The mistake is on the second line, where you take the derivative of the right part, without taking those "x times" into account. They are also dependent on x...
When i saw the title I thought, "definitely a hidden division by zero", then I saw the (x times) being applied to any real number and thought "I can at least credit you with that".
This is the exact reason we define differentiable functions in an open, uncountable intervals. We get abnormalities like this if we include countable sets in our domain.
Hmm. The rational numbers set is countable. There was a nice proof of it, mapping natural numbers onto table of rational fractions represented as Z^2 - going through diagonals. And yet, you can find the derivative for |x^x| for negative x for which it has real values only for a certain subset of rational numbers. Of course using a limit that only operates within its domain; can't get a regular lim[h->0] with h going to zero an arbitrary path.
Here's the mistake: On the left side, you're taking a true derivative, so d/dx (x^2) is 2x On the right side, you're actually taking a *partial derivative*, it's as if you were differentiating a 2-variable function, xy, with respect to x - the result is y, but it just so happens that y has the same value as x. So of course the value is not equal, you're not doing the same operation on both sides. Also, if you would take the *total differential* of the function xy, the result is x + y. If y is the same as x, the result becomes 2x, satisfying the equation
I get the joke, but this is why another math youtuber never does the "divide both sides by x" trick, but, instead, moves terms to set the equation -= 0. So instead of jump solving 2x = x, you go to 2x - x = 0, so x(2-1) = 0, which shows that must = 0 Of course, that's not really the problem here, which is instead in the initial equality x^2 = x + x... Sometimes I wish I had taken a course in discrete math....
@Tiffany Wart well, 2x=x does imply to 2=1 as..let say you take the value of x to be 2 ....So, it gonna be 2*2=2 which will be equal to 4=2...which is ofcourse not equal....So, ur bound to take the value of x as 1...which will be 2*1=1. Which Is 2=1 .....And sorry if I had committed any mistake, I am in my 10th year like a newbie....So, I don't know about differentiation...
I think what is wrong here is that on the left side it is a smooth continuous curve but on the right that same curve is being expressed as discrete sum of some terms. But we know that a smooth curve cannot be expressed as sum of discrete terms. Instead it must be expressed as an integral of some other function or basically it's own derivative.
I'm tired of those using equation (a+b)(a-b)=b(a-b), they don't even realize (a-b) is zero meaning it's not dividable, and finally a new one with better logic
Doing this again, paying attention to Integer parts. x=[x]+{x}, where x>0. []=integer part, {}=fractional part. x²=([x]+{x})²=[x]²+2[x]{x}+{x}² Take the derivative with respect to x, keeping away from integers (where [x] is a constant, so derivative=0) and you get 2x=0+2[x]+2{x}, x∉N 2x=2([x]+{x}) 2x=2x Divide by x, since x0 2=2
Let's say n is a positive integer. You can define adding x with -n terms to mean adding -x with n terms. Then it actually still works and you still have n^2 = n+n+...+n (n times) for all integers n.
Well, you could work with infinitesimals to make it "correct": In this context, taking the derivative means evaluating (f(x+dx)-f(x))/dx at dx=0 (which is an easy substitution after cancelling). So, we have f(x+dx)=(x+dx+x+dx+...+x+dx) (x+dx times), so f(x+dx)-f(x)=dx+...+dx (x times)+(x+dx+x+dx+...) (dx times)=dx*x+(x+dx)*dx, so the derivative is x+x=2x.
everybody missed the real april's fool in the end of the video where he said that it true for x is an integer (he even put an "Z"'), when in fact it is true only for x that is *Natural*
What about this? Define dalta_n(a_n)=a_(n+1)-a_n We have delta_n(a_n+b_n)=a_n+1+b_n+1-(a_n+b_n) =a_n+1-a_n+b_n+1-b_n =delta_n(a_n)+delta_n(b_n) by induction:delta_n(a_(n,1)+a(n,2)+...+a_(n,n))=delta_n(a_(n,1))+delta_n(a_(n,2))+...+delta_n(a_(n,k)) which means delta_n(n+n+...+n(n times))=delta_n(n)+delta_n(n)+...+delta_n(n)(n times) delta_n(n^2)=(n+1-n)+(n+1-n)+...+(n+1-n)(n times) (n+1)^2-n^2=1+1+...+1(n times) 2n+1=n 2(n+1)=n+1 2=1 (We also have for all n that are positive integer in every row)
The explanation is halfway wrong. Yes it because they are integers, but that doesn't make the function a constant. Remember that differentiable functions are defined ON AN OPEN INTERVAL. Meaning uncountable sets like real numbers. Because the definition of a differential is from the definition of a limit, which requires the domain to have a number between two given points on a domain, hence the "closer and closer but never getting there" concept of directional derivatives. And obviously a domain of just whole numbers disqualifies the function from being differentiable.
