nice one. just because two things have equal squares, doesn't mean those two things are equal. they're equal in magnitude, but not necessarily equal in value.
If there's ever a nonsensical answer and there's a square root of a squared value... that's pretty much always where the issue is. Either that or a division by zero hidden somewhere (which is a popular route to 1=2 I believe)
These sorts of tricks almost always fall into one of two categories: Division by zero, or improperly taking square roots. Once the "π² - 9" showed up, I pretty much knew where this was going, and just waited for the square-root step to come along, and bingo, "there's your problem!" "Take the square root of both sides" is pretty much always a red flag in these sorts of "proofs", to be honest...
If I were teaching an introductory proofs class, this is the first lesson I'd show in "spot the problem" proofs. Show the students how common these two tricks are to create fake facts.
@@MaximxlsThere's also one with extraneous rots and also false geometric proofs like the one in 3blue1brown's channel where he proves all triangles are equilateral.
My favourite "find the mistake" was one with integrals that I think started with evaluating the integral from 0 to 1 of 1/x specifically by integration by parts, and the [x*1/x] from 0 to 1 they say it's equal to 1 "because the x and 1/x cancel", but 1 from 0 to 1 gives 1 - 1 = 0.
Just because a^2 = b^2 doesnt mean a = b, since squaring is a non injective function. For example, (-1)^2 = (1)^2, but -1 is not equal to 1. Futher a^2 = b^2 can be simplified as a^2 - b^2 = 0 (a + b) * (a - b) = 0 here either a + b = 0 or a - b = 0 so a = -b or a = b
In the beginning, you assumed that x is equal to three plus pi divided by two, so x is actually a smaller number than pi, for example, something around 3.07. So when we take the square root of the expression (3-pi), it must be the absolute value. Let's take it out, which becomes pi-3. As a result, after solving, we reach the initial equation.
When I actually did higher level math classes, I didn't just write 'line by line' like this, but instead implications. This is because one line could imply up or down, but not necessarily the opposite direction. For instance, [a^2 = b^2] [|a| = |b|] but, you can only assume the general case in one direction [a^2 = b^2] [a = b] (incorrect) since it can otherwise be proven, in most cases, by showing that both numbers inhabit the same side of 0. People (of low and common intelligence) will tend to jump to the conclusion that this false proof is a gimmick and just annoying trickery yet it is actually the opposite: it exposes the trashy syntax commonly used and learned in elementary to high school. "line by line" is not explicit enough to show which direction the implications work.
Another way to think about this is that the square of a number is the square of its negative. Because x is the average of 3 and π, it is the same distance from both of them, therefore, (x-3) and (x-π) will be the additive inverse of each other.
That's so logical and interesting property. Taking average of 2 numbers and subtracting each number from average respectively gives additive inverse. Fascinating! Is there a name for this property?
@@ingiford175 Adding a set of numbers together and dividing by the count of those numbers gives the mean, also knows as the average. In this instance pi and 3 are added together and divided by two, thus getting the average.
The mistake is that id (x-3)^2 = (x-pi)^2, it means that either (x-3) = (x-pi) OR (x-3)=-(x-pi), in this case the correct alternative is that x-3 = -(x-pi), which leads to 2x = 3+pi (the initial assumption).
*When taking the square roots, you can't just cancel out the square root, you need to put a plus-minus sign on one of the sides, becase x-π and -(x-π) both square to become (x-π)².*
In an astrophysics module at uni, during a derivation the professor said "pi is approximately 3, 3 is approximately 1, so we can just cancel it". Since then pi = 3 hasn't seemed so bad.
The mistake is that roots, and powers, do not cancel each other out. squaring, and square-rooting, results in the absolute value of whats inside the root. So for sqrt( (-a)^2) we get |-a| = a and for for sqrt( (a)^2) we get |a| = a
First error: At step two you multiply with (pi - 3), wich makes the equation true for pi = 3. Second error: Square and Squareroot don't cancel one another, but sqrt(x^2) is |x|. So you remove solutions. If you take x-3 = -(x-pi) and solve for x, you get the equation from the start.
That's very interesting. I missed that bit. If the solution is correct and pi = 3 then pi - 3 = 0 and at that step you have multiplied both sides by zero, effectively making everything afterwards kinda moot anyway.
I found it!! Why did you take (x-π)² instead of (π-x)² ? If you would have taken this (π-x)² , then it would come as x=(3+π)/2 ( Just like the equation that you provided earlier)
That's why we create and put "" sign at the beginning of each row. The sign will explain for itself, it makes up a link throughout the math and clearly shows that the first row equals the 6th, but stop it to becomes the 7th (as others have explained).
