**Gil's theorems** are a set of mathematical proofs discovered by Gilbert Martinez between October 2013 and May 2019. Martinez himself states, "I'd rather not think of these as legitimate math-changing proofs, but rather as a look at my evolution of my knowledge of mathematics."

## Theorem 1Edit

The first theorem, discovered in October 2013 and finalized in March 2014, states that, "Any integral can be generalized using summations."

Martinez's first example of this was found by taking the graph of *y*=*x* and setting the intervals from [0, 0] to [0, ∞). Over an extensive process, Martinez discovered that the resulting answers perfectly aligned with the graph of *y*=½*x*^{2}, thus allowing him to produce the first step:

- $ \sum^{\infty\,}_{b=0}\,\int^{b}_{0}\! x \, \mathrm{d}x = \sum^{\infty\,}_{b=0}\,\frac{1}{2}\,b^2 $

With this in mind, Martinez dropped the summations from the equation, since both sides were equal to each other. This results in the following:

- $ \int^{b}_{0}\! x \, \mathrm{d}x = \frac{1}{2}\,b^2 $

This means that, given the function *f*(*x*)=*x* of the real variable *x* and an interval of [0, *b*], where *b* can be any whole value greater than or equal to 0, of the real line, the definite integral is equal to *b*^{2} divided by 2.

Martinez didn't expand on this idea until March 2014, when he began to generalize other integrals that involved sine waves and semicircles. Here are some examples.

- $ \int^{b}_{-b}\! \sqrt{b^2 - x^2}\, \, \mathrm{d}x = \frac{b^2\pi\,}{2}\, \iff\, b>0 $

- $ \int^{b}_{b-1}\! x \, \mathrm{d}x = b-\frac{1}{2}\, $

- $ \int^{b}_{b-\pi\,}\! \sin{x}\, \, \mathrm{d}x= 2 \iff\, b = n * \pi\, $

Through this, Martinez has indirectly proposed that any integral can be generalized.

## Theorem 2Edit

The second theorem, discovered and finalized around November 2013, utilizes factorials and product notation.

Around November 2013, Martinez discovered product notation. This is what he learned of product notation:

- $ \prod^{n}_{x=m}\,f(x) $

Like summation notation, the values of *x* can range from *m* to *n* within the function *f*(*x*). However, *m* cannot be 0, nor can it be greater than *n*.

Upon experimentation, Martinez discovered that a factorial can be represented using product notation. This is what he came up with:

- $ n! = \prod^{n}_{x=1}\,x \iff\, n > 0 $

## Theorem 3Edit

The third theorem, discovered and finalized on May 4, 2014, regards the division by 0 error. Martinez's approach concerns his own observations of the graph of *f*(*x*) = *x*^{–1}. His stance is one of logic and deductive reasoning.

Martinez takes note of the fact that smaller inputs produce larger outputs, and larger inputs produce smaller outputs. He notes that, since an input of 1/4 produces an output of 4, then an output of ∞ must have resulted from an input of 1/∞. He also notes that, while 1/∞ may as well be 0, it is still greater than 0. He assumes the following:

- $ \frac{1}{\infty\,}\, = 0.\bar{0}\,1 \approx\, 0 $

I.e., Martinez assumes that the fraction of 1/∞ is equal to a decimal with infinitely many 0's that ends in a 1. Through this, he affirms that 1/∞ is not equal to 0.

He then proposes that, in order to achieve an answer for 1/0, the function must produce a number that is larger than ∞, which is impossible.

Martinez makes the same argument for 1/–∞, stating that the function must produce a number that is smaller than –∞, which is also impossible.

He ends the proof by stating, "If we put in (–∞)^{–1} for f(x) = x^{–1}, we end up with –∞; if we put in ∞^{–1} for f(x) = x^{–1}, we end up with ∞. Would 0^{–1} produce a number either bigger than ∞ or smaller than –∞? It is impossible for it to be both, yet it also is impossible for it to be either one. This is why 0^{–1} has no definite answer; it is undefined."

## Theorem 4Edit

The fourth theorem, discovered in late May 2014 and finalized on June 30, 2014, regards the 0^{0} problem. Martinez's approach to the problem involves the following expression:

- $ 1+2x+3x^2+4x^3+...=\frac{1}{(1-x)^2}\,\iff\,x<1 $

**View 1 — Using Summation Notation**Edit

This view was discovered in May 2014.

Martinez noticed that the expression could be written as a summation and remain true. In order to figure out the components of the summation, he used powers of *x*.

