Square Root of 2 and 3, Minus One, Squared.

Factual Questions
1

I know I recently did a thread on random experiments I have conducted with calculators. But I just had to share this one too. It’s so weird–and so counterintuitive, as you will see.

I will try to summarize as best as I can. When you take the square root of 2, you get 1.4142… of course. But when you subtract one from that, and square it, then take the inverse of it, you get 5.82842… Clearly, anyone who even recently took math can see, when you subtract 3 from that, you get the square root of 8. Why?

Also, similiarly, when you take the square root of 3, and do all that I just said (plus subtract one from the inverse), you get the square root of .75. Again, Why?

And why when you subtract one? Usually subtracting one from an irrational number, you just get an otherwise useless number.

BTW, I have tried this with other square roots. But so far these are the only two I can get to work with this. What gives?

:slight_smile:

2

Not so hard.

1/ (2^.5 -1)^2 - 3= 8^.5

1/ (2 - 2(2^.5) + 1) - 3 = 8^.5

1/ (2 - 8^.5 + 1) - 3 = 8^.5

1/ (2 - 8^.5 + 1) - 3 = 8^.5

1/ (3 - 8^.5) - 3 = 8^.5

1/ (3 - 8^.5) = 3 + 8^.5

If you multiply the left by (3 + 8^.5)/(3 + 8^.5), you prove the equation.

3

Use a little algebra (including the thing you learn in school where you rationalize the denominator when it contains a root), and you’ll see why the first one works out exactly.

I’m confused by exactly what you’re trying to do with the second one, can you explain more precisely?

4

(sqrt2 - 1)^2 = 2 - 2sqrt2 +1 = 3 - 2sqrt2
1/(3 - 2sqrt2) = (3 + 2sqrt2)/(9 - 4x2) = (3 + 2sqrt2)/1 = (3 + 2sqrt2)
So the inverse is just the conjugate
Clearly 3 + 2sqrt2 - 3 = 2sqrt2 which is the reduced form of sqrt8

(sqrt3 - 1)^2 = 3 - 2sqrt3 +1 = 4 - 2sqrt3
1/(4 - 2sqrt3) = (4 + 2sqrt3)/(16 - 4x3) = (4 + 2sqrt3)/4
This can be split up as
(4/4) + (2sqrt3)/4 = 1 + (1/2)sqrt3
1 + (1/2)sqrt3 - 1 = (sqrt3)/2 which is the square root of 3/4

In general
(sqrtA - B)^2 = (A+B^2) - 2BsqrtA
The inverse is [(A+B^2) + 2BsqrtA] / [(A+B^2)^2 - 4AB^2]
If you subtract (A+B^2)/[(A+B^2)^2 - 4AB^2] you get the square root of (4AB^2) / [(A+B^2)^2 - 4AB^2]^2

So what values of A and B make (A+B^2)/[(A+B^2)^2 - 4AB^2] an integer?

5

Take the square root of 3 (1.732…), subtract one (.732…). Square this and take the inverse (1.866…). Then Subtract one, and you get 0.8660… which is obviously the square root of .75, or three-quarters.

HTH.

6

As an added note
[(A+B^2)^2 - 4AB^2] = A^2 + 2AB^2 + B^4 - 4AB^2
= A^2 - 2AB^2 + B^4
= (A-B^2)^2
This simplification is similar to the one for generalized Pythgorean triplets
So if A-B^2 = 1 then we are guarantied that the rational part is an integer.
#1 A=2 B=1 and 2 - 1^2 = 1
#2 A=3 B=1; 3 - 1^2 = 2 and 2^2 = 4 (notice your denominator)
3 + 1^2 = 4 (that’s the numerator) and 4/4 = 1 which you pointed out needs to be subtracted from the inverse.

To sum up if (A+B^2)/[(A-B^2)^2] is an integer, it works. Try this
sqrt5 - 2
square it
inverse
-9
you have the square root of 80

7

What OP does is: Start with a square root, perform some arithmetic of which the final step is a squaring, and observe that a rational number (8 when he starts with √2, 3/4 when he starts with √3) emerges.

If you trace through the steps, the key is to have a certain subexpression be zero. OP achieves this by subtracting a specific value (3 for √2, 1 for √3) in his penultimate step. More generally, he needs to subtract (x+1)/(xx-2x+1) for √x.

Unfortunately, the rational numbers aren’t always as fine-looking as for the √2 and √3 cases. For √5, he’ll want to subtract 3/8 and finish with (√5)/8; it probably gets worse from there.

8

Now I’m fascinated if we can come up with all possible A, B | (A+B^2)/[(A-B^2)^2] is an integer.
Clearly for any B, we pick A to be B^2 + 1
The algorithm is
sqrtA - B
square it
inverse
subtract A+B^2
result is sqrt(4AB^2)
Therefore there are an infinite number of these

9

You must not have read my post above yours. sqrt5 is straight forward.
And he subtracts 9 not 3/8 if he wants to stick with integers like his second example he allows rational numbers.

10

Correct (it took me more than 5 minutes for my post) and correct. I used 1 for the fixed first subtraction. You figured out that was a loser. Well done!