The equivalence is already broken between the 1st and the 2nd line (if π=3). And also, all this derivation following the 2nd line is a distraction. I mean, by looking at the 2nd equation, we immediately see that π=3 is a solution 😉; no need to go on further.
I didn't really like this problem. Even if there would be no mistake with absolute value, and we would get PI=3, it would simply mean that there is no soultion for x, because regardless of its value the equation is false. In order to show that PI equals 3 somehow, we should treat PI as a variable, and then there are infinitely many soultions to the original equation (linear equation with 2 variables x and PI). What I mean is that the context of this problem is very confusing
You can't square root both sides.( In particular, from the definition of x, x is halfway between 3 and pi, so x-3 is positive and x-pi is negative so they cant possibly be equal.) I liked this one, it's better than the usual 'hidden division by zero' trick.
Why switch to a value for x? You can stay in exact form, knowing {pi} is 3+ (a bit bigger than 3). Then x-3 is = ({pi} + 3)/2 -3 = ({pi} +3 -6)/2 = ({pi} - 3)/2 > 0 since {pi} = 3+ > 3 And x - {pi} is = ({pi} + 3)/2 - {pi} = ({pi} + 3 - 2{pi}) / 2 = (3 - {pi})/2 < 0 since {pi} = 3+ > 3
Replace π with θ or 'a' or any other constant. The actual step where the problem lies is when you multiply both sides by (π-3) because if π=3 then π-3=0 and you are multiplying both sides by zero
Also would like to point out something else, if pi=3, you cannot multiply both sides by (pi-3) in the 2nd step because multiplying by zero would just result in 0=0
the issue here is, if we are going to proof a statement like p = q be true and then we multiply both side by a constant (a - b) we got (a-b)*p = (a-b)*q, then the statement will be true anyway when a equals to b since 0 = 0 but this do not imply we have proven the original statement p = q is correct let put a real number on it (k - 1) * 1 = (k - 1) * 2, the statement is true when k = 1, but 1 is not equal to 2 so when we are doing such multiply like (pi - 3), we need exclude the possibility of pi equals to 3, otherwise the rest move would be meaningless
I actually think it is in the second step, multiplying both sides by pi-3 adds another possible solution to the equation where pi (being used as a variable) is equal to 3 making the whole equation a 0=0 true statement.notice if you just replace pi with 3 in the third step you get 6x-6x=9-9.
pi=3 technically incorrect, but when you rounded it up to make calculating easier. because you can just APPROXIMATE laziest way when doing choice test.
When your take the square root of a square involving variables, the only thing you can say is the absolute values of the quantities are equivalent, not the quantities themselves: i.e. |x - 3| = |x - pi|. This leads to you having to check 2 cases: x-3 = x-pi [ equivalent to -(x-3) = -(x-pi) ] -(x-3) = x - pi [ equivalent to (x-3) = -(x-pi) ] The first equation leads to the false solution that 3 = pi The second equation returns the original assumption for x = (pi + 3)/2
As a nuclear chemist in NASA, headquarters in Malmo. Me and my collegue found this rational misstake quite easily with the help of LeChatlier’s Principle
(x-3/2) and x-π/2) are actually the negative of eachother, so it's natural for their squares to be equal ( (x-3/2)² = (x-π/2)² ) but not the other way around.
The mistake is really in logic. To say this is true would be to say that π is a variable not a constant. This does not prove the value of π since there’s no approximation that is made. It’s basically assuming π is not a constant. And also because of the square root of the squares generating extraneous solutions.
@@ShanthiVeluswamy it is true that's why we have to assume that this is true which means after further calculation we have to reject the case of pi minus 3 equals to 0.
Without looking at the rest of the comments, this is where the mistake lies: x is the arithmetic mean of 3 and п from the start, so it is in the interval (3,п). Thus x - 3 is positive and therefore equal to |x - 3| = sqrt((x - 3)^2), but x - п is not equal to sqrt((x - п)^2), but its negative. This is where the mistake lies.
How do you know that x-3 is positive though? X is only positive if pi is greater than 3. The problem I have with this explanation is that your assuming that pi is greater than 3 to prove that pi is not 3.
@@raphaelnouh1442 п > 3 is a mathematical fact. п = 3 is a contradiction which proves that the "proof" is wrong because the working out is wrong. I showed, using the fact that pi is greater than 3, that his working out was wrong. Second Order Arithmetic + "п = 3" is an inconsistent theory as it proves its own axioms incorrect, thus you shouldn't work in it.
