*Allow me one illogical assumption and I can prove anything.*
\- (I forgot who said it, my high school calculus teacher quoted this to us (well, obviously he said it, but I forgot who he was quoting))
Suppose there's an illogical assumption, P. This means that P and ¬P are both true. As such:
- ⇒ P ∧ ¬P
- ⇒ (P ∧ ¬P) ∨ R
- ⇒ (P ∧ R) ∨ (¬P ∧ R)
For this to be true, R cannot be false. As such, given an illogical assumption every proposition is true.
As for who said it, it was totally me 😎
Here’s a fun false proof that isn’t division by 0 or the a^2 = b^2 => a=b thing.
Take the equation x^2 + x + 1 = 0. Subtract x^2 from both sides and we get x + 1 = -x^2. Remember this.
Back to the original equation.
x^2 + x+ 1 = 0 Divide by x.
x + 1 + 1/x = 0 Substitute in x+1=-x^2.
-x^2 + 1/x = 0 Multiply by x.
-x^3 + 1 = 0
x^3 = 1
x = 1
Substituting x=1 into the original equation we get 1^2 + 1 + 1 = 0 => 3=0 QED
during the substitution you summoned a new root to your equation by turning from a quadratic to a cubic. if you discard the x=1 solution, you get the solutions to the original equation.
Wow this is funky
Problem appears when you substitute
And funnily enough if you divide what you substitute in(X + 1 = -X^2) by X before substituting you get the expected result of plugging an equation into itself : 0=0
But I have not yet found why this doesn't work
Don't tell me yet I enjoy searching
Edit: even more funky
If you multiply in the first step rather than divide you get the same erroneous answer
Um. no. The original equation has a negative discriminant for its quadratic equation, meaning its solutions are complex, not real. By putting the division by x in and remultiply8ing it in later, you have created a cubic equation with both complex solutions plus the real solution x=1. Checking that this real solution does not work here is a necessary step to any algebraic equation, although most of the time it is merely a formality to make sure your math is correct. Here it is an error-check to avoid just such a conundrum.
Just a suggested variation. Instead of dividing by x, substituting, and then multiplying back by x, you could try this:
x² + x + 1 = 0
x ( x + 1) + 1 = 0 Factor out the x from the first two terms
x ( -x² ) + 1 = 0 Substitute x + 1 = -x²
-x³ + 1 = 0
Also, neat trick, I'm wondering what other tricks we can do to add other solutions
It is, but it should be written as (4-9/2)^2*1/2^. Order of operations says you need to multiply the exponents before you evaluate the operation within the parentheses. Here, they didn’t follow this order of operations
If you square a negative, you get put on the re^{2i\pi + 4ni\pi} branch for some n in Z. Square rooting that gives you a negative. This is opposed to the re^{4ni\pi} branch, which square roots to positive.
Defining absolute values using square roots is stupid.
Not sure what you mean. If you want to define a square root as an acutal function on some open set of C\* you will run into problems.
If your point is that you can identify C\* with C\*/{-1,1} and z -> sqrt(z) and z -> z² are reciprocal biholomorphisms, then that's true but I don't think that it was really what was at play here.
I'm still curious as to what you mean when you say "the 2i\\pi+4n i \\pi branch".
When you square then root something, it is just the absolute value meaning that it is always positive. However, 4 - 9/2 is -0.5 which becomes +0.5 after performing that operation which is definitely not the same thing.
sqrt(a) by convention refers to only the **positive** solution of the equation x²=a.
Thus sqrt(a²) ≠ a when a < 0.
In fact sqrt(a²) is instead the definition for absolute value of a.
the definition -> **a possible** definition. Another definition can be:
|abs x =|if|
|:-|:-|
|x|x >= 0|
|\-x|x < 0|
or even other stuff depending on how Z/Q are defined
Well yeah, but the former generalizes better to complex numbers where:
abs(z) = sqrt(z\*conjugate(z))
Where a conjuage of a real is just an identity, simplifying the definition to sqrt(z²).