I.e.,

- $ 1x^0+2x^1+3x^2+4x^3+...=\frac{1}{(1-x)^2}\,\iff\,x<1 $

He noted that each exponent was 1 below the value of the term, i.e. *n*^{n–1} for every term *n*. With this, he produced the following summation:

- $ \sum_{n=1}^{\infty\,}\,nx^{n-1}=\frac{1}{(1-x)^2}\,\iff\,x<1 $

Martinez began to test values of *x* that were below 1, starting with –1.

- $ \sum_{n=1}^{\infty\,}\,n(-1)^{n-1}=1-2+3-4+...=\frac{1}{(1+1)^2}\,=\frac{1}{4}\, $

He found this to correlate to an equation described by Leonhard Euler, thus affirming that his summation was correct.

He then let *x* be equal to 0.

- $ \sum_{n=1}^{\infty\,}\,n0^{n-1}=0^0+0+0+0+...=\frac{1}{(1-0)^2}\,=\frac{1}{1}\,=1 $

Martinez noticed that, since all terms after 0^{0} couldn't influence the answer at all, 0^{0} must be responsible for the sum being equal to 1. Therefore, he proved that 0^{0} = 1.

**View 2 — Without Using Summation Notation**Edit

This view was discovered in November 2014. It is a simplified version of View 1.

Martinez took the equation

$ 1+2x+3x^2+4x^3+...=\frac{1}{(1-x)^2}\,\iff\,x<1 $
and began to ascribe values to *x* that were less than 1, thus retaining the validity of the equation.

Martinez first set *x* equal to –1 and then set it in place of *x* in the equation.

- $ 1+2(-1)+3(-1)^2+4(-1)^3+...=\frac{1}{(1-(-1))^2}\, $

He then simplified the equation.

- $ 1-2+3-4+...=\frac{1}{2^2}\,=\frac{1}{4}\, $

Like before, Martinez found it to align with Euler's equation. With this, Martinez then set *x* to 0 and made the subsequent changes.

- $ 1+2(0)+3(0)^2+4(0)^3+...=\frac{1}{(1-0)^2}\, $

Simplified, this becomes:

- $ 1+0+0+0+...=\frac{1}{1}\,=1 $

This makes sense because adding an infinite amount of 0s to 1 will not change the value of the equation, hence it remains as 1.

Martinez then returned to his summation from View 1 and decided that the two must be equal, since they produce the same answer.

- $ 0^0+0+0+0+...=1+0+0+0+... $

Therefore,

- $ 0^0=1 $

## Theorem 5Edit

The fifth theorem, discovered on July 31, 2014, is a geometric proof. Using a ruler and compass, as well as a scientific calculator, Martinez was looking at dividing areas of circles by 2.

The first circle, which he labeled Circle *A*, had a radius of 2 inches. Therefore, the area of Circle *A*, labeled *A*_{A}, comes out to 4π in^{2}.

He divided *A*_{A} by 2, producing 2π in^{2}. He found the radius of Circle *B* to be equal to the square root of 2 inches.

Through this process, Circle *C* has a radius of 1 inch, and Circle *D* has a radius of root 2 ÷ 2 inches, and Circle *E* has a radius of half an inch, etc.

Martinez then found the differences between each radius, such that *r*_{n+1} – *r*_{n} is the difference between two consecutive radii. He assigned the letters *H* through *M* to each difference *D*.

*D*_{H}=*r*_{A}–*r*_{B}= .8284271247...*D*_{I}=*r*_{B}–*r*_{C}= .5857864376...*D*_{J}=*r*_{C}–*r*_{D}= .4142135624...*D*_{K}=*r*_{D}–*r*_{E}= .2928932188...*D*_{L}=*r*_{E}–*r*_{F}= .2071067812...*D*_{M}=*r*_{F}–*r*_{G}= .1464466094...

He then divided each difference by the preceding difference, such that *D*_{n} ÷ *D*_{n+1} is the quotient of two consecutive differences. He assigned the letters *N* through *R* to each quotient *Q*.

*Q*_{N}=*D*_{I}÷*D*_{H}*Q*_{O}=*D*_{J}÷*D*_{I}*Q*_{P}=*D*_{K}÷*D*_{J}*Q*_{Q}=*D*_{L}÷*D*_{K}*Q*_{R}=*D*_{M}÷*D*_{L}

In each case, he found that *D*_{n} ÷ *D*_{n+1} = root 2 ÷ 2 inches. He also found the same for the differences of each difference, and the differences of those differences, etc.

