfectly correct.
topicmirsad 1 day ago
This has been flagged as spam show
@Steven01001 He didn't add any variables in step 2, and in any other step. First he was multiplying both with (B-A), then he subtracted A^2 from both sides, then he added AB to both side. Those are all perfectly correct algebraic and mathematical operations.
The ONLY problem here is that division by zero in last step,and nothing else. Getting to that step is perfectly correct.
topicmirsad 1 day ago
This has been flagged as spam show
@Steven01001 He didn't add any variables in step 2, and in any other step. First he was multiplying both with (B-A), then he subtracted A^2 from both sides, then he added AB to both side. Those are all perfectly correct algebraic and mathematical operations.
The ONLY problem here is that division by zero in last step,and nothing else. Getting to that step is perfectly correct.
topicmirsad 1 day ago
@topicmirsad You are correct about that step equaling zero, but in the next step he divides both sides by (B-C-A) which is algebraically correct, but results in the wrong answer. If (B-C-A) were anything but zero, the solution would be valid. In this case, it is not. The solution set he provides is valid for the original equation, but not for the final equation, which I repeat is algebraically correct.
Steven01001 2 days ago
@Steven01001 In this last step dividing by (B-C-A) is not even algebraicly correct. If we disregard given set of values and just say that A does not eqal B. Then, for this equation to be valid A(B-C-A)=B(B-C-A), (B-C-A) must be eqal to zero. In any other case eqation is not valid, and therefore you can't cross it out in any case. Only case in wich you would be able to cross them out is when A=B, but we started with assumption that A does not eqal B, just as in this video
topicmirsad 1 day ago
@topicmirsad It's a proof, it doesn't matter what the actual numbers are. The last step is valid, but the proof isn't. He made an error in step two. Everything beyond that is null.
Step two should be:
C(B-A) = B-A
((B-A) / (B-A))(B-A) is not (B-A)^2. It is B-A. From then on, he is running with 1=2.
bryanhaw1 1 day ago
@bryanhaw1 Las step is not valid. (B-C-A) is eqal to 0 in any case, because A and B are different numbers. It would be valid if A would eqal to B, but in this case they are different numbers
topicmirsad 1 day ago
@bryanhaw1 And step 2 is perfectly correct because firs line is C=B-A, and not C=B-A/(B-A). You mistaken the separator line for a division sign...
topicmirsad 1 day ago
@bryanhaw1
It's a proof that A = B when C = B - A. But it works only if (B - C - A) is unequal 0, because you are dividing by it.
You assume that C = B - A.
Subtract C from both sides and you get 0 = B - C - A.
So you assume that (B - C - A) = 0 and you can conclude from this that A = B, but only if (B - C - A) is unequal 0.
So you can say that A = B only if (B - C - A) = 0 AND (B - C - A) is unequal 0.
So you haven't proven anything, even if you ignore what the actual numbers are.
timsr 1 day ago
Bullllcraaap
lioncub2001 2 days ago
@topicmirsad
Search wiki for division by zero.
Look under Fallacies based on division by zero section, It has exactly this example explained.
But Midenbinder already knew that or he wouldn't just crossed it like that, He would divide both sides by B-C-A just like he did till that very point.
This way it is less obvious that one of the golden rules in mathematics has been broken.
odise25 2 days ago
It is not an issue whether the equation is true or not.
Issue is the conclusions you draw from that equation.
Example:
x*0 = y*0
We know this equation must be true because 0=0 is solution.
Can we draw a conclusion that x=y from this equation?
I hope your answer is no, because anyone that answers yes needs to go back to elementary school and learn some math.
odise25 2 days ago 19
@odise25 Why would i need to search Wiki? I perfectly know what is divison by zero, but some other people here don't understand that "division by zero" is causing this weird result at the end, and all i'm trying to do is explain it to some people here.
topicmirsad 1 day ago
@topicmirsad
I posted that to you because I thought you had problem explaining this fallacy to others.
