# Secret pattern of irrational numbers

Are their any pattern in irrational numbers?, I know that there's been no hint as to the appearance of their infinite digits, but I discovered a pattern within themselves

$$\sqrt1 = 1$$

$$\sqrt2 = 1.414....$$

$$\sqrt3 = 1.732....$$

$$\sqrt(4) = 2$$

$$\sqrt(5) = 2.236$$

$$\sqrt(6) = 2.449$$

$$\sqrt(7) = 2.645$$

$$\sqrt(8) = 2.828$$

$$\sqrt(9) = 3$$

$$\sqrt{10} = 3.162$$

$$\sqrt{11} = 3.316$$

$$\sqrt{12} = 3.464$$ etc...

It turns out that the numbers are increasing with the same rate

Notice that in the section of $$1 → 4$$, it seems that the numbers increased by $$0.3 (1/3)$$, while in the section of $$4 → 9$$ there is an increase rate of $$0.2 (1/5)$$, while in the section of $$9 → 16$$, an increase by $$\sqrt(10)-3 , (5/7)$$ and also the section of $$16 → 25$$, an increase by $$\sqrt(17)-4, (2/9)$$

So this process keeps on happening and continues forevermore

To get the exert value, notice it's an A.P. I'll skip some details now to write the terms of the progression

$$1, (x+2)/3 ,4, (x+6)/5 ,9, (x+12)/7 , 16, (x+20)/9 ,25,............$$

I'll try to find an approximate formula for the square root of a number into fraction, using the idea of its nearest perfect square

Let's say a number $$x$$ lies in the number plane $$x£Z$$ as $$0,1,2,3,.......,n,.........,x,...........,m,.........,∞$$ Every number $$x$$ is always bounded by two perfect square, the one before it say $$n$$ and the after it say $$m$$

Now let's write out the set of square roots $$0,1,1.414..,.....,2,....,3,......,\sqrt(n),......,\sqrt(x),........,\sqrt(m),........,∞$$ Since $$\sqrt(x)$$ let's between $$\sqrt(n)$$ and $$\sqrt(m)$$, there must be a rate to with $$\sqrt(n)$$ increases from, even to $$\sqrt(x)$$ and until $$\sqrt(m)$$

By checking the progression, let's skip the calculations to write the basic term, Therefore $$:$$

$$\sqrt(x) ≈ (x + (n+m-1)/2)/(m-n)$$

Since $$n$$ and $$m$$ are closest perfect squares, say $$n=k^2$$ and $$m=(k+1)^2$$

$$\sqrt(x) ≈(x+(k^2+(k+1)^2-1)/2)((k+1)^2-k^2)$$

Which then becomes

$$\sqrt(x) ≈ (x+k^2+k)/(2*k+1)$$ Where $$k^2 ≤ x$$

In fact the statement can be true $$\sqrt(x) = \lim(x→∞,k^2→x) (x+k^2+k)/(2*k+1)$$

We can use this to approximate the square root of many numbers Examples:

$$\sqrt(128) = 11.313....$$

$$\sqrt(128) ≈ (128+11^2+11)/(2*11+1) ≈ 260/23 ≈ 11.304.....$$ because $$11^2<128$$

$$\sqrt(571) = 23.895....$$

$$\sqrt(571) ≈ (571+23^2+23)/(2*23+1) ≈ 1123/47 ≈ 23.893....$$ because $$23^2<571$$

$$\sqrt(75) = 8.660.....$$

$$\sqrt(75) ≈ (75+8^2+8)/(2*8+1) ≈ 147/17 ≈ 8.647....$$ because $$8^2<75$$

$$\sqrt(1004) = 31.685.......$$

$$\sqrt(1004) ≈ (1004+31^2+31)/(2*31+1) ≈ 1996/63 ≈ 31.682....$$ because $$31^2<1004$$

Question: What are the meaning of this, If there is a pattern within themselves, why no pattern in their digits?