11

Speaking of the square root of 5, you’re familiar with the properties of the Golden Ratio, right? Take the square root of 5, add 1, and then divide by 2. You should get 1.61803… That number is called the golden ratio. Now take the reciprocal of that number and see what happens. After you’ve done that, square it and see what happens.

12

Man, this stuff is hard to follow when you are restricted to ASCII. This is why I with the board would support MathML. Sure, we can get close using [noparse]


[/noparse] blocks, but it takes forever and still doesn't look right. (See below)



[quote="Chronos, post:11, topic:698899"]

Speaking of the square root of 5, you're familiar with the properties of the Golden Ratio, right?  Take the square root of 5, add 1, and then divide by 2.  You should get 1.61803....  That number is called the golden ratio.  Now take the reciprocal of that number and see what happens.
[/QUOTE]


[spoiler]

2 2(√̅5-1) 2√̅5 - 2 2√̅5 - 2 √̅5 - 1
------ = ----------- = --------- = ------- = ------
√̅5 + 1 (√̅5+1)(√̅5-1) 5 - 1 4 2

√̅5 - 1 √̅5 - 1 + 2 2 √̅5 + 1
------ = ---------- - — = ------ - 1 ≈ 0.61803…
2 2 2 2



[/spoiler]


[quote="Chronos, post:11, topic:698899"]

Now take the reciprocal of that number and see what happens.
[/QUOTE]

[spoiler]

(√̅5 + 1)[sup]2[/sup] (√̅5+1)(√̅5+1) 5 + 2√̅5 + 1 6 + 2√̅5 3 + 2√̅5
--------- = ------------ = ----------- = ------- = -------
2[sup]2[/sup] 4 4 4 2

3 + 2√̅5 2√̅5 + 1 + 2 2√̅5 - 1
------- = ----------- = ------- + 1 ≈ 2.61803…
2 2 2


[/spoiler]

Pretty nifty, **Chronos**. I never knew that 1/Φ = Φ - 1 and Φ[sup]2[/sup]=Φ + 1.
13

Connecting the topics of golden ratio and getting integers after starting with square root:

Go to The Wolfram Alpha calculater and enter something like

n=7, ((sqrt(5) + 1)^n - (1 - sqrt(5))^n)/(2^n * sqrt(5))

That will give you the 7th Fibonacci number … exactly!
(Google calculater will give you the same answer faster but, AFAIK, there’s no way to introduce a variable as I’ve done above (n) to allow getting a different Fibonacci number with minimum edit.)

14

Let me take a more abstract view of what’s going on:

The square root of 8 is 2 * sqrt(2), so what you’re noting is that when you start with x = sqrt(2), then apply some rational function to it (subtract one, then square, then invert), you end up with something of the form a + bx where a and b are nice (in this case, specifically, 3 + 2x).

The square root of .75 is 1/2 * sqrt(3), so, again, the observation is that when you start with x = sqrt(3), then apply some rational function to it (subtract one, then square, then invert), you end up with something of the form a + bx, where a and b are nice (in this case, specifically 1 + 1/2 * x).

If the relevant niceness is just that these coefficients are rational, this is immediate even without looking at the particular details:

Consider the collection of all numbers a + bx for rational a and b. So long as x^2 is among these, this will be closed under all polynomial operations (that it is closed under addition and negation is obvious, and that it is closed under multiplication follows from our condition on x^2). Furthermore, this is a two-dimensional space over the rationals, so for any irrational value y within it, we must have that y and y^2 form a basis for the whole space; thus, 1 = ay + by^2 for suitable a and b, which is to say, 1/y = a + by. Thus, this space is closed under reciprocation as well, making it closed under all rational operations.

This explains the “rational functions applied to x produce rational linear combinations of 1 and x” observations of the OP for sqrt(2) and sqrt(3). More generally, by the same reasoning, if x is a root of an nth degree polynomial (with rational coefficients), then rational functions of x are equal to polynomial functions of x of degree less than n.

We may further ask why it turned out that the coefficients of interest were not merely rational but in fact positive in each of the OP’s cases and integers in three of the four cases. Here, some further delving into the details as Saint Cad carried out may be necessary.

15

Obligatory ‘Yay, my first post!’ :smack:

Into a thread that’s been quiet for the best part of a year.

Anyway, now that’s out of the way, onwards…

Your 1/((sqrt2-1)^2) number or roughly 5.8284271247 seems to have some significance to a little something I’m working on.

Boiled down I’m looking for infinite lists of Pythagorean triangles, and most of the functions I’ve explored, the gradient tends toward 0, i.e. my slices of cake get thinner and thinner. It’s a consequence of how I went about looking for them.

So instead this time I’m exploring triangles where the gradient tends toward 1, very easy to do with lots (and I mean lots) of scrolling on a spreadsheet.

The first column (a) increases by 1 every row, second column (b) is just a+1, third column © uses Pythagoras’ theorem to give the length of the hypotenuse. The gradient tends toward 1 as the difference between a and b (always 1) becomes smaller and smaller when compared a and b as they go up and up and up. Anyway I’m digressing here.

