2
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It is possible to find rational numbers that approximate square roots in $\mathbb{R}$. For example,

$$ \frac{7}{5} < \sqrt{2} < \frac{3}{2}$$

but we can also solve the equation $x^2 \equiv 2$ in $7$-adic numbers since

$$ 3^2 \equiv 2 \mod 7$$

so at least we know the first digit $\sqrt{2} = \dots 3$. Can we find rational numbers that approximate $\sqrt{2}$ in both spaces at once:

$$ x \approx \sqrt{2} \text{ in } \mathbb{R} \times \mathbb{Q}_7$$

To clarify there are two different norms here:

  • the norm over $\mathbb{R}$ - what we call "absolute value" - is an non-archimedian norm . It's just plain old: absolute value.

  • the norm over $\mathbb{Q}_7$ is such that $|| 7^k|| = \frac{1}{7^k}$. This leads to p-adic numbers

So I would like a solution that is close in both norms: $$ || x^2 - 2 ||_\mathbb{R} < \frac{1}{7^2} \quad\text{ and }\quad || x^2 - 2 ||_{\mathbb{Q}_7} < \frac{1}{7^2}$$

  • 7
    Perhaps I'm being stupid, but it's not clear to me what you would accept as meaning $\approx$ here. You can't say you want $x_1 \lt \sqrt2 \lt x_2$ because $\Bbb Q_7$ cannot be made into an ordered field. And you can't simply say that you want $x$ kinda sorta close to $\sqrt 2$ because then $x=1$ solves the problem kinda sorta, but I guess you don't want that? Maybe? It seems to me that you should be asking for a _sequence_ of numbers $x_1, x_2, x_3\ldots$ so that $x_i^2$ converges to $2$ in both fields, but you didn't ask for that, so I'm not sure what you do mean. – 2017-01-01
  • 0
    @MYUSERNAMEISALIE that makes sense. – 2017-01-01

1 Answers 1

1

This looks good. Start with $p/q = 3/1.$ at each stage, multiply the column vector by $$ \left( \begin{array}{cc} 75 & 106 \\ 53 & 75 \end{array} \right) $$ The factored row is $p^2 - 2 q^2.$

1
3
1
7   =     7^1

2
331
234
49   =     7^2

3
49629
35093
343   =     7^3

4
7442033
5262312
2401   =     7^4

5
1115957547
789101149
16807   =     7^5

6
167341537819
118328336166
117649   =     7^6

7
25093418970021
17743726716857
823543   =     7^7

8
3762841454738417
2660730709175388
5764801   =     7^8

9
564250564277972403
398985400289290201
40353607   =     7^9

10
84611244751512691531
59829184928429302434
282475249   =     7^10

11
12687736958776957922829
8971584841462370333693
1977326743   =     7^11

12
1902568265103283099583633
1345318921924856544936912
13841287201   =     7^12

13
285296425606781026232085147
201735037194838245148200949
96889010407   =     7^13

14
42781145863161430953115686619
30250838346772262776415583966
678223072849   =     7^14

15
6415174804494967175783728396821
4536213606755475548746300188257
4747561509943   =     7^15

16
961976752653202946350887449716817
680220285144893926472510119150788
33232930569601   =     7^16

17
144251606674348977182402631358744803
102001289276486800642035293771300401
232630513987207   =     7^17

18
21631007163883774156735938491663702731
15295431849477005838819986494861004634
1628413597910449   =     7^18

19
3243641313335845680670113955330044196029
2293600768396615468218503727172751592293
11398895185373143   =     7^19

20
486394779950229665681419941730064983485233
343933047236545981191903819170448711811512
79792266297612001   =     7^20

21
72936511503341098932448300461822437213412747
51573901880103120870508043349477097510580749
558545864083284007   =     7^21

22
10937071962041513232207475129681255127127515419
7733677750684812308707863175687371485604431766
3909821048582988049   =     7^22
  • 0
    these look like good 7-adic approximations, I am asking for a number that approximates $\sqrt{2}$ in both $ \mathbb{ Q}_7$ as well as $ \mathbb{ R}$. For example $49/1024$ is about $0.05$ over the reals but has 7-adic size $1/49$ – 2017-01-02
  • 0
    @cactus314 I had in mind each given $p/q$ – 2017-01-02
  • 0
    oh I was scared by $\sqrt{2}\approx \frac{3 }{ 1}$. they get better as you go along – 2017-01-02