Why was the term “discrete” used in discrete logarithm? Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Trying to better understand the failure of the Index Calculus for ECDLPWhat is so special about elliptic curves?Why is the discrete logarithm problem assumed to be hard?What is the difference between discrete logarithm and logarithm?Calculating the discrete logarithmWhy is NON DISCRETE logarithm problem not hard as the DISCRETE logarithm problem (so computationally hard)?How to construct a hash function into a cyclic group such that its discrete log is intractable?Discrete logarithm key sizes for very short term usageDiscrete Logarithm NotationDescribing Discrete Logarithm Assumption

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Why was the term “discrete” used in discrete logarithm?



Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Trying to better understand the failure of the Index Calculus for ECDLPWhat is so special about elliptic curves?Why is the discrete logarithm problem assumed to be hard?What is the difference between discrete logarithm and logarithm?Calculating the discrete logarithmWhy is NON DISCRETE logarithm problem not hard as the DISCRETE logarithm problem (so computationally hard)?How to construct a hash function into a cyclic group such that its discrete log is intractable?Discrete logarithm key sizes for very short term usageDiscrete Logarithm NotationDescribing Discrete Logarithm Assumption










2












$begingroup$


Is there anything especially "discrete" about a discrete logarithm? This is not a question of what is a discrete logarithm or why the discrete logarithm problem is an "intractable problem" given certain circumstances. I'm just trying to determine if there's some additional meaning to the term "discrete" as it's used in name discrete logarithm?



The definition of "discrete" is "individually separate and distinct". Could it be that the term "discrete" is a reference to the least non-negative residues of a modulus or the order of points for a particular cyclic group on an elliptic curve?










share|improve this question









$endgroup$







  • 1




    $begingroup$
    Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
    $endgroup$
    – Mikero
    2 hours ago















2












$begingroup$


Is there anything especially "discrete" about a discrete logarithm? This is not a question of what is a discrete logarithm or why the discrete logarithm problem is an "intractable problem" given certain circumstances. I'm just trying to determine if there's some additional meaning to the term "discrete" as it's used in name discrete logarithm?



The definition of "discrete" is "individually separate and distinct". Could it be that the term "discrete" is a reference to the least non-negative residues of a modulus or the order of points for a particular cyclic group on an elliptic curve?










share|improve this question









$endgroup$







  • 1




    $begingroup$
    Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
    $endgroup$
    – Mikero
    2 hours ago













2












2








2





$begingroup$


Is there anything especially "discrete" about a discrete logarithm? This is not a question of what is a discrete logarithm or why the discrete logarithm problem is an "intractable problem" given certain circumstances. I'm just trying to determine if there's some additional meaning to the term "discrete" as it's used in name discrete logarithm?



The definition of "discrete" is "individually separate and distinct". Could it be that the term "discrete" is a reference to the least non-negative residues of a modulus or the order of points for a particular cyclic group on an elliptic curve?










share|improve this question









$endgroup$




Is there anything especially "discrete" about a discrete logarithm? This is not a question of what is a discrete logarithm or why the discrete logarithm problem is an "intractable problem" given certain circumstances. I'm just trying to determine if there's some additional meaning to the term "discrete" as it's used in name discrete logarithm?



The definition of "discrete" is "individually separate and distinct". Could it be that the term "discrete" is a reference to the least non-negative residues of a modulus or the order of points for a particular cyclic group on an elliptic curve?







discrete-logarithm terminology






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 2 hours ago









JohnGaltJohnGalt

22816




22816







  • 1




    $begingroup$
    Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
    $endgroup$
    – Mikero
    2 hours ago












  • 1




    $begingroup$
    Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
    $endgroup$
    – Mikero
    2 hours ago







1




1




$begingroup$
Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
$endgroup$
– Mikero
2 hours ago




$begingroup$
Traditional logarithm: answer is a real or complex number. Discrete logarithm: answer is an element of a finite set $mathbbZ_n$.
$endgroup$
– Mikero
2 hours ago










2 Answers
2






active

oldest

votes


















6












$begingroup$

The word discrete is used as an antonym of 'continuous', that is, it is the normal logarithmic problem, just over a discrete group.



