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In the 10-adic number system, ....99999=-1

Proof: let ....9999 be x x+1=0 x=-1 ....99999=-1

Therefore,....99999=-1 Now when we plug this into ✓-1 we get:

✓....99999= ✓9*...11111= 3✓...1111

So in conclusion, we can say that ✓-1=3✓....11111 in the 10-adic number system.Here's another proof strengthening this conclusion:

i²=-1 Here,i=3✓....11111 according to what we got earlier.therfore,

(3✓....11111)²=-1

3²×(✓...11111)²=-1

9*....111111=-1

.....99999999=-1

I am not sure if this is a new discovery, something already known or if I came to a wrong conclusion.please let me know.

lisyarus
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PARTH PATEL
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  • It is not clear what number $\ldots 99999$ refers to, so we can't draw any reasonable conclusion out of your calculations. This is something you would have to address as a start. – Dean Miller Jun 21 '25 at 12:33
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    @DeanMiller The OP clearly states that it's a 10-adic number: https://en.m.wikipedia.org/wiki/P-adic_number – lisyarus Jun 21 '25 at 12:38
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    @PARTHPATEL Please, use MathJax to format your formulas: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – lisyarus Jun 21 '25 at 12:40
  • NB: p-adics are usually defined for prime p because they have nicer properties this way, but the ring of p-adic integers still makes sense for non-prime p. – lisyarus Jun 21 '25 at 12:41
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    @lisyarus My bad. Thanks for pointing that out. – Dean Miller Jun 21 '25 at 12:45
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    This site is nothing but a witch hunt for closing questions these days. – lisyarus Jun 21 '25 at 12:52
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    Suppose $-1$ had a square root in the $10$-adic integers, ending in $ab$: $$\ldots ab^2=\ldots 999$$ In particular, we get $$(10a+b)^2\equiv 99\bmod 100$$

    So $b=3$ or $7$. If $b=3$, we get $$10a+b\equiv 20ab+b^2\equiv 60a+9\bmod 100$$ so $60a\equiv 90\bmod 100$.

    If $b=7$, we get $$10a+b\equiv 20ab+b^2\equiv 40a+49\bmod 100$$ so $40a\equiv 50\bmod 100$.

    But these are both impossible, because the l.h.s. is a multiple of $20$, but the r.h.s. isn't. Therefore $\sqrt{-1}$ doesn't exist!

     – TonyK Jun 21 '25 at 13:15
  • Why close?!${}$ – TonyK Jun 21 '25 at 13:16
  • It was reopened. – José Carlos Santos Jun 21 '25 at 13:33
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    You haven't found a square root of -1 in the 10-adics, but from my memories I believe there exists one in either the 5-adics or the 7-adics. You'll have to do more work, like explicitly write out the equation(s) for squaring a number in those rings, and show they have a solution, but I think you might find that a fun challenge. – JonathanZ Jun 21 '25 at 14:02
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    Note that even if one found some nice representation of a square root of $-1$ in the $10$-adic numbers, it is important to keep in mind that number (of which there would be two if any) is not identical to the square root(s) of $-1$ sitting in the complex number field. The square root symbol is treacherous because it pretends to be independent of ambient field. Read https://math.stackexchange.com/a/4007515/96384 – Torsten Schoeneberg Jun 22 '25 at 02:01

2 Answers2

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Your result is true (if interpreted properly, i.e. in some quadratic extension of $\mathbb Z_{10}$), but it is of little value because $\sqrt{\dots 11111}$ generally doesn't exist in the ring of p-adics integers $\mathbb Z_p$ (or the field of p-adics $\mathbb Q_p$, but this only makes sense for prime $p$).

Specifically in $\mathbb Z_{10}$, suppose that $(\dots a_2 a_1 a_0)^2=\dots 111$. This means that $a_0^2=1\mod 10$, so $a_0=1$ or $9$. Then, $20a_0a_1+a_0^2=11\mod 100$. If $a_0=1$, we get $20a_0a_1+a_0^2=20a_1+1=11\mod 100$, thus $20a_1=10\mod 100$, which has no solutions. Similarly, if $a_0=9$, we get $180a_1+81=11\mod 100$, thus $80a_1=-70=30\mod 100$, which also has no solutions.

lisyarus
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In the 10-adic number system, the infinite sequence ...99999 does equal -1. This is one of the beautiful and counterintuitive properties of p-adic numbers.

Let $x = ...99999$ (we can treat ...99999 as a convergent infinite series).

We can express this as: $$x = 9 + 90 + 900 + 9000 + 90000 + ...$$ $$x = 9(1 + 10 + 10² + 10³ + 10⁴ + ...)$$

Now, the crucial difference from ordinary real numbers is that in the 10-adic metric, the geometric series 1 + 10 + 10² + 10³ + ... actually converges.

Using the geometric series formula: $$1 + 10 + 10² + 10³ + ... = 1/(1-10) = 1/(-9) = -1/9$$

Therefore: $$x = 9 × (-1/9) = -1$$

You can also see this by noting that if $x = ...99999$, then: $$x + 1 = ...99999 + 1 = ...000000 = 0$$

So $x = -1$.

This same principle works for any prime $p$: in the $p$-adic numbers, $...((p-1)(p-1)(p-1)...) = -1$.

Regarding root of $-1$

However, $\sqrt{-1}$ may not exist in the 10-adics. When $p$-adic square roots exist, the usual rules like $\sqrt{(ab)} = \sqrt a \sqrt b$ don't always apply.

The step $√(9 × ...11111) = 3√(...11111)$ assumes properties that don't hold generally in p-adic contexts.

arjo
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