@@newwaveinfantry8362 I know pi is not three, I just think you can’t really use “pi isn’t three” to prove pi isn’t three. If you wanted to do that you might as well have just said “pi is 3.14…, therefore it cannot be 3.”
pi + 3 < 2pi => X < pi => x - pi < 0 => sqrt(x -pi) = |x-pi| = Pi - x, not the other way around. So the mistake is right after the square root, where you assume x - 3 = x - pi
1:47 I am going out on a limb and say the mistake is that you cannot add x² to both sides of the equation - and then start splitting the variable (x) on both sides. It should be separated, only on one side of the equation.
Funny how during your proof that pi=3 you inserted 3,14 for pi on the step to eliminate the square root 😅 However, there's another thing to point out: A critical mistake already happens at step 2, where both sides of the equation are multiplied by (pi-3), which - as it turns out by the end - equals to multiplying both sides by zero! So when ignoring this, no matter how the calculation goes on, anything could be proven.
Here pi is the variable i.e, we are trying to find pi...and we have multiplied by pi-3 on both sides which clearly implies that we have introduced an extraneous root for pi which 3. If anyone thinks that what I said is false then think about it this way...by using the argument shown in the video(which is false) you can show that unknown value 'a' is equal to a number 'b' by just multiplying the equation by (a-b)
A general misconception is that any x when squared and rooted we get x. But ans is lxl (mod x) so it doesn't have only 1 answer people are taking it seriously but as a student who is preparing for jee examinations its pretty obvious
1:18 Square root opens with mod always , so this time X - 3 = +( X-Pi ) is not a solution as 3 is not equal to pi , so we use X-3 = -(X - Pi ) to get back to original answer of x=pi+3/2
When you root (x-pi)^2, you don't get x-pi, you get |x-pi|. As it turns out, x-pi is negative, so when we take the absolute value, you get pi-x. (It's the same concept for x-3, but x-3 is positive so you can just drop the absolute values). Once we fix this mistake, we get the original solution of x=(pi+3)*1/2
I did not spot the problem, didn’t remember the trick, but your explanation made sense because we know pi and you were able to logically deduce the answer to begin with (approx 3.0). I assume you would need different explanation if pi was unknown, say Y.
Exactly as I thought You need to remember that √(u)² = |u| So you should get |x - 3| = |x-π| You can solve this piece wise and you'll get two different cases in the end The π = 3 case And the original x = (π+3)/2
Yeah, (x-3)^2 does equal (x-pi)^2, but this does not imply x-3=x-pi because, in fact, x-3=pi-x. And (x-pi)^2 does equal (pi-x)^2 because one is the negative version of the other. Can't just take the positive square root once you've squared stuff, you might need the secondary root as you do here.
This is how I imagine Greek philosophy to have been like... arguing for the sake of arguing. I think you enjoy doing math so much that now you're creating problems instead of solving them.
This is quite tough tho. But as you can see , if (x-3)^2 = ( x - n) ^2 ( sorry i have no pi symbol) than it could be x - 3 = x - n or x-3 = - ( x - n ) so pi wont be equal to x at this because we there's two situation of x so we could easily know that the first one is wrong and the second is the correct. For the second one, as x - 3 = -(x - n) => x - 3 = n - x => 2x = n + 3 => we got the first equation which is x = (n+3) / 2. So, the mistake in this one is that the sqrt (x+3)^2 isn't not equal to x+3, it is |x+3|, same to the sqrt of (x+n) ^2, it is |x+n| . From |x+3| = |x+n|, we have two situation as I said before which is x - 3 = x-n or x-3 = -(x-n) ( it could be reverse)
The step where the writer concludes sqrt((x - π)^2) = x - π is incorrect, because sqrt((x - π)^2) = |x - π|, and since x - π < 0, |x - π| = -(x - π) = π - x. This is a classic mistake many students make. Teachers often mistakenly teach that sqrt(x^2) = x for all x, which is false. sqrt(x^2) = |x| for all x is the correct statement.
Not a divide by zero but if pi = 3 is a valid solution then the multiplication of pi-3 on each side in step two just multiplied both sides by zero and everything else after is meaningless, including getting the solution that pi = 3 in the first place. Just another way this is wrong besides the sq rt pointed out in the video.
The mistake is in the square roots. Substituting x = (3 + pi)/2 at the line (x - 3)^2 = (x - pi)^2 you get ((pi - 3)/2)^2 = ((3 - pi)/2)^2 - which is clearly correct - but after taking square roots it isn't. You should end up wth x - 3 = pi - x or x = (3 + pi)/2
i think there's another mistake, we multiplied both sides by (pi-3) in the beginning, but we came to that the pi=3, but then pi-3=0, so we just multiplied both sides by 0 and that makes no sense and breaks the equation
Also in a parallel universe where pi is actually 3, you cant multiply your term with pi - 3 on both sides. If pi was actually 3, you'd multiply with zero which trolls your entire proof.