The standard logarithmic problem is over the infinite group $mathbbR^*$, this group is called 'continuous', because for any element $x$, there are other elements that are arbitrarily close to it.



The discrete logarithmic problem is over a finite group (for example, $mathbbZ_p^*$); in contrast to $mathbbR^*$, we don't have group elements arbitrarily close together; we call this type of group 'discrete'.






share|improve this answer









$endgroup$








  • 1




    $begingroup$
    yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
    $endgroup$
    – Geoffroy Couteau
    1 hour ago






  • 1




    $begingroup$
    @JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
    $endgroup$
    – Mark
    1 hour ago






  • 1




    $begingroup$
    my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
    $endgroup$
    – Geoffroy Couteau
    1 hour ago






  • 1




    $begingroup$
    @GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
    $endgroup$
    – Mark
    1 hour ago






  • 1




    $begingroup$
    @Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
    $endgroup$
    – Geoffroy Couteau
    1 hour ago


















3












$begingroup$

While I agree completely with poncho's answer, this other viewpoint might be useful.
Specifically, I think a better comparison isn't between $mathbbZ_p^*$ and $mathbbR^*$, but with $mathbbZ_p^*$ and $S^1$. We can view $S^1 cong zinmathbbC mid $. It's not hard to show that any $zin S^1$ is able to be written as $z = exp(2pi i t)$ for $tinmathbbR$ (we don't strictly need the factor $2pi$ here, but it's traditional). Due to $exp(x)$ being periodic, it's in fact enough to have $tin[0,1)$.



This has an obvious group structure, in that:
$$exp(2pi i t_0)exp(2pi i t_1) = exp(2pi i (t_0+t_1))$$
If we're making the restriction that $t_iin[0,1)$, then we have to take $t_0+t_1mod 1$, but this is fairly standard.



More than just having an obvious group structure, we actually have that any $mathbbZ_p^*$ injects into it.
Specifically, we always have:
$$
phi_p:mathbbZ_p^*to S^1,quad phi_p(x) = exp(2pi i x/(p-1))
$$

Here, $p-1$ in the denominator is because $|mathbbZ_p^*| = p-1$.
We can define the discrete logarithm problem for both of these groups in the standard way (here, it's important to restrict $t_iin[0, 1)$ if we want a unique answer).
Then, we can relate these problems to each via the aforementioned injection.
Through this image, we see that $S^1$ is "continuous" in the sense that it takes up the full circle, but the image of $mathbbZ_p^*$ in $S^1$ will always be "discrete" --- there will always be "some space" between points (they can't get arbitrarily close).






share|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    The word discrete is used as an antonym of 'continuous', that is, it is the normal logarithmic problem, just over a discrete group.



    The standard logarithmic problem is over the infinite group $mathbbR^*$, this group is called 'continuous', because for any element $x$, there are other elements that are arbitrarily close to it.



    The discrete logarithmic problem is over a finite group (for example, $mathbbZ_p^*$); in contrast to $mathbbR^*$, we don't have group elements arbitrarily close together; we call this type of group 'discrete'.






    share|improve this answer









    $endgroup$








    • 1




      $begingroup$
      yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago






    • 1




      $begingroup$
      @JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
      $endgroup$
      – Mark
      1 hour ago






    • 1




      $begingroup$
      my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago






    • 1




      $begingroup$
      @GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
      $endgroup$
      – Mark
      1 hour ago






    • 1




      $begingroup$
      @Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago















    6












    $begingroup$

    The word discrete is used as an antonym of 'continuous', that is, it is the normal logarithmic problem, just over a discrete group.