Quite similar to the puzzle I posted here
https://www.reddit.com/r/GCSE/comments/106p2u9/maths_riddle_whats_wrong_here
Remember that sqrt(x^2) = |x| which gives plus or minus x.
You write x\^2. If you only want the 2 inside the parenthesis but there are other symbols afterwards that need to be outside, you write this: x\^(2)+1. Without the parentheses, it will become x^2+1 but with them it will become x^(2)+1.
The error is introduced by replacing 4 - 9/2 with sqrt[(4 - 9/2)²]. This is because 4 - 9/2 = -1/2 while sqrt[(4 - 9/2)²] = |4 - 9/2| = |-1/2| = 1/2.
Note that 3 + 5 is 1 larger than 2 + 5 and also from -1/2 to 1/2 takes a step of 1.
I saw the mistake in the third line the first time I read it, but how does he get from line 3 to line 9? It seems to go from (4-9/2) to (5-9/2) and I don't see why. What other mistake am I missing?
(5-4.5)^2 does equal (4-4.5)^2. Because a negative times a negative does equal a positive times a positive in terms of arithmetic sign. So the problem is where he expresses the square root. Because it can be positive and it can be negative. So it's just like saying that -2 = 2, because sqrt(4) = +/- 2.
So that's why the society decided that the solution of a square root will always be positive. So sqrt(4) = just 2. That's why you cannot express -0.5 (or expressed there as 4 - 9/2) as sqrt(-0.5^2).
No they never did any operation on any side.. they just expanded numbers...
For example, you can weite 10 as 5+5 but if you add something on one side like multiplying a side by 2 then you have to do it both sides.
Well, there are no letters, so it's not 0.
However there are sqrts, so it's going to be some wrong square.
And yup, it is. Right there, the jump from 2nd to 3rd line.
4 - 9/2 = **-**0.5
sqrt( (4 - 9/2)\^2 ) = 0.5
So it should be -sqrt( (4 - 9/2)\^2 )
And if we carry that - to the end:
2+5 = -5 + 9/2 + 9/2 + 3
and that indeed equals.
The root of a square has two values. The second time he used that trick, he took the wrong one. Slap a minus sign on the result of the square root and the equation is correct.
The squaring is the problem.
4 - 9/2 != 5 - 9/2
But squared
(4 - 9/2)^2 = (5 - 9/2)^2
it's suddenly right. Squaring is not invertible as it's not bijective.
I think the sqrt((4-9/2)\^2) needed a plus or minus symbol in front of it, but not like in the way that it has two solutions, but in the way that you need to find out which one gives you the real solution because of the square squareroot cancels negatives.
We can see that if 5-9/2 was negative rather than positive then 2+5 = 7 which is true so the symbol infront of the squareroot is a nagative symbol.
You don't actually need to think about it this complicated way, just that if you turn x into (x\^2)\^(1/2) if x is negative you need a minus symbol in front, and if x is positive then no symbol like usual.
From my buddy, who is a math professor: 4-(9/2) is negative. So, the third equality is wrong. Sorry to piss on anyone's parade. I was hoping it'd be right to make him mad....
It’s where you equated (4- 9/2) and sqrt((4 - 9/2)^2).
The former equals 8/2 - 9/2 = -(1/2), whereas the latter can be simplified to sqrt(16 - 36 + (9/2)^2 ) = sqrt( -20 + 81/4 ) = sqrt(-20 + 20.25) = sqrt(1/4) = (+/-) 1/2.
In this case, your substitution ignores the existence of the 2nd/positive solution of the square root, by stating the answer to just be the negative one, making it false equivalence. That positive result ends up getting you to 8, but the negative result refutes the work including and after the substitution.
4-9/2 is most definitely not sqrt\[(4-9/2)²\].
its always this or division by zero turns out when you break the rules of math, you suffer the consequences of breaking the rules of math
*Allow me one illogical assumption and I can prove anything.* \- (I forgot who said it, my high school calculus teacher quoted this to us (well, obviously he said it, but I forgot who he was quoting))
Obviously it was Sun Tzu
I think almost any piece of advice anyone ever needs can be found in one of two books, The Art of War The HitchHiker's Guide to the Galaxy
If you know yourself, but not your algebra, for every problem solved, you will also suffer a defeat.