There is another case you need to chek. when you do the Square root you need to add a case when you put minus in one of the sides and then you get the correct answer
1:42 so you can't actually take the sqrt there. basically the sqrt has a branch cut and by squaring and taking the square root the left-hand side and right-hand side end up on different branches. if a^2 = y, then y = +a or -a now in our case one of the sides is +a and the other is -a
The error is logical, you cannot interpret Pi as variable and constant at same time. This is cause for presented statements cannot be considered as proof. Note, in presence of this error, no matter how correct were manipulations itself. Actually, you have found one correct solution (Pi=3, x=3) of initial equation x = (Pi+3)/2 with two variables, which satisfies all presented equalities and not always wise transformations )) The fun question is: "In what geometry Pi=n, n=3"? ))
Very quickly saw the problem. Whenever you’re trying to prove something and someone uses a square root without specifying whether it’s the positive or negative root, you’re gonna have a bad time.
No, the mistake is the step where you right x^2-6x+9=(x-3)^2 because if can also be equal to (3-x)^2 by which youget the equation that is in the starting, or in better words, you get what's true
when you remove square then it opens with mod and as x is smaller then 3 but greater then pi so it will open like 3-x=x-pi edit- i flipped the values i know, but still answer is correct xD
Since the value of pi is never used, this "proof" works with any constant 🤣 One mole = 3 apparently Swapping signs in the right (wrong) place can give any kind of broken results you'd like.
If pi is 3 then the second row just says 0=0 from there every manipulation essentialy just states that something is equal to itself. It is the old I multiplied both sides by 0 and added 1. So the actual mistake is not the square root, even tho it might be a mistake, since you cannot prove equality by just multiplying by 0 since if you could I could just say 1*0=2*0 1=2 which is clearly false as I have to devide by 0 to reach that conclusion.
as an engineering student I see no mistake with pi=3 hahahahahaha
pi=3
e=3
@@HonkousBonkous pi^2=g
@longinuspodbipieta3764 Now, as a physicist that one hurts
@@longinuspodbipieta3764 pipig
@@longinuspodbipieta3764sqrt( g ) = 3
I love how you say 'how exciting'. It has such a fun math teacher vibe.
"How. Exciting. 🙂"
nice one. just because two things have equal squares, doesn't mean those two things are equal. they're equal in magnitude, but not necessarily equal in value.
I like that one! A very clever setup.
Using square roots to 'undo' squared values is a good way to hid a - quantity...
its common for jee students
If there's ever a nonsensical answer and there's a square root of a squared value... that's pretty much always where the issue is. Either that or a division by zero hidden somewhere (which is a popular route to 1=2 I believe)
Or integration is involved and someone forgot to put in a "+ c"
These sorts of tricks almost always fall into one of two categories: Division by zero, or improperly taking square roots. Once the "π² - 9" showed up, I pretty much knew where this was going, and just waited for the square-root step to come along, and bingo, "there's your problem!"
"Take the square root of both sides" is pretty much always a red flag in these sorts of "proofs", to be honest...
If I were teaching an introductory proofs class, this is the first lesson I'd show in "spot the problem" proofs. Show the students how common these two tricks are to create fake facts.
There is also a really sneaky one with complex exponents
Absolutely. Very very old tricks...😜
@@MaximxlsThere's also one with extraneous rots and also false geometric proofs like the one in 3blue1brown's channel where he proves all triangles are equilateral.
My favourite "find the mistake" was one with integrals that I think started with evaluating the integral from 0 to 1 of 1/x specifically by integration by parts, and the [x*1/x] from 0 to 1 they say it's equal to 1 "because the x and 1/x cancel", but 1 from 0 to 1 gives 1 - 1 = 0.
People will stop falling for this if we just start teaching that √x^2 = |x| and that the square root is in fact not the inverse of square
What if √(-1)² ?
Just because a^2 = b^2 doesnt mean a = b, since squaring is a non injective function.
For example, (-1)^2 = (1)^2, but -1 is not equal to 1.
Futher a^2 = b^2 can be simplified as
a^2 - b^2 = 0
(a + b) * (a - b) = 0
here either a + b = 0 or a - b = 0
so a = -b or a = b
at least explain what non injective is I do know though (not a one to one function) but others dont
But would it mean | a | = | b | ?
@@feelsdankman211Yes, absolutely...
In the beginning, you assumed that x is equal to three plus pi divided by two, so x is actually a smaller number than pi, for example, something around 3.07. So when we take the square root of the expression (3-pi), it must be the absolute value. Let's take it out, which becomes pi-3. As a result, after solving, we reach the initial equation.
Love it.