    The standard logarithmic problem is over the infinite group $mathbbR^*$, this group is called 'continuous', because for any element $x$, there are other elements that are arbitrarily close to it.



    The discrete logarithmic problem is over a finite group (for example, $mathbbZ_p^*$); in contrast to $mathbbR^*$, we don't have group elements arbitrarily close together; we call this type of group 'discrete'.






    share|improve this answer









    $endgroup$








    • 1




      $begingroup$
      yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago






    • 1




      $begingroup$
      @JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
      $endgroup$
      – Mark
      1 hour ago






    • 1




      $begingroup$
      my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago






    • 1




      $begingroup$
      @GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
      $endgroup$
      – Mark
      1 hour ago






    • 1




      $begingroup$
      @Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago













    6












    6








    6





    $begingroup$

    The word discrete is used as an antonym of 'continuous', that is, it is the normal logarithmic problem, just over a discrete group.



    The standard logarithmic problem is over the infinite group $mathbbR^*$, this group is called 'continuous', because for any element $x$, there are other elements that are arbitrarily close to it.



    The discrete logarithmic problem is over a finite group (for example, $mathbbZ_p^*$); in contrast to $mathbbR^*$, we don't have group elements arbitrarily close together; we call this type of group 'discrete'.






    share|improve this answer









    $endgroup$



    The word discrete is used as an antonym of 'continuous', that is, it is the normal logarithmic problem, just over a discrete group.



    The standard logarithmic problem is over the infinite group $mathbbR^*$, this group is called 'continuous', because for any element $x$, there are other elements that are arbitrarily close to it.



    The discrete logarithmic problem is over a finite group (for example, $mathbbZ_p^*$); in contrast to $mathbbR^*$, we don't have group elements arbitrarily close together; we call this type of group 'discrete'.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 2 hours ago









    ponchoponcho

    94.2k2148247




    94.2k2148247







    • 1




      $begingroup$
      yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago






    • 1




      $begingroup$
      @JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
      $endgroup$
      – Mark
      1 hour ago






    • 1




      $begingroup$
      my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago






    • 1




      $begingroup$
      @GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
      $endgroup$
      – Mark
      1 hour ago






    • 1




      $begingroup$
      @Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago












    • 1




      $begingroup$
      yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago






    • 1




      $begingroup$
      @JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
      $endgroup$
      – Mark
      1 hour ago






    • 1




      $begingroup$
      my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago






    • 1




      $begingroup$
      @GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
      $endgroup$
      – Mark
      1 hour ago






    • 1




      $begingroup$
      @Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
      $endgroup$
      – Geoffroy Couteau
      1 hour ago







    1




    1




    $begingroup$
    yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
    $endgroup$
    – Geoffroy Couteau
    1 hour ago




    $begingroup$
    yes, being discrete is not the "core reason" why dlp can be hard - although note that if we are to ever use the crypto we build on a computer, things better be discrete - at best, we can only approximate a continuous primitive in a discrete way.
    $endgroup$
    – Geoffroy Couteau
    1 hour ago




    1




    1




    $begingroup$
    @JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
    $endgroup$
    – Mark
    1 hour ago




    $begingroup$
    @JohnGalt The hard part of the DLP is the modular reduction "hiding" how many times you've "wrapped around". Without modular reduction, you can do some sort of "binary search" to get accurate lower/upper bounds to the discrete log rather efficiently.
    $endgroup$
    – Mark
    1 hour ago




    1




    1




    $begingroup$
    my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
    $endgroup$
    – Geoffroy Couteau
    1 hour ago




    $begingroup$
    my point is just the following: when you manipulate concrete number on your computer, they will have to come from some discrete structure, since you cannot represent numbers with arbitrary precision (hence you cannot accurately represent elements of a continuous set such as the reals). Hence, you will need your problem to be hard over a discrete structure anyway, even though being discrete is not what makes it hard.
    $endgroup$
    – Geoffroy Couteau
    1 hour ago