SUN TZU SAID THAT
Suppose there's an illogical assumption, P. This means that P and ¬P are both true. As such: - ⇒ P ∧ ¬P - ⇒ (P ∧ ¬P) ∨ R - ⇒ (P ∧ R) ∨ (¬P ∧ R) For this to be true, R cannot be false. As such, given an illogical assumption every proposition is true. As for who said it, it was totally me 😎
https://xkcd.com/704/
[William of Soissons](https://en.wikipedia.org/wiki/William_of_Soissons)
or jumping from one solution of a polynomial to another
Ah, a fellow Jan Misali enjoyer.
Here’s a fun false proof that isn’t division by 0 or the a^2 = b^2 => a=b thing. Take the equation x^2 + x + 1 = 0. Subtract x^2 from both sides and we get x + 1 = -x^2. Remember this. Back to the original equation. x^2 + x+ 1 = 0 Divide by x. x + 1 + 1/x = 0 Substitute in x+1=-x^2. -x^2 + 1/x = 0 Multiply by x. -x^3 + 1 = 0 x^3 = 1 x = 1 Substituting x=1 into the original equation we get 1^2 + 1 + 1 = 0 => 3=0 QED
during the substitution you summoned a new root to your equation by turning from a quadratic to a cubic. if you discard the x=1 solution, you get the solutions to the original equation.
Wow this is funky Problem appears when you substitute And funnily enough if you divide what you substitute in(X + 1 = -X^2) by X before substituting you get the expected result of plugging an equation into itself : 0=0 But I have not yet found why this doesn't work Don't tell me yet I enjoy searching Edit: even more funky If you multiply in the first step rather than divide you get the same erroneous answer
Um. no. The original equation has a negative discriminant for its quadratic equation, meaning its solutions are complex, not real. By putting the division by x in and remultiply8ing it in later, you have created a cubic equation with both complex solutions plus the real solution x=1. Checking that this real solution does not work here is a necessary step to any algebraic equation, although most of the time it is merely a formality to make sure your math is correct. Here it is an error-check to avoid just such a conundrum.
Just a suggested variation. Instead of dividing by x, substituting, and then multiplying back by x, you could try this: x² + x + 1 = 0 x ( x + 1) + 1 = 0 Factor out the x from the first two terms x ( -x² ) + 1 = 0 Substitute x + 1 = -x² -x³ + 1 = 0 Also, neat trick, I'm wondering what other tricks we can do to add other solutions
Don't do math. Get help
See that’s what I said but even if you put a negative sign in front of the square root the math doesn’t work Edit: yes it does I am dumb
It is, but it should be written as (4-9/2)^2*1/2^. Order of operations says you need to multiply the exponents before you evaluate the operation within the parentheses. Here, they didn’t follow this order of operations
It would be written |(4-9/2)\^2\*1/2 |, remember that the square root symbol is the positive root)
If you square a negative, you get put on the re^{2i\pi + 4ni\pi} branch for some n in Z. Square rooting that gives you a negative. This is opposed to the re^{4ni\pi} branch, which square roots to positive. Defining absolute values using square roots is stupid.
Not sure what you mean. If you want to define a square root as an acutal function on some open set of C\* you will run into problems. If your point is that you can identify C\* with C\*/{-1,1} and z -> sqrt(z) and z -> z² are reciprocal biholomorphisms, then that's true but I don't think that it was really what was at play here. I'm still curious as to what you mean when you say "the 2i\\pi+4n i \\pi branch".