@@AndyMathyou used the wrong term. It’s supposed to be how exciting.
@@IllChooseAName bot works for him
When I actually did higher level math classes, I didn't just write 'line by line' like this, but instead implications. This is because one line could imply up or down, but not necessarily the opposite direction. For instance, [a^2 = b^2] [|a| = |b|] but, you can only assume the general case in one direction [a^2 = b^2] [a = b] (incorrect) since it can otherwise be proven, in most cases, by showing that both numbers inhabit the same side of 0. People (of low and common intelligence) will tend to jump to the conclusion that this false proof is a gimmick and just annoying trickery yet it is actually the opposite: it exposes the trashy syntax commonly used and learned in elementary to high school. "line by line" is not explicit enough to show which direction the implications work.
Another way to think about this is that the square of a number is the square of its negative. Because x is the average of 3 and π, it is the same distance from both of them, therefore, (x-3) and (x-π) will be the additive inverse of each other.
That's so logical and interesting property. Taking average of 2 numbers and subtracting each number from average respectively gives additive inverse. Fascinating! Is there a name for this property?
@@iqurramAverage?
@@ingiford175
Adding a set of numbers together and dividing by the count of those numbers gives the mean, also knows as the average.
In this instance pi and 3 are added together and divided by two, thus getting the average.
The mistake is that id (x-3)^2 = (x-pi)^2, it means that either (x-3) = (x-pi) OR (x-3)=-(x-pi), in this case the correct alternative is that x-3 = -(x-pi), which leads to 2x = 3+pi (the initial assumption).
he already proved it there was no need for you to do that
*When taking the square roots, you can't just cancel out the square root, you need to put a plus-minus sign on one of the sides, becase x-π and -(x-π) both square to become (x-π)².*
In an astrophysics module at uni, during a derivation the professor said "pi is approximately 3, 3 is approximately 1, so we can just cancel it". Since then pi = 3 hasn't seemed so bad.
The mistake is that roots, and powers, do not cancel each other out. squaring, and square-rooting, results in the absolute value of whats inside the root.
So for sqrt( (-a)^2) we get |-a| = a
and for for sqrt( (a)^2) we get |a| = a
First error: At step two you multiply with (pi - 3), wich makes the equation true for pi = 3.
Second error: Square and Squareroot don't cancel one another, but sqrt(x^2) is |x|. So you remove solutions.
If you take x-3 = -(x-pi) and solve for x, you get the equation from the start.
That's very interesting. I missed that bit. If the solution is correct and pi = 3 then pi - 3 = 0 and at that step you have multiplied both sides by zero, effectively making everything afterwards kinda moot anyway.
That's what I thought, in step 2, if you multiply by pi-3 where pi=3 you would be multiplying both sides by zero.
I would say it's when you add the x squared to both sides which enables you to get to the point where the sign can be different for the square roots.
I found it!!
Why did you take (x-π)² instead of (π-x)² ?
If you would have taken this (π-x)² , then it would come as x=(3+π)/2 ( Just like the equation that you provided earlier)
Extraneous solutions are not solutions.
That's why we create and put "" sign at the beginning of each row. The sign will explain for itself, it makes up a link throughout the math and clearly shows that the first row equals the 6th, but stop it to becomes the 7th (as others have explained).
The equivalence is already broken between the 1st and the 2nd line (if π=3). And also, all this derivation following the 2nd line is a distraction. I mean, by looking at the 2nd equation, we immediately see that π=3 is a solution 😉; no need to go on further.
From the first to the second line, isn’t it just a multiplication by 2?
I didn't really like this problem. Even if there would be no mistake with absolute value, and we would get PI=3, it would simply mean that there is no soultion for x, because regardless of its value the equation is false. In order to show that PI equals 3 somehow, we should treat PI as a variable, and then there are infinitely many soultions to the original equation (linear equation with 2 variables x and PI).
What I mean is that the context of this problem is very confusing
That was well done.
These things nearly always devolve to a division by zero, or a failure to use ±√.
You can't square root both sides.( In particular, from the definition of x, x is halfway between 3 and pi, so x-3 is positive and x-pi is negative so they cant possibly be equal.) I liked this one, it's better than the usual 'hidden division by zero' trick.
Why switch to a value for x?
You can stay in exact form, knowing {pi} is 3+ (a bit bigger than 3).
Then x-3 is
= ({pi} + 3)/2 -3
= ({pi} +3 -6)/2
= ({pi} - 3)/2
> 0
since {pi} = 3+ > 3
And x - {pi} is
= ({pi} + 3)/2 - {pi}
= ({pi} + 3 - 2{pi}) / 2
= (3 - {pi})/2
< 0
since {pi} = 3+ > 3
Replace π with θ or 'a' or any other constant. The actual step where the problem lies is when you multiply both sides by (π-3) because if π=3 then π-3=0 and you are multiplying both sides by zero
when square root 1:18 when pi equals=3 x must equal 2 so when you square root (x-3)^2 which means (-1)^2 you take it out as x-3 which is -1
am i right?