It's not hard to conclude that something similar to this would exist for $$\sqrt[3]()$$, $$\sqrt[4]()$$, etc, then I'll suggest something more complicated than this exists of all Algebraic-irrational number

Since this numbers here belong to the same group, roots , is it possible that patterns can be found within transcendental numbers of same group

• Since $\sqrt{x}$ is a nice, differentiable function, a linear approximation is pretty good on small intervals. So yes, there would be similar patterns for $\sqrt[3]{x}$ and $\sqrt[4]{x}$ etc., or for $\log(n)$ for that matter. Apr 14, 2020 at 12:25
• Note that not all the numbers you listed are irrational... $\sqrt{4} = 2$ and $\sqrt{9} = 3$, for instance. In addition, there are $2^{\aleph_{0}}$ irrationals between any of the irrationals that you listed, so I think you are more interested in checking whether there is a pattern in the square roots of positive integers. Is this correct? Apr 14, 2020 at 12:35
• I've seen a pattern within the digits of square root of different number, Apr 14, 2020 at 12:55
• "I've seen a pattern within the digits of square root of different number" No, you haven't. Your "patterns" have NOTHING to do with the DIGITS, but with the number themselves. You need to learn to distinguish between properties of the numbers and properties of their representations. Eg: the property "24 is even" is a property of the number 24 itselft (it does not matter if I write it with decimal digits, or binary, or hexa, or with roman numerals). Instead, "141 is palidromic" is a property of its representation (decimal in this case); if I write in in binary (or in roman) that's false. Apr 14, 2020 at 17:28
• This is not about a "pattern in irrational numbers", but rather about a pattern in the square-root function. Aug 2, 2020 at 20:51

Let $$k^2$$ be the largest square less than or equal to $$x$$. Since $$\sqrt{x}$$ flattens out for large $$x$$, drawing a secant line between the points at $$k^2$$ and $$(k+1)^2$$ should give a good approximation of $$\sqrt{x}$$, i.e.$$\sqrt{x} \approx k + \frac{x-k^2}{(k+1)^2 - k^2}\left(k+1 - k\right) = k + \frac{x - k^2}{2k+1} =\frac{x+k^2+k}{2k+1}$$ But how good of an approximation is it? Your calculations suggest it's a good one, but let's use a little calculus and get more precise: Notice that the approximation is exact for $$x=k^2$$ or $$x=(k+1)^2$$. We will find the maximum of the difference on the interval $$[k^2,(k+1)^2]$$ by looking for where the derivative is $$0$$: $$\begin{eqnarray} \frac{d}{dx} \left(\sqrt{x} - \frac{x+k^2+k}{2k+1}\right) &=& 0\\ \frac1{2\sqrt x} - \frac{1}{2k+1} &=& 0\\ \sqrt{x} &=& k+\frac12\\ x &=& (k+\frac12)^2 = k^2 + k +\frac14 \end{eqnarray}$$ hence the most this approximation can be off by is at $$x=k^2 + k +\frac14$$, at which point: $$\begin{eqnarray} \sqrt{x} - \frac{x+k^2+k}{2k+1} &=& k+\frac12 - \frac{k^2 +k +\frac14 + k^2 + k}{2k+1} = k+\frac12 - \frac{2k^2 + 2k +\frac14}{2k+1}\\ &=& \frac{2k^2+k+k+\frac12 -(2k^2 + 2k +\frac14)}{2k+1} = \frac1{4(2k+1)} \end{eqnarray}$$ Hence even for $$k=1$$, the approximation is not off by more than $$1/12$$, which is pretty good, and it gets better for larger $$k$$.

In general, you can get a similarly good approximation by secants to any function that flattens off for large $$x$$, so the same behavior will be observed for higher order roots as well. This is not a special property of the individual numbers $$\sqrt{6}$$, $$\sqrt7$$, etc., so much as it is a special property of the square root function.

First: "patterns" in the (decimal) digits representation of numbers have practically no relationship with your approximations, which only deal with the values.

Your approximation is actually a linear interpolation.

In general, if we know the values of a function in two points $$a,b$$ we can attempt to approximate an intermediate value by

$$f(x) \approx f(a) + (x-a) \frac{f(b)-f(a)}{b-a}$$

Applying this to $$f(x)=\sqrt{x}$$ with $$a=k^2$$ and $$b=(k+1)^2$$ we get

$$\sqrt{x}\approx \frac{x+k^2+k}{2k+1}$$

With respect to the increase rate, it can be obtained easily with Calculus: Letting $$y_n= \sqrt{n}$$ then

$$\frac{y_{n+1}}{y_n}=\frac{\sqrt{n+1}}{\sqrt n}=\sqrt{1+1/n}\approx 1+\frac{1}{2n}=1+\frac{1}{2 y_n^2}$$