Therefore, given that *A* and *m* are the area of each circle *n*, and *l* is each difference *D*_{m+1} – *D*_{m},

- $ \frac{D_m}{D_{m+1}}\, = \frac{\sqrt{2}\,}{2}\, \and\, \frac{E_l}{E_{l+1}}\, = \frac{\sqrt{2}\,}{2}\, \and\, ... \iff\, A_{n+1} = 2A_n \and\, D_m = r_{n+1} - r_n \and\, E_l = D_{m+1} - D_m \and\, ... $

On August 1, 2014, Martinez became interested in what would happen with other factors. He looked at what would happen if each circle was one-third the area of the previous circle. He found the ratio of the differences to always equal root 3 ÷ 3. The same happened with 4, 5, etc. This allowed him to revise his formula even further, as well as simplify it. He used *z* to denote the scale factor. He also clarified that *variable*_{n+1} was the next variable after *variable*_{n}.

- $ \frac{D_{n+1}}{D_n}\, = \frac{\sqrt{z}\,}{z}\, \iff\, A_{n+1} = \frac{A_n}{z}\, \or\, zA_{n+1} = A_n \and\, D_n = r_n - r_{n+1} \and\, z > 0 $

Martinez then looked at the quotient of differences when the gap was 2. He found it to equal one half. When it was 3, it was equal to root 2 ÷ 4. When it was 4, it was equal to one fourth. He found this to correlate to the graph of (root 2 ÷ 2)^{x}, and so he revised the formula once more. He let *x* represent the gap between differences. This resulted in the final form of the mathematical proof of the theorem.

- $ \frac{D_{n+x}}{D_n}\, = (\frac{\sqrt{z}\,}{z}\,)^x \iff\, A_{n+1} = \frac{A_n}{z}\, \or\, zA_{n+1} = A_n \and\, D_n = r_n - r_{n+1} \and\, z > 0 $

The inverse of the formula is

- $ \frac{D_n}{D_{n+x}}\, = (\sqrt{z}\,)^x \iff\, A_{n+1} = \frac{A_n}{z}\, \or\, zA_{n+1} = A_n \and\, D_n = r_n - r_{n+1} \and\, z > 0 $

Martinez combined these two, since they led to the same thing, at 11:22 PM PDT to produce the final formula.

- $ \frac{D_{n+x}}{D_n}\, = (\frac{\sqrt{z}\,}{z}\,)^x \or\, \frac{D_n}{D_{n+x}}\, = (\sqrt{z}\,)^x \iff\, A_{n+1} = \frac{A_n}{z}\, \or\, zA_{n+1} = A_n \and\, D_n = r_n - r_{n+1} \and\, z > 0 $

This is a generalization of consecutive circles scaled up or down by a factor of *z*, as long as *z* is greater than 0.

## Theorem 6Edit

This theorem, "discovered" early in 2015, is Martinez's solution to the Three Square Geometry Problem. It makes use of the Pythagorean theorem and basic trigonometry.

The problem is described as follows:

- There are three congruent squares connected in a row. Draw three lines connecting the bottom right corners of the squares to the top left corner of the leftmost square. Label the angles formed at the bottom right corners as α, β, and γ, respectively. Prove that the sum of angles α β and γ is equivalent to 90°.

Martinez first set the sides of the squares equal to *x*. He then focused on the leftmost triangle formed with the angle α that was contained in the leftmost square, which was a right isosceles triangle. Since he knew that the lengths of the sides were equivalent, he used the Pythagorean theorem to find the length of the hypotenuse.

- $ x^{2}+x^{2}=c^{2} $
- $ 2x^{2}=c^{2} $
- $ x\sqrt{2}\,=c $

Martinez then focused on the triangle whose legs were *x* and 2*x*.

- $ x^{2}+(2x)^{2}=c^{2} $
- $ x^{2}+4x^{2}=c^{2} $
- $ 5x^{2}=c^2 $
- $ x\sqrt{5}\,=c $

Martinez then focused on the triangle whose legs were *x* and 3*x*, which was the rightmost triangle.

- $ x^{2}+(3x)^{2}=c^{2} $
- $ x^{2}+9x^{2}=c^{2} $
- $ 10x^{2}=c^2 $
- $ x\sqrt{10}\,=c $

With the sufficient information needed to use trigonometric ratios, Martinez began to find the measures of the angles α β and γ. Martinez's ratio of choice was sine, though he has stated that any ratio will work as long as it is used consistently.

