# Theorems of mine(?)

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)

• ($25/3$ is a rational number.) – user66081 May 5 '17 at 15:34
• When you do mathematics at least tell clearly what are you working with: are $\;\alpha,\,\beta\;$ , etc. integer numbers? Natural ones? Polynomials?.... And what you did seems to be pretty trivial working with rational numbers (if that's really what they are), but kudos for the self effort put in working out those things. – DonAntonio May 5 '17 at 15:38
• They are real numbers I thought it was obvious and didn't write it. Oh and you are right @user66081 I fixed it now :D – alienCY May 5 '17 at 17:53

## 1 Answer

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.

• Well it is trivial so I guess nobody else proved it or there is no need to be considered a theorem? If I can I want them to be my theorems (Telling people that you made a couple of theorems sounds nice :D ) And just for the record it was harder to think it rather than proving it. – alienCY May 5 '17 at 17:20
• @alienCY: nobody prevents you from telling people that, but maybe it is advisable to avoid telling it to mathematicians... but keep thinking stuff up. – user66081 May 5 '17 at 17:56