You are arguing with more than 10 ppl at once and i thought you need help.
Just point them in right direction (in this case wiki) and everything should be clear even for people with weak math skills.
Most people like to see some evidence for claims like "division by zero is not allowed in math".
odise25 1 day ago
@odise25 Sorry ... I was thinking that you were talking to me, that i need to learn about division by zero ...
Thanks for help, but as you can see, i doubt that even that will open some eyes here :D
topicmirsad 1 day ago
@SuperPandaWarrior
i don't really understand what you're saying lol... is it math is it english i dunno... all i know is im not wrong it's been proven that you cant divide by zero. for instance (a / 0). In ordinary (real number) arithmetic, the expression has no meaning, as there is no number which, multiplied by 0, gives a (a≠0), and so we say that division by zero is undefined.
you can't say that im wrong... its math godddd lol
anywayz look it up big boy ;P
ZeroCool0090 2 days ago
@samk109 What is wrong? Wich one of my comments you think is wrong?
topicmirsad 2 days ago
@NelsonIncredulous Don't think so ...
topicmirsad 2 days ago
@SloMoHeap Why? Run out of arguments?
Again, i said for eqation to be true (B-C-A) must be eqal to 0 and therefore you can not cross it out.
topicmirsad 2 days ago
a load of fucking rubbish
Theklstuff 2 days ago
When you substracted CB you shouldn't do this like B^2 -CB-AB, but like B^2-AB-CB. Sequence of actions is matter.
MrOrnion 2 days ago
The error is in the factorization at the end i.e it's an illegal factorization. Because it's an equation, you have to do the same operation on both sides. If you factorize by B on one side, you need to factorize by B on the other side. If you do that in the second to last step, it'll work out perfectly.
i.e b (-ca/b -a^2/b +a) = b (b - a - c) -----------> 0 = 0. And the world is right again.
zedtransform 2 days ago
so u r a sixth grader right
and had a E last test haha
or is it me
equap 2 days ago
One more thing and then I'll be on my way. If you solve the last equation for "C" you should get the original equation, which you don't, therefore and even though the algebra is correct, the original equation and the last one are not similar. I again say that in some cases, multiplying by variables as done here in step 2 can create extraneous solutions. It was fun, thanks all!
Steven01001 2 days ago
@Steven01001 A(B-C-A)=B(B-C-A)
AB-AC-A^2=B^2-BC-BA
-AC+BC=B^2-BA-AB+A^2
C(B-A)=B(B-A)-A(B-A), when we divide eqation with (B-A) we get
C=B-A
Thank you, i'll be here all week ... :D
topicmirsad 2 days ago
@hiragica Thank you, much appreciated.
Steven01001 2 days ago
If you don't believe me then solve the last equation for C and using a graphing calculator, graph out the first equation and your new one and you will see that they are totally different.
Steven01001 2 days ago
lol no /-A^2 shows that he's adding -A^2 to both sides. it only shows what his next step is gonna be. A.(B-C-A) = A(5-1-4) = A(zero) u cant divide by zero -.-.
ZeroCool0090 2 days ago 30
@ZeroCool0090 wrong, 1/3 = .3 repeating, .3 repeating x 3 = .9 repeating, therefore .9 repeating = 1, therefore 1 - .9 repeating must = 0, but it actually equals 0.0 repeating with a 1 at the end of the infinite amount of 0's, therefore anything divided by 0 is infinity :3
SuperPandaWarrior 2 days ago
@SuperPandaWarrior how do you have a 1 @ the END of an INFINITE amount of zeros?
Grivo5000 2 days ago
@Grivo5000 ahhh, it doesnt matter how, the fact of the matter is that it IS there
SuperPandaWarrior 2 days ago
It got me a 4.0 during my engineering degree.
