2=1!
2=1!
I am puzzled by this: (1/x)*x=1 for all values of x including when x=0. This means that we can set x=0 in the equation which we get (infinity)*0=1. In the same manner, we can write (1/x)*2*x=2. But the lefthand side of the equation can be written as (1/x)*(x+x)=2. Now let's set x equal to zero noting that 0+0=0. we have (infinity)*(0+0)=infinity*0=1. So 2=1!
What is wrong?
Resolving this problem is necessary otherwise one cannot evaluate the first derivative.
What is wrong?
Resolving this problem is necessary otherwise one cannot evaluate the first derivative.
Re: 2=1!
1/0 is undefined, not infinity.bahman wrote: ↑Sat Feb 27, 2021 1:00 am I am puzzled by this: (1/x)*x=1 for all values of x including when x=0. This means that we can set x=0 in the equation which we get (infinity)*0=1. In the same manner, we can write (1/x)*2*x=2. But the lefthand side of the equation can be written as (1/x)*(x+x)=2. Now let's set x equal to zero noting that 0+0=0. we have (infinity)*(0+0)=infinity*0=1. So 2=1!
What is wrong?
Resolving this problem is necessary otherwise one cannot evaluate the first derivative.
Your brain probably did a lim x → 0 (1/x) and you arrived at infinity, but the limit only exists if x → 0 (approaching from the left) or x → 0+ (approaching from right) not both.
https://www.wolframalpha.com/input/?i=lim+x%3E0+1%2Fx
Re: 2=1!
So the first derivative also undefined. Please note that the first derivative is defined as the limit df(x)/dx when dx → 0. I am however interested in the exact value rather than the limit value.Skepdick wrote: ↑Sat Feb 27, 2021 6:14 am1/0 is undefined, not infinity.bahman wrote: ↑Sat Feb 27, 2021 1:00 am I am puzzled by this: (1/x)*x=1 for all values of x including when x=0. This means that we can set x=0 in the equation which we get (infinity)*0=1. In the same manner, we can write (1/x)*2*x=2. But the lefthand side of the equation can be written as (1/x)*(x+x)=2. Now let's set x equal to zero noting that 0+0=0. we have (infinity)*(0+0)=infinity*0=1. So 2=1!
What is wrong?
Resolving this problem is necessary otherwise one cannot evaluate the first derivative.
But lim x → 0+ (1/x) can never be infinity since x is a nonzero value.Skepdick wrote: ↑Sat Feb 27, 2021 6:14 am Your brain probably did a lim x → 0 (1/x) and you arrived at infinity, but the limit only exists if x → 0 (approaching from the left) or x → 0+ (approaching from right) not both.
https://www.wolframalpha.com/input/?i=lim+x%3E0+1%2Fx
 Terrapin Station
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Re: 2=1!
Nothing a good psychiatrist can't tackle if you have a few decades.bahman wrote: ↑Sat Feb 27, 2021 1:00 am I am puzzled by this: (1/x)*x=1 for all values of x including when x=0. This means that we can set x=0 in the equation which we get (infinity)*0=1. In the same manner, we can write (1/x)*2*x=2. But the lefthand side of the equation can be written as (1/x)*(x+x)=2. Now let's set x equal to zero noting that 0+0=0. we have (infinity)*(0+0)=infinity*0=1. So 2=1!
What is wrong?
Re: 2=1!
What is wrong with OP?Terrapin Station wrote: ↑Sat Feb 27, 2021 4:26 pmNothing a good psychiatrist can't tackle if you have a few decades.bahman wrote: ↑Sat Feb 27, 2021 1:00 am I am puzzled by this: (1/x)*x=1 for all values of x including when x=0. This means that we can set x=0 in the equation which we get (infinity)*0=1. In the same manner, we can write (1/x)*2*x=2. But the lefthand side of the equation can be written as (1/x)*(x+x)=2. Now let's set x equal to zero noting that 0+0=0. we have (infinity)*(0+0)=infinity*0=1. So 2=1!
What is wrong?
Re: 2=1!
[/quote]
1/0 is undefined, not infinity.
[/quote]
According to my limited understanding, It is undefined in classical arithmetic, but not necessarily in the context of number theory, if "0" is understood as Robinson's "h"  in which case 1/0 does equal infinity?