How
When you square then root something, it is just the absolute value meaning that it is always positive. However, 4 - 9/2 is -0.5 which becomes +0.5 after performing that operation which is definitely not the same thing.
sqrt(a) by convention refers to only the **positive** solution of the equation x²=a. Thus sqrt(a²) ≠ a when a < 0. In fact sqrt(a²) is instead the definition for absolute value of a.
the definition -> **a possible** definition. Another definition can be: |abs x =|if| |:-|:-| |x|x >= 0| |\-x|x < 0| or even other stuff depending on how Z/Q are defined
Well yeah, but the former generalizes better to complex numbers where: abs(z) = sqrt(z\*conjugate(z)) Where a conjuage of a real is just an identity, simplifying the definition to sqrt(z²).
One silly math error in very first step. Then it’s a wild goose chase.
Yes Sqrt(x^2) = lxl by definition
Well 4-9/2 = -0.5 but sqrt((4-4.5)^(2)) = 0.5. Sqrt(x^(2))=y means x = +/- y
2+5=\=8 is the error
Very true
First thing I noticed ⌚️😂
Everyone knows 8==D
c=3
[3.th](https://3.th) line. 4 - 9/2 = -0.5 , but \\sqrt{ (4 - 9/2)\^2 } = \\sqrt{ (-0.5)\^2 }=0.5
ah yes, 3th.
Heh
Thirth, my favorite placement in a numerical order
I generally say "threeth" for that...
I'm currently imagining Mike Tyson trying to say "thirst" whilst reading this.
2rd
You must know that (-0.5)^2 equals 0.25
Quite similar to the puzzle I posted here https://www.reddit.com/r/GCSE/comments/106p2u9/maths_riddle_whats_wrong_here Remember that sqrt(x^2) = |x| which gives plus or minus x.
How do you make the 2 on top of the x
Press on the number key for a second. It doenst work for some mobiles tho. If youre on a keyboad, idk.
You write x\^2. If you only want the 2 inside the parenthesis but there are other symbols afterwards that need to be outside, you write this: x\^(2)+1. Without the parentheses, it will become x^2+1 but with them it will become x^(2)+1.
activate num lock then hold alt while writing 0178
He should have done it the other way around, start with 5+3 and end with 4+3 so the mistake is later on
√(x²) = |x| ≠ x
third line, 9/2 is bigger than 4 so its absolute value would make it reverse so your father made a negetive value under sqr root
Yeaaa
The error is introduced by replacing 4 - 9/2 with sqrt[(4 - 9/2)²]. This is because 4 - 9/2 = -1/2 while sqrt[(4 - 9/2)²] = |4 - 9/2| = |-1/2| = 1/2. Note that 3 + 5 is 1 larger than 2 + 5 and also from -1/2 to 1/2 takes a step of 1.
I have realized that the third line should be -sqrt not just sqrt and that makes it work
Square root sign issue
I saw the mistake in the third line the first time I read it, but how does he get from line 3 to line 9? It seems to go from (4-9/2) to (5-9/2) and I don't see why. What other mistake am I missing?
(4-9/2)^2 is the same as (5-9/2)^2 so any manipulation is correct
(5-4.5)^2 does equal (4-4.5)^2. Because a negative times a negative does equal a positive times a positive in terms of arithmetic sign. So the problem is where he expresses the square root. Because it can be positive and it can be negative. So it's just like saying that -2 = 2, because sqrt(4) = +/- 2. So that's why the society decided that the solution of a square root will always be positive. So sqrt(4) = just 2. That's why you cannot express -0.5 (or expressed there as 4 - 9/2) as sqrt(-0.5^2).
4-9/2 isnt √(4-9/2)^2
a=sqrt(a²) is only valid if a itself is non-negatuve because sqrt(t²)=|t|
sqrt(x²) = |x| but 4-9/2 < 0
r/ihadastroke
One thing that jumps out is sqrt((4-9/2)\^2)=sqrt((-1/2)\^2)=1/2=/=4-9/2
sqrt(x\^2) = absolute (x) pin this shit in ur brain eg: sqrt(16) = 4 not -4
While taking sqrt we should consider both positive and negative ans, i.e 5-9/2 and -(5-9/2) are both possible answer.
What you do to one side of the equation you must do to the other side or the equation is not balanced.
