0 = 1

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>>9087088
> 0(0-10)=0
> => 0 = 0 or 10
> => 0=10

>>9087094
10 doesn't work as a value for A but 1 does.
A(A-1)=0
A=0 and A=1 both are valid.
A is constant though, not a variable.
0=A=1

>>9087088
> A=0 or 1
> A=0 and 1
> Derp.

>>9087117
>A=0 and A=1 both are valid.
No, they are not

>>9087120
yes they are

>>9087122
No, only one of them is, since A is a constant. Unless, of course, 0 = 1, but proving 0 = 1 by assuming 0 = 1 is logically fallacious

>>9087124
Where's the assumption? A is defined then it's shown that A evaluates to 0 as well as 1.

Okay so I know sqrt(4)=+/-2 doesn't imply -2=+2 but it is at least true that (-2)^2=sqrt(4)=(+2)^2. Same magnitude different sign.
So with A maybe 0=/=1 but 0^2=A=1^2 doesn't work either. Different magnitudes.

Try B:
B=sqrt(2+sqrt(2+...))
B^2=2+B
B^2-B-2=0
(B-2)(B+1)=0
B=2 or -1
Is it actually possible to correctly evaluate these things?

I know this is bait but you're dividing 0 by 0 and you can't do that shithead

>>9087180
>Where's the assumption? A is defined then it's shown that A evaluates to 0 as well as 1.
No, it evaluates to one of the two. You do not know at that point to which unless you can rule one out

>Is it actually possible to correctly evaluate these things?
Yes, limits

You need limits anyway if you want to cleanly define infinite sums (or in this case infinitely stacked square roots)

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>>9087088
Quite logical.

>>9087088
Math proofs are a meme. To only way to prove something is to feel it. Numbers aren't real, they're abstractions. That's why these things happen in math. These things don't happen in physics or chemistry.

>>9087180
>>9087196
Oh, and by the way, the limit for the first one is definitely not 1, and the limit of the second one is definitely not -1, as you can easily check with the epsilon-delta criterion

>>9087088
A= 0, so when you factor out an A in line 4, you're dividing by 0.

>>9087241
what if I don't factor out anything and just brute force the first two solutions?

>>9087241
>Factoring out = division
No

The mistake is that the solution is 0, and "A=0 or A=1" does not disagree with that

>>9087088
Fuck off with your math bullshit

>>9087088
you are defining A to be a constant
but you treat it as a variable.

All you have demonstrated is that 0 and 1 are invariant under the map [math]x\mapsto x^2[/math]

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>>9087088
OP is retarded.

A can be 0 or 1, both solutions are valid. Plug them in and you'll see.

1 = sqrt(0 + 1), this is true
0 = sqrt(0 + 0), this is also true

But this doesn't mean that 0 = 1. If you have a quadratic, say x^2 - 5x + 6 = 0, you can factor to get (x - 3)(x - 2), so x = 3, or x = 2. What you're doing is saying 3 = 2 because they are both solutions to x^2 - 5x + 6 = 0. In other words, you're an idiot.

>>9087278
but what about 2 and -1? >>9087180

>>9087289
>f(x) = x^2 - 5x + 6
>f(2)=f(3) =/=> 2=3
A has no variables though.

P -> Q is not the same as Q -> P. You're welcome.

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>>9087088
I simplified your example.

Same solution.

You are adding an "artificial" solution when you square A.

Similar to Cardano's method for the cubic.

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Found this in a book when I googled it. Looks like it depends on how you set the problem up.

Pic related

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>>9087088
x = 1
sqrt(x^2) = 1 or -1
x = -1

its a roundabout way of finding which numbers are unaffected by squaring, those numbers being 0 and 1. this doesn't mean they are equal

what a bunch of garbage.

A^2-A=0 is fine, but going to A(A-1) = 2 is incorrect because you've substituted 1 in place of A.

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>>9087088
I think you forgot a sollution
[math]A^2 = A \implies A(A-1) = A(A^2-1) = 0[/math]
So clearly:
[math]A = 0 \vee 1 \vee -1 [/math]

>>9087297
>let x=2
>x^2=4
>x^2+6=10
>x^2+6=5x
>x^-5x+6=0
>...
>2=3

2A^4 - A^2 - A = 0
;)

>>9087088
You're misusing the null factor law/zero product principle.

If x*y=0 then we can conclude x=0 OR y=0. NOT x=y=0.

>>9089260
/thread

>>9087088
>Prove me wrong
Easy. A is not a variable so you can't treat it like that, faggot.

>>9088868
so infinite nested radical is discontinuous at 0

Can someone come up with a good notation for iterated nesting? Maybe something like [math]\bigodot\limits_x^5\ f(x)[/math] to represent [math]f(f(f(f(f(x)))))[/math] so the OP thing would be [math]\bigodot\limits_n^\infty\ \sqrt{1+n}[/math]

>>9090450
Most people just use [math] f^5(x) [/math]

>A^2-A
>A(A-1)

A = 0, so by factoring out an A you are dividing by zero.

oh it's another thread where people who never took calculus try to act like math is broken

Why don't you guys go find girlfriends or something? this is sad

>>9087088
If you admit A = 0 then the step from "A(A - 1) = 0" to "A = 0 or 1" is an invalid inference.

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>>9090450
I've developed my own "recursion" notation.
You need a symbol similar to the "dx" used in integrals to specify where the recursion is happening.
I use a box.

>>9090524
great, now use >>9090487 like people who matter do

>>9087088
patrickJMT's proof of 0 = 1 is more valid than this memery

>>9090528
not all the f's are the same function

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>>9090450
>>9090487
>>9090524
>>9090528
>>9090546
Example for the stubborn.

>>9087298
That's a bad way to say it. The inclusion of Q -> P ambiguates the process being DECIPHERED.

What you need to say is that the inclusion of Q -> P makes it circular and this is a pattern of linear regression. The act of remembering. Properly. As in for yourself.

Please don't be a stupid jewish german and please remember that...suddenly I forget.

>>9090450
rad(1+n)/ ( side + side + ... );