You’re close with this reasoning, except consider the rational numbers. The rationals are countably infinite, and for every rational number q there exist a number between q and 0, say q/2. Forgive my lack of vocabulary, but it’s more of a “continuity” issue with the limit
1:00 is that the only problem though? I’m not great at math, but to me it sounds like you are saying “this is true for all integers”, which I don’t think is right.
No, he has multiple problems in this video. The thing is though, derivatives depend on limits, which considers a number h approaching 0 and getting arbitrarily small. Because his expansion is only valid if x is an integer, you can’t actually take the derivative in a valid manner here (at least not how he did it). There’s other issues like ignoring the “X times” you’re adding up the x, which itself is a function of x so it needs to be accounted for in the derivative.
Why does “x times x” have to be an integer x? Multiplication works with rationals. If x was 4.2, then 4.2 times 4.2 is, well, 4.2 “4.2” times: 4.2 + 4.2 + 4.2 + 4.2 + 4.2(.2)
It doesn’t have to be, but what you’re describing is a more generalized form of multiplication. The original doesn’t work with numbers that aren’t positive integers, and that’s part of the problem with expanding the expression the way he did. Also, how would you multiply something by adding it “.2” times? That’s the issue, what you did essentially is distribution, but that doesn’t exactly solve the generalization issue
That is not even a problem, x could very well be an integer and you would conclude the same. Problem is when you differentiate *x + x + x ... + x (x times)* without taking into account the (x times) which is of course x dependant.
This is the cute one that I have not seen before but I think your reason why it is fallacious is incorrect. The basic problem here is that you are misapplying the derivative. When you expand out the xs on the right side they are still continuing to change in in size because X is a variable. But they are also changing in number and there's no proper consideration for that. The derivative is the rate of change and both the size of the x's and the number of exes are changing. On the right side you only take the derivative of the size of X not the number of Xs.
This is wrong because if you add up something you can only add the single derivertiv‘s ub, if the number of the things we add up has nothing to do with the variable
There are multiple flaws in his video, and the one you just proposed still has a similar flaw. The problem with yours is considering x0=x, then treating it as a constant during differentiation. It’s a function of X, so the product rule must be applied
@@orisphera I believe so, yes. There’s an issue with his expansion of the multiplication to addition, an issue with taking the derivative, and the fact that going from 2x=x implies 2=1 removes the solution x=0, so in a general sense, yes there are more than two flaws
No im pretty sure that the first line is completely correct with all numbers, but i think the mistake comes when u take the derivative, idk why its wrong but it is
product rule should be d/dx f(x)g(x)=g(x)*f’(x)+f(x)*g’(x) but hes doing it like it is f’(x)g’(x) as hes just turning multiplication to addition and back to nultiplicatin gain
this is not true... x^2=x+x+x+x+x+x.. x times makes sense for natural numbers but it doesnt make sense for other rational numbers.(like 1.5) thus not even an identity to begin with. as we can only differentiate an identity, your diffentiating this thing here is illogical.
The explanation for why it's not correct isn't true. Just because x is an integer doesn't mean it's constant. By that logic, x would always be constant in any function, because it's just some real number. Therefore the derivative of any function should just be zero, but we know this isn't true. The actual mistake is that since we have to restrict x to be an integer, x^2 is not a continuous function. Therefore it is not differentiable, which means we get nonsense for the derivative.
"The explanation for why it's not correct isn't true. Just because x is an integer doesn't mean it's constant." Huh? Where did he say x is constant because it's an integer? "The actual mistake is that since we have to restrict x to be an integer, x^2 is not a continuous function." Any function out of a discrete space in continuous ;) However, you can't define the derivative on a discrete space, so sure.