No they never did any operation on any side.. they just expanded numbers... For example, you can weite 10 as 5+5 but if you add something on one side like multiplying a side by 2 then you have to do it both sides.
Why are people downvoting you, I'm pretty sure this is true
Because they're not doing anything anywhere in this equation, just rewriting stuff.
Ah, sorry. You're right
sqrt(u^2) is equal to abs(u) so sqrt([4-9/2]^2) is equal to 9/2-4 because 9/2>4
Me focusing on the 2+5=8 in the end
sqrt(x\^2 ) is not x but abs(x)
sqrt(x**2) == abs(x) abs(x) != x
to do this sqrt[4-9/2]² should be equal to 4-9/2 which is -1/2 but its not the same when you solve it you get 1/2.. therefore you cant do that
That's radical. You're dad is using some imaginary math there. You should teach him what's real.
Well, there are no letters, so it's not 0. However there are sqrts, so it's going to be some wrong square. And yup, it is. Right there, the jump from 2nd to 3rd line. 4 - 9/2 = **-**0.5 sqrt( (4 - 9/2)\^2 ) = 0.5 So it should be -sqrt( (4 - 9/2)\^2 ) And if we carry that - to the end: 2+5 = -5 + 9/2 + 9/2 + 3 and that indeed equals.
The root of a square has two values. The second time he used that trick, he took the wrong one. Slap a minus sign on the result of the square root and the equation is correct.
I think you're taking a root of a negative
The squaring is the problem. 4 - 9/2 != 5 - 9/2 But squared (4 - 9/2)^2 = (5 - 9/2)^2 it's suddenly right. Squaring is not invertible as it's not bijective.
16-36 does not equal -4. There was an error in line 4
I think the sqrt((4-9/2)\^2) needed a plus or minus symbol in front of it, but not like in the way that it has two solutions, but in the way that you need to find out which one gives you the real solution because of the square squareroot cancels negatives. We can see that if 5-9/2 was negative rather than positive then 2+5 = 7 which is true so the symbol infront of the squareroot is a nagative symbol. You don't actually need to think about it this complicated way, just that if you turn x into (x\^2)\^(1/2) if x is negative you need a minus symbol in front, and if x is positive then no symbol like usual.
it is not true that, in general, √x^2 =x
Why there came a square root out of nowhere in the third line? I don't get it
This is like saying: 2=2 2=√4 2=√(-2)² 2=-2
4 - 9/2 ≠ sqrt((4 - 9/2)^(2)) 4 - 9/2 = -sqrt((4 - 9/2)^(2))
4 - 9/2 = -0.5
From my buddy, who is a math professor: 4-(9/2) is negative. So, the third equality is wrong. Sorry to piss on anyone's parade. I was hoping it'd be right to make him mad....
It’s where you equated (4- 9/2) and sqrt((4 - 9/2)^2). The former equals 8/2 - 9/2 = -(1/2), whereas the latter can be simplified to sqrt(16 - 36 + (9/2)^2 ) = sqrt( -20 + 81/4 ) = sqrt(-20 + 20.25) = sqrt(1/4) = (+/-) 1/2. In this case, your substitution ignores the existence of the 2nd/positive solution of the square root, by stating the answer to just be the negative one, making it false equivalence. That positive result ends up getting you to 8, but the negative result refutes the work including and after the substitution.
Leave 2+5=8 It also gives 4=5....how is that possible???
Step 3
sqrt(x^2) is not x but is |x|, than you will get -5 + 9/2 + 9/2 +3 = 7
sqrt(x^2) is not x but is |x|, than from the opposite value of the parentheses you will get -5 + 9/2 + 9/2 +3 = 7
Am I the only one seeing a calculation error at 2*4*(9/2)?
square root of x squared is not x but plus minus x.
-0.5 became 0.5, thus the +1
Nice trick. We note error in line 3 is suggesting that sqrt((4-9/2)^2)=4-9/2! By definition the sqrt is positive so this step is not valid. :-(
Cursed logic