Steven01001 2 days ago
dude, your math is pretty poor:)
adyvulea 2 days ago
The last equation where (B-C-A) is on both sides, he crossed them out as if he had divided both sides by (B-C-A) which is algebraically correct, except in this case, "with the given constants",... (B-C-A)=0 which makes division not mathematically correct. Everyone seems to see this, but fails to see that step 2 is where this wrong solution originated. A=4, B=5, C=1 is a solution set for C=B-A, but not for A(B-C-A)=B(B-C-A)
Steven01001 2 days ago
@Steven01001 best answer
hiragica 2 days ago
all the steps are correct ... but you can NOT divide by 0 .. (B-C-A)
hiragica 2 days ago
@Steven01001 Why A=4, B=5, C=1 is not a solution for A(B-C-A)=B(B-C-A)? When you plug them in you get 0=0. What is wrong with that?
If you plug them in first eqation you get 1=1, let's now subtract 1 from both sides of this eqation and we get 0=0? Is that not the same?
topicmirsad 2 days ago
@0815Nolifer Agreed, the only error is at the end, but as I said... His algebra is correct except for not realizing that when you multiply by variables (as in step 2) it can create extraneous solutions as it has in this case with the constants he used. Why else would his final solution be wrong after plugging if the constants when his algebra is correct up to that point? Look it up, you'll see what I'm talking about. Multiplying by (B-A) in step 2 caused the false result in the end.
Steven01001 2 days ago
@Steven01001 Multiplying both sides of equation with same "things" ALWAYS gives correct solutions, and eqation we started with is the same as the one we got after we multiply eqation with "something" (number, algebraic phrase, ...)
The last eqation is same as the first one. Try to solve A(B-C-A)=B(B-C-A) for C, and you'll get C=B-A.
topicmirsad 2 days ago
@ThatRubikGuy I won't argue that the last result creates the error when the given values are substituted, but as I said, step 2 is where the error is created which makes the solution set valid in the original equation, but invalid in the final equation.
Steven01001 3 days ago
@Steven01001
Just substitute A B and C and check if it is correct. Only error is at the end.
0815Nolifer 3 days ago
There's nothing right about this. His derivations are all completely incorrect. Going to step 2, the right side should not be squared. Assuming that was even correct, going to step 3 does not introduce division by -A^2. I mean, where did that even come from? And it continues from there...
bondsbw1 3 days ago
look up ( algebraic Extraneous solutions: multiplication wikipedia ) and you will understand where you went wrong. Otherwise, your math is correct.
Steven01001 3 days ago
If you substitute the proposed values for the variables in step 2, it will give you the wrong answer.
Steven01001 3 days ago
@Steven01001
No it won't. Stop being stupid please.
C=1, B=5 and A=4
CB-CA-A*A=B*B-2AB
1*5-1*4-4*4=5*5-2*5*4
5-4-16=25-40
1-16=-15
-15=-15
The invalid part is at the end when he divides by zero.
ThatRubikGuy 3 days ago
Step 2 is invalid and can happen when multiplying both sides by variables.
Steven01001 3 days ago
Comment removed
Trias88 3 days ago
Guys, you're missing it here. Because the letters have set values, this formula will only prove true if you can plug in the values and work it out the same way. You can't without dividing by zero.
eagles4christ 3 days ago
Lol at the math noobs who think he's NOT dividing by zero... to "cancel" something out you MUST divide it by itself. You don't magically make B-C-A disappear, you DIVIDE BOTH SIDES to make it happen. Sorry for the caps, just I would assume that highschool / younger kiddies need capital letters to understand what to focus on.
mercscout 3 days ago
@mercscout Which part are you talking about near the end? I dont see him divide by zero, he has A(b-c-a)=b(b-c-a) and he divided BOTH SIDES by (b-c-a), unless your trying to state b-c-a=0, and he is diving by zero
dabestplayer 3 days ago
@dabestplayer
Yes B-C-A does equal zero. And he is dividing by it. 5-4-1=0
ANY algebra trick you can see on youtube titled "OMG I BROKE MATH" is either dividing or multiplying by zero. That's why when you divide by something, you need to make condition it's not zero. When you have an equation and you want to divide it by X, you are making a statment that x=/=0, then you solve the equation for x=0.
