Prove axioms.

Anatoly · 2019-06-20 13:33:16

Who doesn’t know, axioms are the basic, defined as true, statements in math. They cannot be proven, but these topic contradicts it, examples would be 1 + 1 = 2, through two points only one line can go.

=== START ===

We have an array of numbers starting with 1, where every next number increases by one (natural set), let’s prove with mathematical induction, that the nth number will be derived by the formula k = n (the nth number is equal n). We’ve got the basis for n = 1, where it’s actually one. Let’s assume k = n, we get k+1 = n+1, which is true.

That’s why the n-th natural number is equal to n.

AllenCaspe9510 · 2019-06-20 13:54:12

Confusion Confirmed

Snowester

PiotrGrochowski · 2019-06-20 14:09:25

Who doesn’t know, axioms are the basic, defined as true, statements in math. They cannot be proven, but these topic contradicts it, examples would be 1 + 1 + 1 = 3, through three points only one circle can go.

=== START ===

We have an array of numbers starting with 1, where every next number increases by 2n+1 (quadratic set), let’s prove with mathematical induction, that the nth number will be derived by the formula k = n² (the nth number is equal n²). We’ve got the basis for n = 1, where it’s actually one. Let’s assume k = n², we get k+2n+1 = (n+1)², which is true.

That’s why the n-th quadratic number is equal to n².

Anatoly · 2019-06-20 14:54:11

PiotrGrochowski wrote:

k+2n+1 = (n+1)²

that is not true, k is not always n^2

PiotrGrochowski · 2019-06-20 15:30:21

Anatoly wrote:

PiotrGrochowski wrote:
k+2n+1 = (n+1)²
that is not true, k is not always n^2

but that's a quadratic sequence? in case of n=3, k=n²=9

HeyNK · 2019-06-20 23:57:42

In a plane, given a line and a point not on it, at most one line parallel to the given line can be drawn through the point, because it may also be thought of, informally, as a binary relation between two line segments xy and zw, more intuitively, that if xy is congruent with a segment that begins and ends at the same point, x and y are the same point, or even the empty set; rather than measuring distances to the point directly as in trilateration - in simple terms, with two receivers at known locations, an emitter can be located onto one hyperboloid if the receivers are placed in a configuration that minimizes the error of the estimate of the position, by measuring signals emitted from synchronized transmitters at known locations. Then, the parameters are refined iteratively, that is, the values are obtained by successive approximation, so it changes from one iteration to the next such that the origin is the center of the hyperbola and the x-axis is the major axis for an arbitrary point (x, y) in canonical form defined in terms of the unit circle, yielding the exponential form of the hyperbolic cosine.

The second case is proven analogously.

MODMERGE:

If we allow for the additional restriction that that the point MAY NOT be on the line, or that the line may not intersect the point,

the intersection of an upright double cone by the unit plane through the vertex with slope greater than the slope of the lines on the point is an arbitrary hyperbola with different x- and y-coordinates (that is, perpendicular to the hypothetical line) with the normal to the baseline greater than a right angle. These locational features seen by considering the trilinear or barycentric coordinates given above for the circumcenter can be found using a generalized method. An alternate method to determine the opposite vertex includes drawing any two of the three perpendicular bisectors to the base of the hyperbola if and only if it behaves well with respect to the operation of taking convex hulls of a non-convex set of an object among obstacles of the configuration space, which is the set of all admissible positions of the object. It follows that when individual objects to be aggregated are characterized via sets, two convex polygons P and Q can be used to model the ordered sequences counterclockwise along the polygon boundary. In other geometric contexts, this axiom also generalizes to 3D surfaces, where it is called an offset surface. In general, presuming certain conditions, one can prove the existence of given curve that satisfies the postcondition for offsets of arbitrary dimension. Assume you have a regular parametric representation of an n-dimensional surface PQ, given by the Jordan product of polygons P and Q. In this case one gets a parallel curve on the opposite side of the curve with the unit normal, but the object's shadow remains that of a perfect sphere. The opposite operation is sometimes called the radius of curvature of the given curve. If the given curve is polynomial, the cutting path is equivalent to linear interpolation.