    1




    1




    $begingroup$
    @GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
    $endgroup$
    – Mark
    1 hour ago




    $begingroup$
    @GeoffroyCouteau This isn't precisely true. Any computable number has a finite-length turing machine that, on input $n$, will output the $n$th digit of the number. This can be viewed as a finite-length representation of the number. As an example, there are formula for $pi$ (such as the BBP formula) that compute the $i$th digit without computing all smaller digits. This is definitely a different notion of "representation" than simply storing the literal value in memory, but the computable numbers are notably not discrete (they contain $mathbbQ$ as a sub-field).
    $endgroup$
    – Mark
    1 hour ago




    1




    1




    $begingroup$
    @Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
    $endgroup$
    – Geoffroy Couteau
    1 hour ago




    $begingroup$
    @Mark you are perfectly right, although for the purpose of getting a high level intuition regarding this specific question (regarding discrete log), I felt like my answer would provide one. We could perhaps work with problems over more general structures through appropriate representations, though I believe this has never been done in Cryptography.
    $endgroup$
    – Geoffroy Couteau
    1 hour ago











    3












    $begingroup$

    While I agree completely with poncho's answer, this other viewpoint might be useful.
    Specifically, I think a better comparison isn't between $mathbbZ_p^*$ and $mathbbR^*$, but with $mathbbZ_p^*$ and $S^1$. We can view $S^1 cong zinmathbbC mid $. It's not hard to show that any $zin S^1$ is able to be written as $z = exp(2pi i t)$ for $tinmathbbR$ (we don't strictly need the factor $2pi$ here, but it's traditional). Due to $exp(x)$ being periodic, it's in fact enough to have $tin[0,1)$.



    This has an obvious group structure, in that:
    $$exp(2pi i t_0)exp(2pi i t_1) = exp(2pi i (t_0+t_1))$$
    If we're making the restriction that $t_iin[0,1)$, then we have to take $t_0+t_1mod 1$, but this is fairly standard.



    More than just having an obvious group structure, we actually have that any $mathbbZ_p^*$ injects into it.
    Specifically, we always have:
    $$
    phi_p:mathbbZ_p^*to S^1,quad phi_p(x) = exp(2pi i x/(p-1))
    $$

    Here, $p-1$ in the denominator is because $|mathbbZ_p^*| = p-1$.
    We can define the discrete logarithm problem for both of these groups in the standard way (here, it's important to restrict $t_iin[0, 1)$ if we want a unique answer).
    Then, we can relate these problems to each via the aforementioned injection.
    Through this image, we see that $S^1$ is "continuous" in the sense that it takes up the full circle, but the image of $mathbbZ_p^*$ in $S^1$ will always be "discrete" --- there will always be "some space" between points (they can't get arbitrarily close).






    share|improve this answer









    $endgroup$

















      3












      $begingroup$

      While I agree completely with poncho's answer, this other viewpoint might be useful.
      Specifically, I think a better comparison isn't between $mathbbZ_p^*$ and $mathbbR^*$, but with $mathbbZ_p^*$ and $S^1$. We can view $S^1 cong zinmathbbC mid $. It's not hard to show that any $zin S^1$ is able to be written as $z = exp(2pi i t)$ for $tinmathbbR$ (we don't strictly need the factor $2pi$ here, but it's traditional). Due to $exp(x)$ being periodic, it's in fact enough to have $tin[0,1)$.



      This has an obvious group structure, in that:
      $$exp(2pi i t_0)exp(2pi i t_1) = exp(2pi i (t_0+t_1))$$
      If we're making the restriction that $t_iin[0,1)$, then we have to take $t_0+t_1mod 1$, but this is fairly standard.