ThatRubikGuy 3 days ago
@dabestplayer I'm not 'trying' to say it. I am saying it. In the first 30 seconds of the video he defines a,b,c and if you plug in their values we get b-c-a=0 :), 0/0 DNE
mercscout 12 hours ago
the algebraic operation used to cancel those things from both sides is division..
BerylBite 3 days ago
in order to remove (b-c-a), you must multiply by reciprocal. reciprocal of 0 is a division by zero. cannot divide by zero. arguement is invalid.
finalzeroone 3 days ago
There is a mistake in your equation:
On the second line, you had (B-A)^2
And on the third one, you had B^2 + A^2 - 2AB
That shouldn't be the case, seeing as it's by the power of 2 and not multiplied by 2. The equation should just be:
CB - CA = B^2 + A^2 and even there, it's wrong. seeing as on the left, the solution is 1 while on the right, the solution is 61.
cheftoniFTW 3 days ago
@cheftoniFTW no that part is actually correct, you dont simply add a square to both sides thats incorrect math, when given (B-A)^2 you have to F.O.I.L it, multiply the first terms then outer then inner than last, so (B-A)^2 equals (B-A)(B-A) first terms multiplied eqal B^2 outer: -AB inner: -AB and last: A^2 which equates into B^2 -2AB + A^2
Airman32292 3 days ago
@Airman32292 But even then, shouldn't the equation then be:
CB - CA = B^2 + A^2 + 2AB (As opposed to - 2AB, seeing as 2 negatives become a positive?) And in both cases, the - 2AB would make the equation as 1 = 21, and the other as +2AB would become 1 = 101?
cheftoniFTW 3 days ago
@cheftoniFTW just to say i dont agree with this problem as a whole, but no it wouldnt be positive, once you do the F.O.I.L method you then have terms that you add together, not multiply, so you add B^2 + -AB -AB + A^2, which is the same as B^2 -AB -AB + A^2 which equals B^2 -2AB + A^2
Airman32292 3 days ago
2x2=😡
2LEGIT2SWAGG 3 days ago
Porn music!?
2LEGIT2SWAGG 3 days ago
squared (B-A)
jamnotakusingshoribl 3 days ago
CB-CA=B^2-2AB+A^2/-A^2 is false you would come up with 1=-14 which is wrong. to prove it. 1*5-1*4=1 so 1=5^2-2*4*5+4^2/-4^2 so reduced it is 25-40+16/16 so reduced even more it is 25-40+1 so -15+1 which = -14 and last time i check -14 doesn't = 1
ajerrils 3 days ago
@ajerrils That's not what he meant rofl, the /-A^2 part refers to the step being taken, not a division. The notation isn't ideal but I'd have thought at least that was obvious.
MakiRole 3 days ago
@ajerrils the /-A^2 is not part of the equation. it's showing what to do next.
eagles4christ 3 days ago
Love how so many people have gone through and said "Oh you've done done all teh steps wrong herp derp". The original statement says C=B-A, so naturally 0=B-A-C or for here 0=B-C-A. Of course if you have that then it's no different to multiplying that part with say A, or B or C; you're just multiplying 0 by something. The statement's perfectly valid, it's just that if you put any values in you'll get 0=0. So many have given thumbs down for illegitimate reasons it's painful.
MakiRole 3 days ago
@MakiRole Lets set aside the many errors in math throughout the problem... in the end in order to cancel out (B-C-A) on each side of the equation you must divide by zero, which you cannot do.
OonslaughtoO 3 days ago
@OonslaughtoO Naturally, I'm just saying it undoes itself before that point too as each side just ends up being 0 anyway.
MakiRole 3 days ago
@OonslaughtoO There is no division by zero, he is simply removing from both sides a similar equation seeing a both (B -C - A)s are equal, they cancel themselves out.
cheftoniFTW 3 days ago
at 1:44 he was saying "shit shit shit shit!!!!........shit shit shit!!!"
quigz125 3 days ago
i dont understand
Allistermills 3 days ago
why so much hate, its a riddle and not like "hey look i found a mistake in the universe" ..