peace · 2019-06-21 06:52:19

2^4=4^2

PiotrGrochowski · 2019-06-21 08:26:41

 ⌠a
(│ xⁿ dx)=aⁿ⁺¹÷(n+1)
 ⌡0

MWstudios · 2019-06-21 12:50:56

so, is this how you play this forum game? Increasing the text length?

drstereos, Minimania, Luka504

Minimania · 2019-06-21 14:37:55

0=1

good **** luck

▼Hidden text

Anatoly · 2019-06-23 07:43:53

Crybaby wrote:

0=1

good **** luck

▼Hidden text
Okay but like, try doing this without googling/youtubing it

Okay

0 = 1 | : 0

0/0 = 1/0
0/0 = 1
1 = 1

It's obviously correct.

Luka504 · 2019-06-23 08:01:20

Anatoly you can't divide by zeros.

peace, RailMat

ASDruska · 2019-06-23 08:07:15

0=/1

easy now do this: 5x0=0 7x0=0 5+7=0

TaskManager · 2019-06-23 08:09:47

Luka504 wrote:

Anatoly you can't divide by zeros.

He saw it in a meme so it must be true!

Anatoly · 2019-06-23 08:37:53

ASDruska wrote:

0=/1
easy now do this: 5x0=0 7x0=0 5+7=0

5 * 0 = 5 donuts sold by zero dollars gives you zero dollars = 0

Luka504 wrote:

Anatoly you can't divide by zeros.

I can.

Minimania · 2019-06-23 08:54:08

He can because 0 = 1

duh

peace · 2019-06-23 09:14:56

0 = 1 is soo wrogn my eyes cant stand to this theyr elookign at somethign impossible

Anatoly · 2019-06-23 09:25:06

Luka504 wrote:

Anatoly you can't divide by zeros.

Crybaby wrote:

He can because 0 = 1
duh

Right!

0 = 1 | : (0 = 1)
1 = 1

TaskManager · 2019-06-23 10:48:16

Anatoly wrote:

I can.

Have you ever heard of limits

Luka504 · 2019-06-23 15:19:07

Anatoly if you claim that you can divide one by zero, you're also claiming that there exists a number which, when multiplied by zero, gives one.
What number would that be?

peace

peace · 2019-06-23 15:43:22

if 0=1 then 0*x=1? but how if you dont have any group of X

Official Everybody Edits Forums

#1 2019-06-20 13:33:16

Prove axioms.

#2 2019-06-20 13:54:12

Re: Prove axioms.

#3 2019-06-20 14:09:25

Re: Prove axioms.

#4 2019-06-20 14:54:11

Re: Prove axioms.

#5 2019-06-20 15:30:21

Re: Prove axioms.

#6 2019-06-20 23:57:42, last edited by Tomahawk (2019-06-21 01:21:28)

Re: Prove axioms.

#7 2019-06-21 06:52:19

Re: Prove axioms.

#8 2019-06-21 08:26:41

Re: Prove axioms.

#9 2019-06-21 12:50:56

Re: Prove axioms.

#10 2019-06-21 14:37:55, last edited by Minimania (2019-06-21 14:38:28)

Re: Prove axioms.

#11 2019-06-23 07:43:53

Re: Prove axioms.

#12 2019-06-23 08:01:20

Re: Prove axioms.

#13 2019-06-23 08:07:15

Re: Prove axioms.

#14 2019-06-23 08:09:47

Re: Prove axioms.

#15 2019-06-23 08:37:53

Re: Prove axioms.

#16 2019-06-23 08:54:08

Re: Prove axioms.

#17 2019-06-23 09:14:56

Re: Prove axioms.

#18 2019-06-23 09:25:06

Re: Prove axioms.

#19 2019-06-23 10:48:16

Re: Prove axioms.

#20 2019-06-23 15:19:07, last edited by Luka504 (2019-06-23 15:21:31)

Re: Prove axioms.

#21 2019-06-23 15:43:22

Re: Prove axioms.

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