The table goes as follows:

a - b - c
1 - 2 - 2.236068
2 - 3 - 3.605551
3 - 4 - 5 - The classic, there should be songs written about this one.
4 - 5 - 6.403124
5 - 6 - 7.810250
6 - 7 - 9.219544
7 - 8 - 10.63015
8 - 9 - 12.04159
9 - 10 - 13.453624
10 - 11 - 14.866068
11 - 12 - 16.278820
12 - 13 - 17.691806
13 - 14 - 19.104973
14 - 15 - 20.518284
15 - 16 - 21.931712
16 - 17 - 23.345235
17 - 18 - 24.758836
18 - 19 - 26.172504
19 - 20 - 27.586228
20 - 21 - 29

That should give an idea I hope. So then when continuing the list I get the following triangles…(and x doesn’t mean times here)

axbxc
3x4x5
20x21x29
119x120x169
696x697x985
4059x4060x5741
23660x23661x33461
137903x137904x195025

With this list, when dividing the a, b or c numbers by their previous one you get… (and I’m bolding the bits which match our 1/((sqrt2-1)^2) number or 5.8284271247 from earlier)

6.6666666667 - 5.25 - 5.80
5.95 - 5.7142857143 - 5.8275862069
5.8487394958 - 5.8083333333 - 5.8284023669
5.8318965517 - 5.824964132 - 5.8284263959
5.8290219266 - 5.8278325123 - 5.8284271033
5.8285291631 - 5.8283250919 - 5.8284271241

They do seem to be tending toward that number, the third column more rapidly than the others, but still, the direction is clear. I would’ve gone further if it weren’t for the fact that I’ve already had to copy/scroll down to row 138,000 just to get those 8 triangles, and based on the evidence, I’d have to continue to around row 803,758 to get the next one.

So yeah this is just a comment about how 5.8284271247… seems to have some significance of sorts, to me anyway.

I guess I’ll be turning those raw numbers into functions of n next, which brings me on to a question, surely there’s a more elegant way of putting 1/((sqrt2-1)^2)? I kinda skipped through a lot of the posts, so there could be one right there somewhere, I’ll look later.

16

If you’re looking for Pythagorean triangles in which the difference between the hypoteneuse and the long legh is one, there is an infinite number of them – every odd number is a short leg in such a Pythagorean triple.

Use n as your index, running from 1 to infinity in steps of one – 1,2,3,4,…

Every odd number will then be given by the short leg, a = 2n + 1.

The longer side of the triangle will then be** b = 2n[sup]2[/sup] + 2n**

And the hypoteneuse will, of course, be ** c = 2n[sup]2[/sup] + 2n + 1**
Plugging in you get, for n, a, b, c:

1 3 4 5
2 5 12 13
3 7 24 25
… and so on. You can verify that the formulas for the sides fulfill the Pythagorean theorem.
As an added plus, in the above table, n is the radius of the inscribed circle in the triangle! (The radius of the circumscribed circle is uninteresting).

17

Check out this paper, published in the June, 1968 issue of The Fibonacci Quarterly: http://www.fq.math.ca/Scanned/6-3/forget.pdf . They demonstrate that this sequence is really rather similar to the Fibonacci Sequence.

Rather than starting with 1,2 and then calculating the next number as 1+2 to get 3, then 2+3 to get 5, then 3+5 to get 8, et cetera… we instead start with 1,2 and then calculate 1+2(2) to get 5, then 2+2(5) to get 12, then 5+2(12) to get 29, et cetera. This gives the list 1,2,5,12,29,70,169,408,985,2378,… which they call “q” and then they notice that any two consecutive numbers in that sequence multiplied together and then doubled will give you 4,20,120,696,4060,23360, et cetera which will always be one of the legs of the Pythagorean triple you described where a and b differ by exactly 1. When n is odd, you get b (so subtract 1 to find a) and when n is even you get a (so add 1 to find b).

Your observation that the end result produces a sequence that tends toward the ratio of 5.8284271247 would have made an excellent Lemma to be added on to Forget & Larkin’s paper. If you get your hands on a time machine, set the controls for 1968 and go find the authors in Sunnyvale, California.

It’s not an entirely surprising result when you consider the fact that the Fibonacci Sequence tends toward the ratio 1.61803398875 but I’d still call it interesting.

FYI, the next six values for “a” are…
137,903
803,760
4,684,659
27,304,196
159,140,519
927,538,920

18

For x = golden ratio (~ 1/68013)

1/x = x - 1

:slight_smile:

x^2 = x + 1

:slight_smile:

It’s been far too long since I did much algebra. If I didn’t goof:

1/x = x-1 <==> x^2 - x - 1 = 0

x^2 = x + 1 <==> x - x^2 + 1 = 0 <==> x^2 -x - 1 = 0 (same as above)

So, the second tidbit shouldn’t come as a surprise!

If we look at y = x^2 - x - 1, the x intercepts are (1± √̅5)/2. Isn’t math fun?
Thanks to BigT for the derivations.