1/0 is undefined, not infinity.
[/quote]
According to my limited understanding, It is undefined in classical arithmetic, but not necessarily in the context of number theory, if "0" is understood as Robinson's "h"  in which case 1/0 does equal infinity?
Re: 2=1!
This is just a trivial consequence of Quine's "Holding true what come may". (ref: confirmation holism and the QuineDunheim thesis).
Take any symbolic expression and assume it to be a theorem of the system. That is, assume the expression evaluates to "true". Then proceed to craft (butcher?) the semantics of the system/symbols in such a way so as to always protect the theorem. If you so choose to make N + 1 = N a theorem, then you shall find a solution to that equation. The solution is ∞.
Another example is that you take the negation of the LNC as a theorem (which blows up all of Logic). So what? Now we have a system in which P and notP is true. Some people would tell you that such system is "inconsistent". OK  it's inconsistent. But it exists and executes just fine on my computer.
Re: 2=1!
(1/x)*x ≠ 1 if/when x = 0bahman wrote: ↑Sat Feb 27, 2021 1:00 am I am puzzled by this: (1/x)*x=1 for all values of x including when x=0. This means that we can set x=0 in the equation which we get (infinity)*0=1. In the same manner, we can write (1/x)*2*x=2. But the lefthand side of the equation can be written as (1/x)*(x+x)=2. Now let's set x equal to zero noting that 0+0=0. we have (infinity)*(0+0)=infinity*0=1. So 2=1!
What is wrong?
Resolving this problem is necessary otherwise one cannot evaluate the first derivative.
(1/x)*x = 0 if/when x = 0
We can not set x = 0 in the equality such to equate (1/0) = inf. as the latter equality is false. Follows: 2 ≠ 1.
The ratio 1/0 is "undefined" in ordinary mathematics & fallaciously places "something" (ie. 1) on (a base of) "nothing" (ie. 0).
Similar to the query "how can something come from nothing?", it begs the question "how can something be on nothing?"
Such a ratio has no practical meaning in the real, physical universe  (usage of) any such ratio is thus explicitly userdefined.
Re: 2=1!
I think we should define different zero as we define different infinity. The problem solved.nothing wrote: ↑Wed Apr 28, 2021 3:54 pm(1/x)*x ≠ 1 if/when x = 0bahman wrote: ↑Sat Feb 27, 2021 1:00 am I am puzzled by this: (1/x)*x=1 for all values of x including when x=0. This means that we can set x=0 in the equation which we get (infinity)*0=1. In the same manner, we can write (1/x)*2*x=2. But the lefthand side of the equation can be written as (1/x)*(x+x)=2. Now let's set x equal to zero noting that 0+0=0. we have (infinity)*(0+0)=infinity*0=1. So 2=1!
What is wrong?
Resolving this problem is necessary otherwise one cannot evaluate the first derivative.
(1/x)*x = 0 if/when x = 0
We can not set x = 0 in the equality such to equate (1/0) = inf. as the latter equality is false. Follows: 2 ≠ 1.
The ratio 1/0 is "undefined" in ordinary mathematics & fallaciously places "something" (ie. 1) on (a base of) "nothing" (ie. 0).
Similar to the query "how can something come from nothing?", it begs the question "how can something be on nothing?"
Such a ratio has no practical meaning in the real, physical universe  (usage of) any such ratio is thus explicitly userdefined.

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 Joined: Wed Jul 08, 2015 1:53 am
 Location: Saskatoon, SK, Canada
Re: 2=1!
The actual statement (1/x)*x = 1 means that x can be anything but...bahman wrote: ↑Sat Feb 27, 2021 4:28 pmWhat is wrong with OP?Terrapin Station wrote: ↑Sat Feb 27, 2021 4:26 pmNothing a good psychiatrist can't tackle if you have a few decades.bahman wrote: ↑Sat Feb 27, 2021 1:00 am I am puzzled by this: (1/x)*x=1 for all values of x including when x=0. This means that we can set x=0 in the equation which we get (infinity)*0=1. In the same manner, we can write (1/x)*2*x=2. But the lefthand side of the equation can be written as (1/x)*(x+x)=2. Now let's set x equal to zero noting that 0+0=0. we have (infinity)*(0+0)=infinity*0=1. So 2=1!