Also would like to point out something else, if pi=3, you cannot multiply both sides by (pi-3) in the 2nd step because multiplying by zero would just result in 0=0
Well, 0 is indeed equal to 0, so I don't see an issue
You would be anyway able to multiply by (π - 3), it's just that you'll get ⇒ from top equation to the bottom and not ⇔
the issue here is, if we are going to proof a statement like p = q be true and then we multiply both side by a constant (a - b)
we got (a-b)*p = (a-b)*q, then the statement will be true anyway when a equals to b since 0 = 0
but this do not imply we have proven the original statement p = q is correct
let put a real number on it
(k - 1) * 1 = (k - 1) * 2, the statement is true when k = 1, but 1 is not equal to 2
so when we are doing such multiply like (pi - 3), we need exclude the possibility of pi equals to 3, otherwise the rest move would be meaningless
I actually think it is in the second step, multiplying both sides by pi-3 adds another possible solution to the equation where pi (being used as a variable) is equal to 3 making the whole equation a 0=0 true statement.notice if you just replace pi with 3 in the third step you get 6x-6x=9-9.
pi=3 technically incorrect, but when you rounded it up to make calculating easier.
because you can just APPROXIMATE laziest way when doing choice test.
When your take the square root of a square involving variables, the only thing you can say is the absolute values of the quantities are equivalent, not the quantities themselves: i.e. |x - 3| = |x - pi|.
This leads to you having to check 2 cases:
x-3 = x-pi [ equivalent to -(x-3) = -(x-pi) ]
-(x-3) = x - pi [ equivalent to (x-3) = -(x-pi) ]
The first equation leads to the false solution that 3 = pi
The second equation returns the original assumption for x = (pi + 3)/2
You've to take modulus while removing square root.
As a nuclear chemist in NASA, headquarters in Malmo. Me and my collegue found this rational misstake quite easily with the help of LeChatlier’s Principle
No, you didn't. But this is perhaps the most creative lie I've ever seen on UA-cam.
(x-3/2) and x-π/2) are actually the negative of eachother, so it's natural for their squares to be equal ( (x-3/2)² = (x-π/2)² ) but not the other way around.
The mistake is really in logic. To say this is true would be to say that π is a variable not a constant. This does not prove the value of π since there’s no approximation that is made. It’s basically assuming π is not a constant. And also because of the square root of the squares generating extraneous solutions.
I figured it was something do with the sqaure roots being positive/negative just not what exactly
"There are no accidents" - Master Oogway
Yeah I am an engineer. How'd you know?
when you multipied pie minus 3 both sides , case also involved that that is zero for lhs equal to rhs
But pi-3 is not 0. So you can multiply both sides by pi-3
@@ShanthiVeluswamy it is true that's why we have to assume that this is true which means after further calculation we have to reject the case of pi minus 3 equals to 0.
I knew there wasn't going to be any mistake as soon as I saw pi=3
Without looking at the rest of the comments, this is where the mistake lies: x is the arithmetic mean of 3 and п from the start, so it is in the interval (3,п). Thus x - 3 is positive and therefore equal to |x - 3| = sqrt((x - 3)^2), but x - п is not equal to sqrt((x - п)^2), but its negative. This is where the mistake lies.
How do you know that x-3 is positive though? X is only positive if pi is greater than 3. The problem I have with this explanation is that your assuming that pi is greater than 3 to prove that pi is not 3.
@@raphaelnouh1442 п > 3 is a mathematical fact. п = 3 is a contradiction which proves that the "proof" is wrong because the working out is wrong. I showed, using the fact that pi is greater than 3, that his working out was wrong. Second Order Arithmetic + "п = 3" is an inconsistent theory as it proves its own axioms incorrect, thus you shouldn't work in it.
@@newwaveinfantry8362 I know pi is not three, I just think you can’t really use “pi isn’t three” to prove pi isn’t three. If you wanted to do that you might as well have just said “pi is 3.14…, therefore it cannot be 3.”
pi + 3 < 2pi => X < pi => x - pi < 0 => sqrt(x -pi) = |x-pi| = Pi - x, not the other way around. So the mistake is right after the square root, where you assume x - 3 = x - pi
Sq root of (x-pie)^2 not equal to x-pie bit is equal to pie-x
as x=(pie+3)/2 so x-pie less than zero means modulus opens with negative sign
1:47 I am going out on a limb and say the mistake is that you cannot add x² to both sides of the equation - and then start splitting the variable (x) on both sides. It should be separated, only on one side of the equation.