At 1:05 sec you told that it is true for x belongs to Z which is not exactly true because if x is negative how can you add x for negative number of times so it is valid only when x is a positive integer.
that's not right, we can still say that x * x = {x + x + x + x + ... (x times)} when x is not an integer. If we do this and define adding x .5 times as adding half an x. so 5 * 2.5 = 5 + 5 + (5*0.5) = 12.5. The real reason it fails is because d/dx({x + x + x + x + ... (x times)}) isn't equal to d/dx(x) * x because of the x times
I'm a little bit sad that he said to divide by x. The way he set it up, he could have evaluated x at 1. Then you get 2 = 1 without division by 0. The problem is really in taking the derivative.
@@groszak1 I get your point, but in this case you would consider infinity as a real number. Actually it's more like a limit and I wouldn't deal with it the same way as a real number.
@@novidsonmychanneljustcomme5753 2x=x on linear scale is x+ln2=x on logarithmic scale. The logarithmic solutions are -∞ and ∞. On a linear scale, they correspond to ±0 and ±∞ respectively.
We have to just imagine this but practicaly that's not wright , The first line x²= x+x+....+x why You written this then , X²=x+(x+x+....+x) then , we get x²=x+x² that's not fare, similarly we get x=x+x+x...+x². Why that's not fare So that's not fare, 👍
if X is not an integer we can say that we have x+x+x+... and +rx, where r is equal to what X has after the decimal point, hmmmm and the derivative of that f=(x mod 1) is, well, still 1 except for the points where x is an integer
I think it would be r not 1 with what you propose. Also, 1=0 mod 1, so yeah, it’d make more sense to say 0. Also, derivatives with modular arithmetic isn’t quite as straightforward
1)it is better to say applying d/dx (x+... x) =d/dx(x) +.... is true only if we add x some constant times.infact we used to prove this fact using induction for some x added constant times. 2) yeah also this x+.. +x = x^2 true, only if x is integer and so above statement (differentiation part) false, even if differentiation done at x = integer as differentiation is done for interval not point and you can't leave non-integer part ,while taking interval Or set for x, under which f(x) is being differentiated.
In case you are not just continuing the joke, watch the explaination. It's probably not what you would expect. If you are continuing the joke, sorry for spoiling it.
Sir i have a doubt... if let x^2=2x... take ln of both sides--->2ln(x)=ln(2x) ...and take derivative of both sides ---> 2/x =1/x---> 2=1. Pls tell me where am i wrong.
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I actually expected a rickroll lol
The mistake is on the second line, where you take the derivative of the right part, without taking those "x times" into account. They are also dependent on x...
Both reason are valid.
Exactly what I was gonna say.
how would you take the derivative to take into account 'x times'
@@ichigo_nyanko You would have to define a non-integer number of additions somehow
@@ichigo_nyanko x*x
When i saw the title I thought, "definitely a hidden division by zero", then I saw the (x times) being applied to any real number and thought "I can at least credit you with that".
Also dividing both sides by x, which assumes x isn’t 0
This is the exact reason we define differentiable functions in an open, uncountable intervals. We get abnormalities like this if we include countable sets in our domain.
Hmm. The rational numbers set is countable. There was a nice proof of it, mapping natural numbers onto table of rational fractions represented as Z^2 - going through diagonals. And yet, you can find the derivative for |x^x| for negative x for which it has real values only for a certain subset of rational numbers. Of course using a limit that only operates within its domain; can't get a regular lim[h->0] with h going to zero an arbitrary path.
except we do not define differentiable functions only on open uncountable intervals
Happy April fool’s day! 🤗🤗
@@anonymous-xe6du sexy rats cats
@@anonymous-xe6du agreed
This comment saved my 1:18 seconds thankyou 💜
The real mistake is not differentiating the x in "x times" on the second line. Do that and you get the correct answer of 2x.
Wdym?
@@jamirimaj6880 d/dx( x+…+x (x times) ) = (1+…+1) (x times) + (x+…+x) (1 time) = x + x = 2x, this is like the product rule
that's not how it works
amongus
@@tavishu and what do mean by (1+1+1+..) x times like (1+1... 1) 1/2 times. Isn't adding this like x times, not something well defined?
