MHB How did we find sqrt(ε_7(2)) ?

  • Thread starter Thread starter evinda
  • Start date Start date
Click For Summary
The discussion focuses on the function ε_p that maps integers to p-adic integers, specifically examining ε_7(2) and its square root. The square root of 2 modulo 7 is identified as 3, while for 7^2 (49) it is 10, and for 7^3 (343) it is 108, illustrating the process of finding these roots. The conversation emphasizes the importance of defining square roots in modular arithmetic without using the traditional square root symbol due to its multivalued nature. A correction is noted regarding a typo in the fourth term of the square root sequence. Overall, the thread provides insights into the methodology of calculating square roots in p-adic contexts.
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hi! (Smile)

Let the function
$$\epsilon_p: \mathbb{Z} \rightarrow \mathbb{Z}_p$$
$$x \mapsto (\overline{x_n} )_{n \in \mathbb{N}_0}$$

where $\overline{x_n} \equiv x \pmod {p^n}$

There are some examples:

$$\epsilon_7 (173)=(173,173,173, \dots, 173, \dots)=(5,26,173,173, \dots)$$

$$\epsilon_7(-1)=(-1,-1,-1, \dots)=(6,48,342,2400, \dots)$$

$$\epsilon_7(2)=(2,2, \dots, 2, \dots)$$

$$\sqrt{\epsilon_7(2)}=(3,10,108,2016, \dots)$$

Could you explain me how we found $\sqrt{\epsilon_7(2)}$ ? (Thinking)
 
Mathematics news on Phys.org
evinda said:
Hi! (Smile)

Let the function
$$\epsilon_p: \mathbb{Z} \rightarrow \mathbb{Z}_p$$
$$x \mapsto (\overline{x_n} )_{n \in \mathbb{N}_0}$$

where $\overline{x_n} \equiv x \pmod {p^n}$

There are some examples:

$$\epsilon_7 (173)=(173,173,173, \dots, 173, \dots)=(5,26,173,173, \dots)$$

$$\epsilon_7(-1)=(-1,-1,-1, \dots)=(6,48,342,2400, \dots)$$

$$\epsilon_7(2)=(2,2, \dots, 2, \dots)$$

$$\sqrt{\epsilon_7(2)}=(3,10,108,2016, \dots)$$

Could you explain me how we found $\sqrt{\epsilon_7(2)}$ ? (Thinking)

A square root of two modulo $7$ is $3$, since $3^2 \equiv 9 \equiv 2 \pmod{7}$. A square root of two modulo $7^2 = 49$ is $10$ since $10^2 \equiv 100 \equiv 2 \pmod{49}$. A square root of two modulo $7^3 = 343$ is $108$ since $108^2 \equiv 11664 \equiv 2 \pmod{343}$. And so on. You can still define square roots modulo $n$, though I would recommend not using the $\sqrt{}$ symbol as it is multivalued and generally a nuisance, just stick with the definition: an integer $x$ is a square root of $a$ modulo $n$ if and only if $x^2 \equiv a \pmod{n}$.

EDIT: I suppose the fourth term should be $2166$, not $2016$. Typo?
 
Last edited:
Bacterius said:
A square root of two modulo $7$ is $3$, since $3^2 \equiv 9 \equiv 2 \pmod{7}$. A square root of two modulo $7^2 = 49$ is $10$ since $10^2 \equiv 100 \equiv 2 \pmod{49}$. A square root of two modulo $7^3 = 343$ is $108$ since $108^2 \equiv 11664 \equiv 2 \pmod{343}$. And so on. You can still define square roots modulo $n$, though I would recommend not using the $\sqrt{}$ symbol as it is multivalued and generally a nuisance, just stick with the definition: an integer $x$ is a square root of $a$ modulo $n$ if and only if $x^2 \equiv a \pmod{n}$.

EDIT: I suppose the fourth term should be $2166$, not $2016$. Typo?

Yes, that was a typo..

I understand, thank you very much! (Smile)
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 26 ·
Replies
26
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
5K
  • · Replies 125 ·
5
Replies
125
Views
19K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K