- $ \sin{\alpha}\,=\frac{x}{x\sqrt{2}\,}\,=\frac{x\sqrt{2}\,}{2x}\,=\frac{\sqrt{2}\,}{2}\, $

- $ \sin{\beta}\,=\frac{x}{x\sqrt{5}\,}\,=\frac{x\sqrt{5}\,}{5x}\,=\frac{\sqrt{5}\,}{5}\, $

- $ \sin{\gamma}\,=\frac{x}{x\sqrt{10}\,}\,=\frac{x\sqrt{10}\,}{10x}\,=\frac{\sqrt{10}\,}{10}\, $

Then, Martinez used the inverse sine function to find the values of the angles α β and γ. He measured them by degrees.

- $ \sin^{-1}{\frac{\sqrt{2}\,}{2}\,}\,=45=\alpha $

- $ \sin^{-1}{\frac{\sqrt{5}\,}{5}\,}\,\approx\,26.565051177078=\beta $

- $ \sin^{-1}{\frac{\sqrt{10}\,}{10}\,}\,\approx\,18.434948822922=\gamma $

Finally, Martinez summed the values to find their sum.

- $ 45+26.565051177078+18.434948822922=90\therefore\,\alpha+\beta+\gamma=90 $

This concluded his proof.

## Theorem 7Edit

This theorem, "discovered" on December 12, 2016, concerns the nature of perfect numbers. Here, Martinez proves that all perfect numbers with a Mersenne prime in its factorization are even. He hopes to someday extend this to prove that there are no odd perfect numbers.

**Part One: Proving the Link between Mersenne Primes and Perfect Numbers**Edit

A perfect number is any number that is equal to the sum of its factors, excluding the number itself. e.g.:

- $ \text{Factors of 6, excluding 6 itself:}\, 1, 2, 3; 1 + 2 + 3 = 6 $

- $ \text{Factors of 28, excluding 28 itself:}\, 1, 2, 4, 7, 14; 1 + 2 + 4 + 7 + 14 = 28 $

A Mersenne prime is a prime number that is one less than a power of 2. e.g. 3 is a prime number, and it is equal to 2^{2} – 1, which is 4 – 1.

- $ n: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,... $

- $ 2^{n}: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192... $

- $ 2^{n}-1: 0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 8191... $

Of those in the list for 2^{n} – 1, the primes are 3, 7, 31, 127, 8191, etc. This corresponds to *n* = 2, 3, 5, 7, 13, etc. Not all prime *n* will give you a Mersenne prime, but every Mersenne prime has prime *n*.

Martinez observed that, for each perfect number, there was only one Mersenne prime among their factors. For the time being, he defined perfect numbers as numbers whose factors sum to twice the original number, e.g.:

- $ \text{Factors of }\,6: 1, 2, 3, 6; 1 + 2 + 3 + 6 = 12 = 2(6) $

- $ \text{Factors of }\,28: 1, 2, 4, 7, 14, 28; 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2(28) $

Noticing a specific pattern, and using the next perfect number (496) as the example, Martinez showed the following way to add the sum of a perfect number's factors. First, he showed the sum of all the factors of 496.

- $ 2(496) = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 + 496 $

He then factored out 31 from the final 5 terms...

- $ 2(496) = (1 + 2 + 4 + 8 + 16) + 31(1 + 2 + 4 + 8 + 16) $

...arranged the problem as so...

- $ 2(496) = (31 + 1)(1 + 2 + 4 + 8 + 16) $

...and wound up with this.

- $ 2(496) = 32(1 + 2 + 4 + 8 + 16) $

He then found the sum of the terms in the parentheses.

- $ 1 + 2 + 4 + 8 + 16 = 31 \therefore\, 32 * 31 = 992 = 2(496) $

From this, Martinez went on to demonstrate a general case.

Let 2^{n} – 1 be prime and *P* represent a perfect number.

- $ \text{Sum of factors of}\,P = 2P = \sum_{k=0}^{n-1}\,2^{k} + (2^{n}-1) \sum_{k=0}^{n-1}\,2^{k} $

Here, the summation adds successive powers of 2 up to 2^{n – 1}, since 2^{n} would be larger than the required Mersenne prime 2^{n} – 1. We'll look at what each summation sums to in a moment. Like before, we combine the factors of the sums, resulting in the following:

- $ 2P = (2^{n} - 1 + 1) \sum_{k=0}^{n-1}\,2^{k} $

Simplified:

- $ 2P = 2^{n} \sum_{k=0}^{n-1}\,2^{k} $

Now to find out what the summation is equal to. Let's lay out the summation, like before:

- $ \sum_{k=0}^{n-1}\,2^{k} = 2^{0} + 2^{1} + 2^{2} + 2^{3} + ... + 2^{n-2} + 2^{n-1} $

Let's multiply this sum by 2. From this, we get the following:

- $ 2 \sum_{k=0}^{n-1}\,2^{k} = 2(2^{0} + 2^{1} + 2^{2} + 2^{3} + ... + 2^{n-2} + 2^{n-1}) = 2^{1} + 2^{2} + 2^{3} + 2^{4} + ... + 2^{n-1} + 2^{n} $

Let's subtract the latter result from the former. Namely...

