So I think that I have made $2$ theorems that are actually quite cool and I was wondering if they have already been proven by somebody else. Moreover are they really considered my theorems if they have not been claimed by anyone else?
Definition:
$γ = \frac αβ$
$α,β,γ,ν\in R$
Theorem $1$:
For every $ν$ incrementation on the numerator there is a $ν \cdot γ^{-1}$ incrementation on the denominator in order for the fraction to not change value.
$$ \frac αβ = \frac{α+ν}{β+ν \cdot γ^{-1}}$$
Limitations: $β\neq0,β \neq -ν\cdotγ^{-1}$
Theorem $2$:
For every $ν$ incrementation on the denominator there is a $ν \cdot γ$
incrementation on the numerator in order for the fraction to not change value.
$$ \frac αβ = \frac{α+ν\cdotγ}{β+ν}$$
Limitations: $β \neq 0 ,β \neq -ν$
Sorry if the definitions are not very clear. English is not my mother tongue and as a result I haven't been taught math in english. I have also not included the proof for my theorems because it's a pain writing math in here. Afterall I believe almost everybody can prove it themselves. No complicated math is necessary.
Oh and remember $γ = \frac αβ$ !! If you haven't noticed :D
An example using my(?) theorems:
Here we have this fraction $\frac {25}3$ if we add 5 to the numerator based on theorem 1 if we want to still have the same value ($\frac {25}3$) we must add $ν \cdot γ^{-1}$. Which is $5\cdot\frac 3{25}$ aka 0.6 so we have $\frac {25}3$ = $\frac {30}{3.6}$. So on this fraction when we add 5 to the numerator we must add 0.6 on the denominator in order to stay the same.
Adding 1 to the denominator will result in us having to add γ which is $\frac {25}3$ to the denominator. That number has a repeating decimal and it's kind of "ugly" so we conclude that we can only add multiples of 3 to the denominator in order for our nominator to be "pretty". Like if we add 3 to the denominator we must add $3\cdot\frac{25}{3}$ aka 25. So $\frac {25}3$ = $50\over6$. (Well.. pretty numbers are obvious aren't they)
Observe that
$$\frac\alpha\beta=\frac{\alpha+\nu}{\beta+\nu\gamma^{-1}}\iff \alpha(\beta+\nu\gamma^{-1})=(\alpha+\nu)\beta\iff$$
$$\stackrel{\text{since}\;\gamma^{-1}=\frac\beta\alpha}\iff\require{cancel}\cancel{\alpha\beta}+\cancel\alpha\nu\frac\beta{\cancel\alpha}=\cancel{\alpha\beta}+\nu\beta\iff \nu\beta=\nu\beta$$
and we get a proof of a rather trivial equality.
The other theorem is equally true...and trivial.