What is wrong?
Division by zero is not defined and is implicitly understood. The full explicit statement is...
(1/x)*x = 1, x is Real and x ≠ 0
It is a convention to skip the exclusionary statements when common to save clutter in easy algebraic expressions when the context is understood. All text books that are appropriately correct tell you this near the beginning of such courses. It is also repeated in other math texts, often as a summary at the very least.
You used 'derivative' colloquially and not to mean Calculus limits here. If you are expecting to extend meaning to division by zero, limits are used to express this but even there it only means that you can get as close to zero as possible but never equal zero. You CAN devise your own system if you'd like but require clear specification of what anything divided by zero means and in a way that is noncontradictory. Calculus suffices though. If you'd like a reason to understand mathematical "Limits" that define the 'derivatives', I CAN give you an example. Hint: Draw a square of ANY unit size and ask how you could represent this size using ONLY a line.
(1/x)*(x+x) = 2 is a valid sentence to mean that x = 2 ONLY. That is, you cannot force x = 0.
Re: 2=1!
The problem one will encounter if/when doing this is they will no longer be describing anything in the real, physical universe.
That is fine if the purposes for doing the same are for another system, but it would have 0 application to reality/physics.
There is no such thing as "infinity" in the real, physical universe. It is a concept/construct concocted by humans thus begins/ends there (ie. is finite).
Re: 2=1!
The continual change of the physical necessitates infinity as proven through the physical.nothing wrote: ↑Sun May 02, 2021 1:49 amThe problem one will encounter if/when doing this is they will no longer be describing anything in the real, physical universe.
That is fine if the purposes for doing the same are for another system, but it would have 0 application to reality/physics.
There is no such thing as "infinity" in the real, physical universe. It is a concept/construct concocted by humans thus begins/ends there (ie. is finite).
Re: 2=1!
1 line divided by a zero dimensional point results in two lines. 0 is a process of halving thus an manifestation of multiplicity of parts.nothing wrote: ↑Wed Apr 28, 2021 3:54 pm(1/x)*x ≠ 1 if/when x = 0bahman wrote: ↑Sat Feb 27, 2021 1:00 am I am puzzled by this: (1/x)*x=1 for all values of x including when x=0. This means that we can set x=0 in the equation which we get (infinity)*0=1. In the same manner, we can write (1/x)*2*x=2. But the lefthand side of the equation can be written as (1/x)*(x+x)=2. Now let's set x equal to zero noting that 0+0=0. we have (infinity)*(0+0)=infinity*0=1. So 2=1!
What is wrong?
Resolving this problem is necessary otherwise one cannot evaluate the first derivative.
(1/x)*x = 0 if/when x = 0
We can not set x = 0 in the equality such to equate (1/0) = inf. as the latter equality is false. Follows: 2 ≠ 1.
The ratio 1/0 is "undefined" in ordinary mathematics & fallaciously places "something" (ie. 1) on (a base of) "nothing" (ie. 0).
Similar to the query "how can something come from nothing?", it begs the question "how can something be on nothing?"
Such a ratio has no practical meaning in the real, physical universe  (usage of) any such ratio is thus explicitly userdefined.
Re: 2=1!
In reality, there is no such thing as a "point", let alone a "zero dimensional point".
Mathematicians use "points" as devices because they are useful for solving mathematical problems.
Dividing a line (l) in two is mathematically expressed as fixing a base of 2 viz. l/2.
One need not any zero dimensional "point" to do this.
0 is not a process of halving  such a suggestion is absurd.
Re: 2=1!
A point is nothing and as nothing is the gap between phenomena. Phenomenon move through nothingness.nothing wrote: ↑Wed May 05, 2021 10:14 amIn reality, there is no such thing as a "point", let alone a "zero dimensional point".
Mathematicians use "points" as devices because they are useful for solving mathematical problems.
Dividing a line (l) in two is mathematically expressed as fixing a base of 2 viz. l/2.
One need not any zero dimensional "point" to do this.
0 is not a process of halving  such a suggestion is absurd.
All mathematics are a description of reality and as descriptions necessitate them as having been grounded in some phenomenon.
Dividing a line between two points results in another line between two points. The point exists as the divisor.