Here's the mistake:
On the left side, you're taking a true derivative, so d/dx (x^2) is 2x
On the right side, you're actually taking a *partial derivative*, it's as if you were differentiating a 2-variable function, xy, with respect to x - the result is y, but it just so happens that y has the same value as x.
So of course the value is not equal, you're not doing the same operation on both sides.
Also, if you would take the *total differential* of the function xy, the result is x + y. If y is the same as x, the result becomes 2x, satisfying the equation
I get the joke, but this is why another math youtuber never does the "divide both sides by x" trick, but, instead, moves terms to set the equation -= 0.
So instead of jump solving 2x = x, you go to 2x - x = 0, so x(2-1) = 0, which shows that must = 0
Of course, that's not really the problem here, which is instead in the initial equality x^2 = x + x...
Sometimes I wish I had taken a course in discrete math....
@Tiffany Wart well, 2x=x does imply to 2=1 as..let say you take the value of x to be 2 ....So, it gonna be 2*2=2 which will be equal to 4=2...which is ofcourse not equal....So, ur bound to take the value of x as 1...which will be 2*1=1. Which Is 2=1 .....And sorry if I had committed any mistake, I am in my 10th year like a newbie....So, I don't know about differentiation...
@@laxmiparida4119 yep, x definitely doesn't equal 0 here.
nowhere to be seen...
Hey see 2x=x
Can be 2=1
>>> 2= x/x
>>> 2=1
Simple
It does work. 2x = x(x=0) ==> 2(0)=0 and that's the only real solution for this equation?@@cheeseburgermonkey7104
Wouldn’t the very first step be the initial mistake, where he states that all values for x will follow the assumption that x^2 = x + x + … (x times)?
Adding x x times is dependent on x. So you have to factor that when taking derivative of x+x+...+x+x.
how do you do it?
@@ichigo_nyanko By taking derivative of x², i guess.
I think what is wrong here is that on the left side it is a smooth continuous curve but on the right that same curve is being expressed as discrete sum of some terms. But we know that a smooth curve cannot be expressed as sum of discrete terms. Instead it must be expressed as an integral of some other function or basically it's own derivative.
epic
first
bri has got to be the most creative mathematician here
lol he didn't invent this
well there're some old vid that did the same
@@drk1888 ok bud
I meant generally throughout all his videos. The x^dx was a good one
i know you from musescore
I'm tired of those using equation (a+b)(a-b)=b(a-b), they don't even realize (a-b) is zero meaning it's not dividable, and finally a new one with better logic
Doing this again, paying attention to Integer parts.
x=[x]+{x}, where x>0. []=integer part, {}=fractional part.
x²=([x]+{x})²=[x]²+2[x]{x}+{x}²
Take the derivative with respect to x, keeping away from integers (where [x] is a constant, so derivative=0) and you get
2x=0+2[x]+2{x}, x∉N
2x=2([x]+{x})
2x=2x
Divide by x, since x0
2=2
Why is [x] a constant ? I don't get it.
@@abelvictor8322 it’s the floor-function =greatest integer less than. So [x]= 4 for 4
When discrete meets calculus:
Not even the integers, just the naturals, since you can’t add something to itself a negative number of times
Let's say n is a positive integer. You can define adding x with -n terms to mean adding -x with n terms. Then it actually still works and you still have n^2 = n+n+...+n (n times) for all integers n.
Use binomial expansion for negative index case
I was thinking how can you add -n numbers -n times.
Well, you could work with infinitesimals to make it "correct": In this context, taking the derivative means evaluating (f(x+dx)-f(x))/dx at dx=0 (which is an easy substitution after cancelling).
So, we have f(x+dx)=(x+dx+x+dx+...+x+dx) (x+dx times), so f(x+dx)-f(x)=dx+...+dx (x times)+(x+dx+x+dx+...) (dx times)=dx*x+(x+dx)*dx, so the derivative is x+x=2x.
everybody missed the real april's fool in the end of the video where he said that it true for x is an integer (he even put an "Z"'), when in fact it is true only for x that is *Natural*
What about this?