- $ (2-1) \sum_{k=0}^{n-1}\,2^{k} $

All the terms get cancelled out, except for two: 2^{n} and 2^{0} (which is equal to 1).

- $ \therefore\,(2-1) \sum_{k=0}^{n-1}\,2^{k} = \sum_{k=0}^{n-1}\,2^{k} = 2^{n} - 1 $

Sure enough, here is our Mersenne prime. Adding this back into the original equation...

- $ 2P = 2^{n}(2^{n}-1) $

Since the sum of the factors of *P* is equal to 2*P*, we should divide by 2 to get our value of *P*.

- $ \frac{2^{n}(2^{n}-1)}{2}\, = (2^{n-1})(2^{n}-1) = P $

Therefore, any perfect number *P* can be expressed as the product of a power of 2 and a Mersenne prime, given that the Mersenne prime + 1 is twice the size of the power of 2.

**Part Two: Using the Result of Part One to Prove ***P* Is Always EvenEdit

*P*Is Always Even

This part should be much faster to get through.

As proved in Part One, any perfect number *P* can be expressed as the product of a power of 2 and a Mersenne prime, given that the Mersenne prime + 1 is twice the size of the power of 2. i.e.:

- $ P = (2^{n-1})(2^{n}-1) \iff\, 2^{n}-1\text{ is prime}\, $

Here are some observations to take into consideration:

- If
*n*> 0, then 2^{n}is always even. - If
*n*> 1, then 2^{n – 1}is always even. - If
*n*> 0, then 2^{n}– 1 is always odd.

Let us now consider the smallest perfect number, 6.

- $ 6 = 2 * 3 $

In this case, 2 = 2^{1}, which is even because 1 > 0; and 3 = 2^{2} – 1 = 4 – 1, which is odd because 2 > 0. As shown here, an odd number multiplied by an even number results in an even number, therefore all perfect numbers derived from Mersenne primes are even; *P* is always even.

Q.E.D.

Now to prove that there aren't any *odd* perfect numbers...

## Theorem 8 (Theta theorem)Edit

This theorem, "discovered" by Martinez on 29 Aug 2018 and refined the next day, regards what he called the theta function (which he described the same day).

The theta function is as follows:

- $ \Theta (n, m) = e^{ni}m = (\cos{n}+i\sin{n})(a+bi) $

As can be seen above, the theta function inputs two variables (*n* representing the angle of rotation and *m* representing the complex number to be rotated).

This theorem regards cases when Θ(*n*, *m*) equals 0. Martinez conjectured that this was only possible when *m* = 0. The proof of this conjecture is as follows:

Assume Θ(*n*, *m*) equals 0 and *m* ≠ 0. This would mean the following:

- $ \Theta (n, m) = e^{ni}m = 0. $

Since *m* is nonzero, that would mean *e*^{ni} = 0, since multiples of 0 must equal 0. Therefore, there must be a way to exponentiate any number to reach 0. However, this is not possible unless the exponentiated number is 0. Thus, *e* must be equal to 0.

However, this is not true.

- $ e = \lim_{n → \infty} (1 + \frac{1}{n})^{n} \neq 0 $

Thus, *e*^{ni} ≠ 0. This implies that there is a way to get a product of 0 from two nonzero numbers. This is impossible.

- $ \therefore \Theta (n, m) = 0 \iff m = 0 \square $

## Theorem 9 (Daleth theorem)Edit

On 27 Sep 2018, Martinez created the daleth function, which would generate any geometric series.

- $ \daleth (a, r) = a \sum_{n=0}^{\infty} r^{n} = \frac{at}{t-s} \iff |r| < 1 \and r = \frac{s}{t} $

Here, *a* is the first term of the series, and *r* is the fraction that is exponentially multiplied to each term.

