19-14=3

When $19-14=3$ ,
$19=17$ , as the Right moves Leftwards,
Unharming the $\mathbb{R}$ line nor logic,
But invalidating the premise itself.
An $\neq$ would convert to $=$ if
We change our original stance.

When $19-14=3$ ,
$14=16$ , as the Left moves Rightwards,
Unvanquishing the linearity of $\mathbb{R}$ ,
By tweaking the premise itself.
An error would automatically rectify if
We decide to move ahead.

When $19-14=3$ ,
$x=\frac{5}{3}x \ \forall\ x \in \mathbb{R}$ , as the range contracts,
Completely deforming and twirling the $\mathbb{R}$ into an Accordion-
A never-ending coiling of an Accordion, $x_n= \left(\frac{5}{3}\right)^n x_0$ .
The coiling becomes so dense that-
Despite reaching $19$ in three integral steps, if we get lost in between,
we’d never reach the terminals $14$ or $19$ till eternity.
A blunder could be blemished by patching over it, but
We must burden the consequences and spiraling regrets upon us.

When $19-14=3$ ,
2 elements from $\{15,16,17,18\}$ will be kicked out,
Bearing holes in the interval $[14,19]$ ,
A random action that untangles a network of possibilities.
If Neighbours : $15$ & $16$ –
We’d kick out the integers, and the set $\{ x \in \mathbb{R}\ |\ floor(x)=15\ OR\ 16 \}$ ,
Creating a single big hole in the $[14,19]$ continuum.
Alternating neighbours : $15$ & $17$ –
Create two holes in the $[14,19]$ continuum.
Cousins like $15.4$ & $16.5$ –
Create two tenth of a hole in the $[14,19]$ continuum,
And hence we need to kick ten set of cousins out,
And we get twenty holes in the continuum
Strangers like $5\pi$ , $\sqrt{67 e^3}\ \log_{10}3$ –
Create uncountable holes in the continuum;
If the Strangers are rational, the hole count would be
$N=\ \sum^m_{i=1}{10^{\psi \left(a_i \right)}}\qquad \forall\ a_i\ \in \mathbb{Q}$
and $\psi(x)$ provides the number of significant figures in $x$ , and
The $[14,19]$ continuum would look like MacBeth written in freakin’ Morse Code.
If irrational, then $\lim_{x \in \bar{\mathbb{Q}}} \psi(x)=\infty$ ,
The number of holes would go berzerk, and it ain’t pretty!
In the course of masking a glitch, we’d end up having
Infinitesimally significant infinite casualties.

When $19-14=3$ ,
Two sets of 2 elements from ${15,16,17,18}$ will be clubbed together,
Forming kinks / hills on the existing $\mathbb{R}$ continuum,
Resulting in a chaotic range of hills in the entire number line.
$15$ will be divisible by $4$ , and $16$ will be divisible by $5$ ,
Blackjack would become a terrible gambling game,
Providing more probabilistic advantage to the Player than the Dealer.
$17$ will stop being a prime number, OR
Both $17$ and $18$ will be prime numbers.
Neighbours: $(15,\ 16),\ (16,\ 17)$ would create one hill on the continuum.
Alternating Neighbours; create two hills on the continuum.
Cousins; create twenty hills on the continuum.
The determining the number of hills would be similar to that of number of holes, and
The $[14,19]$ continuum would look like an ECG of a Methhead doing Gangnam Style.
In the course of covering a screw-up, we’d end up producing
A plethora of consequences along with it.

A mistake, an error, a fuck-up has transformed a simple relation
Into a messy, chaotic, scary monstrosity.
Actually, justification of the said mistake has created the chaos.
However, it has been agreed from the start that $19-14=3$
Is a mistake, and I have been trying to justify the consequences.

But, is it still a mistake just because it leads to chaos?
Because of how convoluted, and topologically inequivalent
The number line would become, does that make it a blunder?
Why do we yet expect things to run smooth and linear,
When the entire existence and reality are consequences of chaos?

Sometimes, alcohol is the solution.
If you have $19\ ltr$ of alcohol, and
I borrow $14\ ltr$ from you,
You’ll be left with $3\ kg$ of alcohol.