Define dalta_n(a_n)=a_(n+1)-a_n
We have delta_n(a_n+b_n)=a_n+1+b_n+1-(a_n+b_n)
=a_n+1-a_n+b_n+1-b_n
=delta_n(a_n)+delta_n(b_n)
by induction:delta_n(a_(n,1)+a(n,2)+...+a_(n,n))=delta_n(a_(n,1))+delta_n(a_(n,2))+...+delta_n(a_(n,k))
which means delta_n(n+n+...+n(n times))=delta_n(n)+delta_n(n)+...+delta_n(n)(n times)
delta_n(n^2)=(n+1-n)+(n+1-n)+...+(n+1-n)(n times)
(n+1)^2-n^2=1+1+...+1(n times)
2n+1=n
2(n+1)=n+1
2=1
(We also have for all n that are positive integer in every row)
EXELLENT IT IS THE MOST BEAUTIFULL TRICK I HEVE EVER SEEN
The explanation is halfway wrong. Yes it because they are integers, but that doesn't make the function a constant. Remember that differentiable functions are defined ON AN OPEN INTERVAL. Meaning uncountable sets like real numbers. Because the definition of a differential is from the definition of a limit, which requires the domain to have a number between two given points on a domain, hence the "closer and closer but never getting there" concept of directional derivatives. And obviously a domain of just whole numbers disqualifies the function from being differentiable.
You’re close with this reasoning, except consider the rational numbers. The rationals are countably infinite, and for every rational number q there exist a number between q and 0, say q/2. Forgive my lack of vocabulary, but it’s more of a “continuity” issue with the limit
when you wake to april foles
2x=x which is only satisfied by x=0 so you can not divide by x on both sides to get 2=1 in general
People on different time zones:🤔🤔
can also be said this way
d/dx x²=d/dx x.x
2x = dx/dx x
2x= 1x
2x=x
2=1
also, it is disingenuous to just differentiate the x on the top, while forgetting to differentiate the "x times" in the bracket.
I'm really bad at maths but I saw something was wrong with the x+x+x... thing almost immediately.
"Its only true if X is an integer" bruh does that mean it could be true😂😂
What does it mean for you "1/3 times"
How many one "1" you need to sum up to get 1/3
By dividing both sides of the equation with x that you eliminating the x = 0 root
That’s not the only issue, but is an issue
1:00 is that the only problem though?
I’m not great at math, but to me it sounds like you are saying “this is true for all integers”, which I don’t think is right.
No, he has multiple problems in this video. The thing is though, derivatives depend on limits, which considers a number h approaching 0 and getting arbitrarily small. Because his expansion is only valid if x is an integer, you can’t actually take the derivative in a valid manner here (at least not how he did it). There’s other issues like ignoring the “X times” you’re adding up the x, which itself is a function of x so it needs to be accounted for in the derivative.
@@andrewkarsten5268 I think I understand, thank you for the explanation.
1¹ = 1
1² = 1
1¹ = 1²
1 = 2
-B.P.R.P
Very nice❤.
Why does “x times x” have to be an integer x? Multiplication works with rationals. If x was 4.2, then 4.2 times 4.2 is, well, 4.2 “4.2” times: 4.2 + 4.2 + 4.2 + 4.2 + 4.2(.2)
It doesn’t have to be, but what you’re describing is a more generalized form of multiplication. The original doesn’t work with numbers that aren’t positive integers, and that’s part of the problem with expanding the expression the way he did. Also, how would you multiply something by adding it “.2” times? That’s the issue, what you did essentially is distribution, but that doesn’t exactly solve the generalization issue
Now that you've shown that 1=2: April 2nd= April 1st, so April 2nd is also an April fool, and by induction the whole month of April is April fool 🤪
May 1st is also April Fool’s (with this logic)
2x only equals x when x=0 therefore you can’t divide both sides by x
*This is known as beauty of mathematics 💜✨*
1:07 🙄😶 I thought the whole thing you did was true....
X should be a natural number X not = 1
if u do the factoring of both sides by factoring x u clearly see its wrong........ x^2/x does not equal 2x
Please how do u make your videos , what software
I love the animations please help me
i was solving for x when i got this:
1=0.27
What about the Rational numbers?
That is not even a problem, x could very well be an integer and you would conclude the same. Problem is when you differentiate *x + x + x ... + x (x times)* without taking into account the (x times) which is of course x dependant.