FunnyRumel2 3 days ago
Niestety chłopie ale z tego że każda liczba podniesiona do potęgi 0 daje 1 nie oznacza że wszystkie liczby na świecie są sobie równe :)
NieWyimaginowany 3 days ago
this is the shittest thing i have ever seen!
l0rd0f0blivi0n 3 days ago
lol porn music
victorito95 3 days ago
Nearly every step is wrong.
immattmac 3 days ago
TBF - CBA
Azzadaarcher 3 days ago
berybite its ok and can fesics law or some one some how tell me why 0/0 its not ok it is undeterminate u can not reduce every thing in 1 equation becose 0 i dont untherstand how fisics can reduce every thing in one ecuation.
dgslayer08 3 days ago
@dgslayer08 What?
Aimfortheeyes 3 days ago
/-(A^2)?hhhh you are veryy crazyy good work enstien ? hhh :D
mohamemdamine 3 days ago
@mohamemdamine why crazy: when you take away A^2 from both sides you have
CB-CA-A^2=B^2-2AB+A^2-A^2
17=-15
but if you take -(A^2) from both sides you get:
CB-CA-(A^2)= B^2-2AB+A^2-(A^2)
-15=-15
so where is my mistake dude?
the trick is coz every number (negative or positive) raises to square power is positive e g. -2^2=4 and 2^2=4
greetings from Poland
niespokojnadusza83 3 days ago
instead of adding the A squared you subtracted it and you supposed to add it cause it was a negative dummy! you guys are so dumb i saw it right away
skinnychick5 3 days ago
=4
4223lati 3 days ago
I like the porno music.
JMB661 3 days ago 8
@JMB661 well,thats random and true story
lesserflamingo 3 days ago
C=B-A/(B-A) is ok :L then you tried to multiply both sides by B-A? so you should get C(B-A)=B-A not C(B-A)=(B-A)2 you cant just say 'oh ill add a little squared sign on it because my balls hurt..'
mafoooo911 4 days ago
I was expecting something funny... without commentary all I see is a mistake with steps 2-5... (B-A) gets squared for no reason, divide the second equation by -A^2 for no reason.. dvide by -AB, IDK what your doing at all here... the whole thing is a giant mistake.
walkingpen 4 days ago
The mistake is in the second equation. B-A/(B-A) is not equal to (B-A)^2 after factors are multiplied. It's equal to B(B-A) - A = B^2 - AB - A
Steven01001 4 days ago
if there is no A B C ???? 2x2=4
RapsyJigo 4 days ago
Why use letters instead of the numbers to write it? It would be 10 times easier to spot your mistake, although I already know you must have made one. Haven't you even been to elementary school? I think they taught the two times table there already!
aabbcc0837 4 days ago
@aabbcc0837
But the point of the video is that the "mistake" shouldn't be so easy to spot...
ImGonnaShout2000 4 days ago
Nerds.. always hated them in school.
KeybladeKorney 4 days ago
数字でもっかいkwsk
yuyu1365 4 days ago
B-C-A=0 and you can't divide with zero! :D
Dani004able 4 days ago
@Dani004able yes ur right.
zetarobot 4 days ago
B-C-A = 0....cannot divide both sides by zero you stupid fuck
MrAp43 4 days ago
Comment removed
aluingoogl 4 days ago
Silly Nerds, if you just realised you made a mistake at the start, 2x2=4 ! All them equations and it was this simple after all ! ;)
ryulovesyou 4 days ago
3rd wrong :))
blackocean5955 4 days ago
3rd line is false. cb-ca =1 so 1= bb-2ab+aa/-aa, so 1=25-40+16/16, so 1=1/16 is the 3rd line result which is a false statement. I am sure later in the video they devide by zero like all the other silly tricks, but they usually come after a false statement.
nodice4me 4 days ago