      More than just having an obvious group structure, we actually have that any $mathbbZ_p^*$ injects into it.
      Specifically, we always have:
      $$
      phi_p:mathbbZ_p^*to S^1,quad phi_p(x) = exp(2pi i x/(p-1))
      $$

      Here, $p-1$ in the denominator is because $|mathbbZ_p^*| = p-1$.
      We can define the discrete logarithm problem for both of these groups in the standard way (here, it's important to restrict $t_iin[0, 1)$ if we want a unique answer).
      Then, we can relate these problems to each via the aforementioned injection.
      Through this image, we see that $S^1$ is "continuous" in the sense that it takes up the full circle, but the image of $mathbbZ_p^*$ in $S^1$ will always be "discrete" --- there will always be "some space" between points (they can't get arbitrarily close).






      share|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        While I agree completely with poncho's answer, this other viewpoint might be useful.
        Specifically, I think a better comparison isn't between $mathbbZ_p^*$ and $mathbbR^*$, but with $mathbbZ_p^*$ and $S^1$. We can view $S^1 cong zinmathbbC mid $. It's not hard to show that any $zin S^1$ is able to be written as $z = exp(2pi i t)$ for $tinmathbbR$ (we don't strictly need the factor $2pi$ here, but it's traditional). Due to $exp(x)$ being periodic, it's in fact enough to have $tin[0,1)$.



        This has an obvious group structure, in that:
        $$exp(2pi i t_0)exp(2pi i t_1) = exp(2pi i (t_0+t_1))$$
        If we're making the restriction that $t_iin[0,1)$, then we have to take $t_0+t_1mod 1$, but this is fairly standard.



        More than just having an obvious group structure, we actually have that any $mathbbZ_p^*$ injects into it.
        Specifically, we always have:
        $$
        phi_p:mathbbZ_p^*to S^1,quad phi_p(x) = exp(2pi i x/(p-1))
        $$

        Here, $p-1$ in the denominator is because $|mathbbZ_p^*| = p-1$.
        We can define the discrete logarithm problem for both of these groups in the standard way (here, it's important to restrict $t_iin[0, 1)$ if we want a unique answer).
        Then, we can relate these problems to each via the aforementioned injection.
        Through this image, we see that $S^1$ is "continuous" in the sense that it takes up the full circle, but the image of $mathbbZ_p^*$ in $S^1$ will always be "discrete" --- there will always be "some space" between points (they can't get arbitrarily close).






        share|improve this answer









        $endgroup$



        While I agree completely with poncho's answer, this other viewpoint might be useful.
        Specifically, I think a better comparison isn't between $mathbbZ_p^*$ and $mathbbR^*$, but with $mathbbZ_p^*$ and $S^1$. We can view $S^1 cong zinmathbbC mid $. It's not hard to show that any $zin S^1$ is able to be written as $z = exp(2pi i t)$ for $tinmathbbR$ (we don't strictly need the factor $2pi$ here, but it's traditional). Due to $exp(x)$ being periodic, it's in fact enough to have $tin[0,1)$.



        This has an obvious group structure, in that:
        $$exp(2pi i t_0)exp(2pi i t_1) = exp(2pi i (t_0+t_1))$$
        If we're making the restriction that $t_iin[0,1)$, then we have to take $t_0+t_1mod 1$, but this is fairly standard.



        More than just having an obvious group structure, we actually have that any $mathbbZ_p^*$ injects into it.
        Specifically, we always have:
        $$
        phi_p:mathbbZ_p^*to S^1,quad phi_p(x) = exp(2pi i x/(p-1))
        $$

        Here, $p-1$ in the denominator is because $|mathbbZ_p^*| = p-1$.
        We can define the discrete logarithm problem for both of these groups in the standard way (here, it's important to restrict $t_iin[0, 1)$ if we want a unique answer).
        Then, we can relate these problems to each via the aforementioned injection.
        Through this image, we see that $S^1$ is "continuous" in the sense that it takes up the full circle, but the image of $mathbbZ_p^*$ in $S^1$ will always be "discrete" --- there will always be "some space" between points (they can't get arbitrarily close).







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        MarkMark

        1835




        1835



























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