[Ve9aPrim3 (OP)]

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(x,y,z)=(f(x)=x)+(f(x)≠x)±(<1/≥1)
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Using closest Wolfram|Alpha interpretation: (f(x)=x)
Input:
f(x) = x
Open code
Geometric figure:Properties
line
Properties as a real function:Domain:
R (all real numbers)
Open code
Range:
R (all real numbers)
Open code
Bijectivity:
bijective from its domain to R
Parity:
odd
Open code

Derivative:Step-by-step solution
d/dx(x) = 1
Open code
Indefinite integral assuming all variables are real:
integral x dx = x^2/2 + constant

[Colin2B]

The meaning is called GIGO in computing terms. Garbage in, gatbage out.
If you try to run an illogical or otherwise faulty program, or put in information which is faulty, they you will get a faulty output.
In this case the program is trying to make sense of your ‘formula’ and failing. At one point it thinks you might be describing a straight line and trying to solve a derivative.

[evan_au]

The Wolfram tools are very good at solving mathematical problems, and even general knowledge problems.
I understand that some of the smartphone voice response systems use these as one of their knowledge sources.

But the provided formula is not a mathematical problem, it is a semantic problem - the formula doesn't mean anything, so Wolfram Alpha can't make any sense of it..

[Ve9aPrim3 (OP)]

The Wolfram tools are very good at solving mathematical problems, and even general knowledge problems.
I understand that some of the smartphone voice response systems use these as one of their knowledge sources.

But the provided formula is not a mathematical problem, it is a semantic problem - the formula doesn't mean anything, so Wolfram Alpha can't make any sense of it..

Indefinite integral assuming all variables are real:
integral x dx = x^2/2 + constant
...
+constant?
Garbage in and garbage out? No way!
Considering the equation is nothing but undefined variables, I'd say they're real enough. All numbers are fundamentally imaginary anyways, and this just allows you to "see" the 3D shadow of the universe.
It's the Universal Constant buddy... I found it.

[Bored chemist]

This still makes no sense
(x,y,z)=(f(x)=x)+(f(x)≠x)±(<1/≥1)

[alancalverd]

Indefinite integral assuming all variables are real:
integral x dx = x^2/2 + constant

Absolutely true. What's the problem?

[Ve9aPrim3 (OP)]

This still makes no sense
(x,y,z)=(f(x)=x)+(f(x)≠x)±(<1/≥1)

IF
(x,y,z)=(f(x)=x)+(f(x)≠x)±(<1/≥1)
TRUE
THEN
0=1
ELSE
0=0

Proof:

f(x)=x
f(-x)=f(yz)
f(yz)=x±(0/1)
f(xyz)=f(x)+f(yz)
∅=∞
0=1

[Bored chemist]

Assuming you are right about this
"IF
(x,y,z)=(f(x)=x)+(f(x)≠x)±(<1/≥1)
TRUE
THEN
0=1"
 and given that 1 isn't zero, we can deduce that this assterion
(x,y,z)=(f(x)=x)+(f(x)≠x)±(<1/≥1)
is false by reductio ad absurdum.

[Ve9aPrim3 (OP)]

Assuming you are right about this
"IF
(x,y,z)=(f(x)=x)+(f(x)≠x)±(<1/≥1)
TRUE
THEN
0=1"
 and given that 1 isn't zero, we can deduce that this assterion
(x,y,z)=(f(x)=x)+(f(x)≠x)±(<1/≥1)
is false by reductio ad absurdum.

Unless we are looking at the current model of the universe upside down and in pieces.
"f(x)" is like balancing an equation on a teeter-totter
"(0=1)OR(0≠0)" is like balancing an equation on a pin.

[Ve9aPrim3 (OP)]

Indefinite integral assuming all variables are real:
integral x dx = x^2/2 + constant

Absolutely true. What's the problem?

No problem... but if this is absolutely true, then 0=1, 0≠0, 1=1, 1≠-1, 0±0, 0=0, 1≠1, 1=-1, 1±1... All True.
Because putting f(x)=x into wolfram..something gave me that integral... And I don't know what an integral is.

I just pictured equations being balanced differently I guess.
1=1
1≠-1
1±1
All these mean the same to me.

[Colin2B]

putting f(x)=x into wolfram..something gave me that integral... And I don't know what an integral is.

That’s fairly obvious

I just pictured equations being balanced differently I guess.
1=1
1≠-1
1±1
All these mean the same to me.

Neither do you know maths.

Posting outside of New Theories will be taken as trolling and a ban, understand? Last warning.

[Colin2B]

OP has requested all further discussion be directed to https://www.thenakedscientists.com/forum/index.php?topic=72397.0

So this topic is now locked