- $ a \sum_{n=0}^{\infty} r^{n} = a + ar + ar^{2} + ar^{3} + ... $

Martinez hypothesized that any rational number could be generated using this function, and after a day of hard work, he proved this hypothesis on 28 Sep 2018. He split his proof into two parts:

- Generation of any integer.
- Generation of any fraction, which would in turn verify part 1.

**Part I**Edit

There are a few ways to generate any integer using the daleth function. The following is the trivial way.

- $ \daleth (1, \frac{x-1}{x}) = \frac{1 * x}{x - (x - 1)} = \frac{x}{x - x + 1} = \frac{x}{1} = x $

For a nontrivial way, we'll have to get more general.

Let *a* = *p* ÷ *q*, *r* = *s* ÷ *t*, and |*r*| < 1.

- $ \daleth (\frac{p}{q}, \frac{s}{t}) = \frac{pt}{q(t-s)} $

Let's suppose this equals a whole number *n*.

- $ \frac{pt}{q(t-s)} = n $

- $ \frac{t}{t-s} = \frac{nq}{p} $

- $ \therefore t = nq \and p = t-s = nq-s $

This will work for multiples of *p* as well.

**Part II**Edit

The approach for this part of the proof will need to be the same as the first part.

Let *a* = *p* ÷ *q*, *r* = *s* ÷ *t*, and |*r*| < 1.

- $ \daleth (\frac{p}{q}, \frac{s}{t}) = \frac{pt}{q(t-s)} $

Let's suppose this equals a fraction *x* ÷ *y*, where *y* ≠ 0.

- $ \frac{pt}{q(t-s)} = \frac{x}{y} $

- $ \frac{t}{t-s} = \frac{xq}{yp} $

- $ \therefore t = xq \and yp = t-s = xq-s \therefore p = \frac{xq-s}{y} $

This will work for multiples of *p* as well. $ \square $

## Theorems 10 & 11 (δ_{N} theorems)Edit

The δ_{N} theorems, "discovered" by Martinez from 03 to 05 May 2019, regards the ratios of consecutive numbers in additive sequences related to and including the Fibonacci numbers.

- $ \delta_{N} = \lim_{n → \infty} \frac{P_{n}}{P_{n-1}} $
- $ \delta_{N} = \frac{N+\sqrt{N^{2}+4}}{2} $

where

- $ P_{n} = NP_{n-1} + P_{n-2} $

**Theorem 10: Is δ**_{N} whole?Edit

_{N}whole?

Assume yes.

- $ \therefore N+\sqrt{N^{2}+4} = 2x \because \delta_{N} = \frac{N+\sqrt{N^{2}+4}}{2} $
- $ \sqrt{N^{2}+4} = 2x-N $

Let's check the result.

- $ (2x-N)^{2} = 4x^{2}-4xN+N^{2} \neq N^{2}+4 \square $

Another observation that proves this result is that, by default, N^{2} + 4 will never be a square. $ \square $

**Theorem 11: Is δ**_{N} rational?Edit

_{N}rational?

**Proof I: Using fractions**Edit

Assume yes.

The following can be trivially observed.

- $ \delta_{N} = N + \cfrac{1}{N + \cfrac{1}{N + \cfrac{1}{N + {_\ddots}}}} $

Because the fraction is infinitely continuous, it can never be expressed as a fraction. $ \square $

**Proof II: Using the Daleth function and theorem**Edit

Assume yes.

- $ \mathbb{N} \in \mathbb{D} \iff \daleth (a, r) = \mathbb{D} $

- $ \therefore \daleth (a, r) = \delta_{N} \and \delta_{N} \in \mathbb{D} $

According to the Daleth theorem, all numbers generated by the function are rational if only whole numbers are used.

Let *a* = *p* ÷ *q*, *r* = *s* ÷ *t*, and |*r*| < 1.

- $ \therefore \frac{pt}{q(t-s)} = \frac{N+\sqrt{N^{2}+4}}{2} $

Here's the results of this equality:

- $ q(t-s) = 2 \therefore t-s = \frac{2}{q} $

Since 2 is prime, and *p*, *q*, *s*, and *t* are whole,

- $ |t-s| = 1 \or 2 \because |q| = 1 \or 2 $

Now for the numerator.

- $ pt = N+\sqrt{N^{2}+4} $

Since *p* and *t* are whole, then $ N+\sqrt{N^{2}+4} $ must be whole. However, as stated in Theorem 10, N^{2} + 4 will never be a square. This implies a non-whole sum can be the product of two whole numbers, which is impossible.

- $ \therefore \delta_{N} \notin \mathbb{D} \square $