The only way to get pi to equal three is to have a circle with a longer diameter than a flat diameter. So you have to have a warped 2-D space.
The other major flaw you find in bad proofs is when you inadvertently divide by zero.
Funny how during your proof that pi=3 you inserted 3,14 for pi on the step to eliminate the square root 😅
However, there's another thing to point out: A critical mistake already happens at step 2, where both sides of the equation are multiplied by (pi-3), which - as it turns out by the end - equals to multiplying both sides by zero! So when ignoring this, no matter how the calculation goes on, anything could be proven.
yeah, canceling square root is the average mistake, instantly saw that issue, so many people like doing that, but don't understand that they can't
For who doesnt understand ... while taking (x-3)² its negative but rhs is positive so we cannot take it so (x-3)² not equal to (x-pi)²
no rhs is negative and lhs is positive that's why
Here pi is the variable i.e, we are trying to find pi...and we have multiplied by pi-3 on both sides which clearly implies that we have introduced an extraneous root for pi which 3. If anyone thinks that what I said is false then think about it this way...by using the argument shown in the video(which is false) you can show that unknown value 'a' is equal to a number 'b' by just multiplying the equation by (a-b)
no, x is the variable.
Or, if x AND pi would be vaiables then we would have 1 equation for 2 variables --> unsolvable.
A general misconception is that any x when squared and rooted we get x. But ans is lxl (mod x) so it doesn't have only 1 answer people are taking it seriously but as a student who is preparing for jee examinations its pretty obvious
These squaring and square rooting teachers always try to make fool us but always fails
1:18 Square root opens with mod always , so this time X - 3 = +( X-Pi ) is not a solution as 3 is not equal to pi , so we use X-3 = -(X - Pi ) to get back to original answer of x=pi+3/2
When you root (x-pi)^2, you don't get x-pi, you get |x-pi|. As it turns out, x-pi is negative, so when we take the absolute value, you get pi-x. (It's the same concept for x-3, but x-3 is positive so you can just drop the absolute values). Once we fix this mistake, we get the original solution of x=(pi+3)*1/2
in engineering school, we were taught e=3, pi=3, and g/3=3. hell, 4 can equal 3 if you try hard enough!
I did not spot the problem, didn’t remember the trick, but your explanation made sense because we know pi and you were able to logically deduce the answer to begin with (approx 3.0). I assume you would need different explanation if pi was unknown, say Y.
bro, I already loved math, now with your videos I love even more
Exactly as I thought
You need to remember that √(u)² = |u|
So you should get
|x - 3| = |x-π|
You can solve this piece wise and you'll get two different cases in the end
The π = 3 case
And the original x = (π+3)/2
whenever there's squares on both sides like this, its best to take it to the other side, and apply a^2-b^2.
Remember a number has a positive and a negative square root, so x^2=y^2 does not always mean x=y.
Yeah, (x-3)^2 does equal (x-pi)^2, but this does not imply x-3=x-pi because, in fact, x-3=pi-x. And (x-pi)^2 does equal (pi-x)^2 because one is the negative version of the other. Can't just take the positive square root once you've squared stuff, you might need the secondary root as you do here.
Nice one❤
The problem here was that if π = 3 then π - 3 = 0. Thus you cannot multiply both sides by π - 3 as that would be dividing by 0
Short answer: if one wants to prove a equality, never manipulate both sides at the same time or one can get some really bizarre results.
At first i thought the floor function was missing
The moment you see something like sqrt (a^2) fire alarms should start to ring in your head.
1:20 You can't cancel oyt the square root like that. You need to add +-
This is how I imagine Greek philosophy to have been like... arguing for the sake of arguing.
I think you enjoy doing math so much that now you're creating problems instead of solving them.
After the square root thing you should write modulus
This is quite tough tho. But as you can see , if (x-3)^2 = ( x - n) ^2 ( sorry i have no pi symbol) than it could be x - 3 = x - n or x-3 = - ( x - n ) so pi wont be equal to x at this because we there's two situation of x so we could easily know that the first one is wrong and the second is the correct. For the second one, as x - 3 = -(x - n) => x - 3 = n - x => 2x = n + 3 => we got the first equation which is x = (n+3) / 2. So, the mistake in this one is that the sqrt (x+3)^2 isn't not equal to x+3, it is |x+3|, same to the sqrt of (x+n) ^2, it is |x+n| . From |x+3| = |x+n|, we have two situation as I said before which is x - 3 = x-n or x-3 = -(x-n) ( it could be reverse)
The step where the writer concludes sqrt((x - π)^2) = x - π is incorrect, because sqrt((x - π)^2) = |x - π|, and since x - π < 0, |x - π| = -(x - π) = π - x. This is a classic mistake many students make. Teachers often mistakenly teach that sqrt(x^2) = x for all x, which is false. sqrt(x^2) = |x| for all x is the correct statement.