(x²-x²)=(x²-x²)
(x+x)(x-x)=x(x-x)
both sides (x-x) will canceled
(x+x)=(x)
2x=1x
both sides x will cancel as been assumed
2=1
Dividing by 0 strikes again!
You cannot divide both sides by (x-x) ... it is zero.
Wrong at last line as
2x = x
= 2x - x =0
= X = 0
So x ≠ 1
= 1 ≠ 2
This is the cute one that I have not seen before but I think your reason why it is fallacious is incorrect.
The basic problem here is that you are misapplying the derivative. When you expand out the xs on the right side they are still continuing to change in in size because X is a variable. But they are also changing in number and there's no proper consideration for that. The derivative is the rate of change and both the size of the x's and the number of exes are changing. On the right side you only take the derivative of the size of X not the number of Xs.
So 2=1 for all integers?
Ok. Good to know.
We know ,1^1=1 ---------------->(i)
1^2=1 ---------------->(ii)
Now, (i) and (ii)=>1^1=1^2
=>1=2
[cause base is same here that is 1]
This is wrong because if you add up something you can only add the single derivertiv‘s ub, if the number of the things we add up has nothing to do with the variable
Here's a version without that flaw:
x²=xx=x0x where x0=x
d/dx x² = d/dx x0x
2x=x0=x
2=1
There are multiple flaws in his video, and the one you just proposed still has a similar flaw. The problem with yours is considering x0=x, then treating it as a constant during differentiation. It’s a function of X, so the product rule must be applied
d/dx does not take into account that x0 = x
@@official-obama yes, my answer was a wordy way to say this. Yours is straightforward, much better
@@andrewkarsten5268 “multiple flaws” - are there more than two?
@@orisphera I believe so, yes. There’s an issue with his expansion of the multiplication to addition, an issue with taking the derivative, and the fact that going from 2x=x implies 2=1 removes the solution x=0, so in a general sense, yes there are more than two flaws
Yes
Why is 1+1+1+...+1 x times equal to x?
Bruh little mistake x^2 is x multiplied 2 times😑😑😑
No im pretty sure that the first line is completely correct with all numbers, but i think the mistake comes when u take the derivative, idk why its wrong but it is
Me too
nice sir
product rule should be d/dx f(x)g(x)=g(x)*f’(x)+f(x)*g’(x) but hes doing it like it is f’(x)g’(x) as hes just turning multiplication to addition and back to nultiplicatin gain
this is not true... x^2=x+x+x+x+x+x.. x times makes sense for natural numbers but it doesnt make sense for other rational numbers.(like 1.5) thus not even an identity to begin with. as we can only differentiate an identity, your diffentiating this thing here is illogical.
happy april foolz day everyone
The explanation for why it's not correct isn't true. Just because x is an integer doesn't mean it's constant. By that logic, x would always be constant in any function, because it's just some real number. Therefore the derivative of any function should just be zero, but we know this isn't true.
The actual mistake is that since we have to restrict x to be an integer, x^2 is not a continuous function. Therefore it is not differentiable, which means we get nonsense for the derivative.
That's not the error. the error is earlier with (x times). He NEVER claimed x was an integer.
@@david_haz that’s only at the very end. I’m talking about the beginning.
"The explanation for why it's not correct isn't true. Just because x is an integer doesn't mean it's constant."
Huh? Where did he say x is constant because it's an integer?
"The actual mistake is that since we have to restrict x to be an integer, x^2 is not a continuous function."
Any function out of a discrete space in continuous ;) However, you can't define the derivative on a discrete space, so sure.
The explanation is correct
At 1:05 sec you told that it is true for x belongs to Z which is not exactly true because if x is negative how can you add x for negative number of times so it is valid only when x is a positive integer.
that's not right, we can still say that x * x = {x + x + x + x + ... (x times)} when x is not an integer. If we do this and define adding x .5 times as adding half an x. so 5 * 2.5 = 5 + 5 + (5*0.5) = 12.5.
The real reason it fails is because d/dx({x + x + x + x + ... (x times)}) isn't equal to d/dx(x) * x because of the x times
That’s not really we’re the mistake is. A bunch of other people explained it in the comments
2x=x implies x=0
Oh man, this was eXtreme!