I paused video at start and am going to say, accidental divide by zero somewhere.
Now let me watch.
Not a divide by zero but if pi = 3 is a valid solution then the multiplication of pi-3 on each side in step two just multiplied both sides by zero and everything else after is meaningless, including getting the solution that pi = 3 in the first place. Just another way this is wrong besides the sq rt pointed out in the video.
The mistake is in the square roots. Substituting x = (3 + pi)/2 at the line (x - 3)^2 = (x - pi)^2 you get ((pi - 3)/2)^2 = ((3 - pi)/2)^2 - which is clearly correct - but after taking square roots it isn't. You should end up wth x - 3 = pi - x or x = (3 + pi)/2
its usually dividing by zero or doing something weird with functions
If you take π=3 in the second step where you are multiplying both side by π-3 then you are multipling 0
i think there's another mistake, we multiplied both sides by (pi-3) in the beginning, but we came to that the pi=3, but then pi-3=0, so we just multiplied both sides by 0 and that makes no sense and breaks the equation
The mistake is that x is by definition some Number between 3 and pi so x-pi is negative wich makes cancelling it in the sqrt equal to -x+pi not x-pi
Also in a parallel universe where pi is actually 3, you cant multiply your term with pi - 3 on both sides. If pi was actually 3, you'd multiply with zero which trolls your entire proof.
No mistake, Pi is three, in base pi/3.
As a physicist, I have no issues in saying π²=g (on earth)
Intuition told me it was an absolute value issue, glad I was right
There is another case you need to chek. when you do the Square root you need to add a case when you put minus in one of the sides and then you get the correct answer
1:42 so you can't actually take the sqrt there.
basically the sqrt has a branch cut and by squaring and taking the square root the left-hand side and right-hand side end up on different branches.
if a^2 = y, then y = +a or -a
now in our case one of the sides is +a and the other is -a
If a^2 = y then a = +sqrt(y) & -sqrt(y)
You forgot to put the root over y
@@Mirror1o11iM oh yeah nice catch
This is like
An equation x=-2
You square both sides than
Take square root both sides and say hey there are 2 solutions
X=2,-2😂😂😂😂😂😂😂😂
How exciting
Really cool, glad i found the mistake!
The engineers must be pissed rn
x is defined to be the halfway point between three and pi. so x-3 = -(x - pi).
The error is logical, you cannot interpret Pi as variable and constant at same time. This is cause for presented statements cannot be considered as proof. Note, in presence of this error, no matter how correct were manipulations itself. Actually, you have found one correct solution (Pi=3, x=3) of initial equation x = (Pi+3)/2 with two variables, which satisfies all presented equalities and not always wise transformations ))
The fun question is: "In what geometry Pi=n, n=3"? ))
Real at my school we have 3.14. tho i have an advanced physics so idk😂
When you do square root, you want to make sure the result is positive. X-pi is negative. You need to write pi-X instead.
At the square roots should be taken the absolute value(first 2 seconds of the video, bye, bye)
I knew where the mistake was and what was done incorrectly, but I couldn't articulate how it would give us this answer
Very quickly saw the problem. Whenever you’re trying to prove something and someone uses a square root without specifying whether it’s the positive or negative root, you’re gonna have a bad time.
No, the mistake is the step where you right x^2-6x+9=(x-3)^2 because if can also be equal to (3-x)^2 by which youget the equation that is in the starting, or in better words, you get what's true
Engineer: good enough
when you remove square then it opens with mod and as x is smaller then 3 but greater then pi so it will open like 3-x=x-pi
edit- i flipped the values i know, but still answer is correct xD
Since the value of pi is never used, this "proof" works with any constant 🤣
One mole = 3 apparently
Swapping signs in the right (wrong) place can give any kind of broken results you'd like.
square roots are the bane of my existence
You have a green screen. How handy!
How exciting!
1:30 method of matching coefficients. π² = 9
If pi is 3 then the second row just says 0=0 from there every manipulation essentialy just states that something is equal to itself. It is the old I multiplied both sides by 0 and added 1. So the actual mistake is not the square root, even tho it might be a mistake, since you cannot prove equality by just multiplying by 0 since if you could I could just say 1*0=2*0 1=2 which is clearly false as I have to devide by 0 to reach that conclusion.
While cancelling the root on both sides you didn’t take mod on both sides which resulted in this error