2x=x -> Only true if x=0, so you can't divide by x and this is why the last step doesn't work. 🙃
I'm a little bit sad that he said to divide by x. The way he set it up, he could have evaluated x at 1. Then you get 2 = 1 without division by 0.
The problem is really in taking the derivative.
2x=x also true if x=∞
@@groszak1 I get your point, but in this case you would consider infinity as a real number. Actually it's more like a limit and I wouldn't deal with it the same way as a real number.
@@novidsonmychanneljustcomme5753 2x=x on linear scale is x+ln2=x on logarithmic scale. The logarithmic solutions are -∞ and ∞. On a linear scale, they correspond to ±0 and ±∞ respectively.
@@groszak1 Interesting perspective, I'll think about it more in detail. ;)
Eff it! *localizes your field at 0*
Hello sir,
May I use your contents to make the same in Hindi?
I’m having a TH-cam channel named Akbar Classes.
Regards
Zareef Akbar 🇮🇳
Okay! You got me good! That was good!
but these proofs were made with the assumption that 1 = 1
We have to just imagine this but practicaly that's not wright ,
The first line x²= x+x+....+x why You written this
then ,
X²=x+(x+x+....+x) then , we get x²=x+x² that's not fare, similarly we get x=x+x+x...+x². Why that's not fare
So that's not fare, 👍
x^2=x+x+x+x___(x-time) is not continues
i have another solution multiply by zero on LHS and RHS so 0=0 hence proved
if X is not an integer we can say that we have x+x+x+... and +rx, where r is equal to what X has after the decimal point, hmmmm
and the derivative of that f=(x mod 1) is, well, still 1 except for the points where x is an integer
I think it would be r not 1 with what you propose. Also, 1=0 mod 1, so yeah, it’d make more sense to say 0. Also, derivatives with modular arithmetic isn’t quite as straightforward
(x^2 - x^2) = (x^2 - x^2)
x(x-x) = (x+x)(x-x)
x = x+x
1x = 2x
1 = 2
If 2x = x
X=0 👍
Ohhh nonono
Happy april fools day!
A subscriber sent me to your channel, could you plis stop copying my old video topics lol No idea who you are but ye
Lol It's funny how nobody realised you're in the comments lol Papa Flammy
🤣🤣
yes but not but yes
Cool
That's illegal
1)it is better to say applying d/dx (x+... x) =d/dx(x) +.... is true only if we add x some constant times.infact we used to prove this fact using induction for some x added constant times.
2) yeah also this x+.. +x = x^2 true, only if x is integer and so above statement (differentiation part) false, even if differentiation done at x = integer as differentiation is done for interval not point and you can't leave non-integer part ,while taking interval Or set for x, under which f(x) is being differentiated.
Bhai log ye chutiyapa mein kyun pad te hai
1apple can't be equal to two apple that the truth
you lost me at 0:17
0⁄0 = 1
You even explained where the mistake was... It's not fun, I wanted to say it 😂
In case you are not just continuing the joke, watch the explaination. It's probably not what you would expect.
If you are continuing the joke, sorry for spoiling it.
Hehe
Sir i have a doubt... if let x^2=2x... take ln of both sides--->2ln(x)=ln(2x) ...and take derivative of both sides ---> 2/x =1/x---> 2=1.
Pls tell me where am i wrong.
X • X = x^2. Not x + x. You can’t fool me on this April fool’s day!
SUS
x^2를 x+x+x+...+x라 두는 것은 x를 자연수라고 정한 것이기 때문에 불연속이고 미분 불가능 한데 미분을 하므로 모순임. Translate on google
Epic second
Spoiler
Best april fool I ever had
You ruined it cause the comment was visible ☹️
yeah ur an opp
Oof I’m so sorry
Huh, I had a better one today from blackbeltbarrister (thought I had to do a new driving test)
@@lyrimetacurl0 Just checked it
I’m thinking of sending it to my brother…
I think that x + x + x … + x (x times) should be x^x 🤔
I think x^x is x times x (x times)
@@dimitris_zaha yup
x+x+x+x... x times is x*x
and this is x²
Wrong hai beta jiii
𝒊 love 𝒊t!