HP Req: Question? Why is Mod = 13 ? - Hewlett Packard

This is a discussion on HP Req: Question? Why is Mod = 13 ? - Hewlett Packard ; While I'm afraid I'm betraying my ignorance, I really don't understand the significance of the number 13 as a default modulo setting. I see it's a prime... Why not 12, or 10, or 8 ( or 3, or 23, or ...

Thread: HP Req: Question? Why is Mod = 13 ?

1. HP Req: Question? Why is Mod = 13 ?

While I'm afraid I'm betraying my ignorance,
I really don't understand the significance
of the number 13 as a default modulo setting.

I see it's a prime...
Why not 12, or 10, or 8 ( or 3, or 23, or 1999..) ?

I Thank anyone for a clue, and while I may not
see the entire grand significance initialy, I'll
an example.

I just think it's a bizarre choice at this point.

John.

2. Re: HP Req: Question? Why is Mod = 13 ?

"John Henry" wrote in message
news:48c40ea0\$0\$3373\$c3e8da3@news.astraweb.com...
>
> While I'm afraid I'm betraying my ignorance,
> I really don't understand the significance
> of the number 13 as a default modulo setting.
>
> I see it's a prime...
> Why not 12, or 10, or 8 ( or 3, or 23, or 1999..) ?
>
> I Thank anyone for a clue, and while I may not
> see the entire grand significance initialy, I'll
> an example.
>
> I just think it's a bizarre choice at this point.
>
> John.
>
>

Maybe that particular HP programmer has triskaidekaphilia.

Bob
--
== All google group posts are automatically deleted due to spam ==

3. Re: HP Req: Question? Why is Mod = 13 ?

"BobW" wrote in
news:K9WdnVFcZbmXj1nVnZ2dnUVZ_szinZ2d@giganews.com :

>
> "John Henry" wrote in message
> news:48c40ea0\$0\$3373\$c3e8da3@news.astraweb.com...
>>
>> While I'm afraid I'm betraying my ignorance,
>> I really don't understand the significance
>> of the number 13 as a default modulo setting.
>>
>> I see it's a prime...
>> Why not 12, or 10, or 8 ( or 3, or 23, or 1999..) ?
>>
>> I Thank anyone for a clue, and while I may not
>> see the entire grand significance initialy, I'll
>> an example.
>>
>> I just think it's a bizarre choice at this point.
>>
>> John.
>>
>>

>
> Maybe that particular HP programmer has triskaidekaphilia.
>
> Bob

P.S.
Ohhh, M'gosh, I like this 40gs! (Really nice..)

4. Re: HP Req: Question? Why is Mod = 13 ?

On 7 Sep., 19:25, John Henry wrote:
> I really don't understand the significance
> of the number 13 as a default modulo setting.
> I see it's a prime...
> Why not 12, or 10, or 8 ( or 3, or 23, or 1999..) ?
>
> I Thank anyone for a clue, and while I may not
> see the entire grand significance initialy, I'll
> an example.

Hello John,

more than thirty years ago, I learned in an algebra course
that the intergers modulo a prime number are a field. Let me
try to remember what this means:

Example: if you calculate modulo 9 (which is not prime)
you get 3 * 3 mod 9 = 0. This means in the integers modulo 9,
the rule "if a product is zero, then one of the factors must be zero"
is not true.

If you calculate modulo a prime, this rule is true and you have a
so called finite field.

One little hint why modulo 9 calculation can be usefull:
take as an example 487 * 327 = 159249
now calculate the cross sum of these numbers by adding their digits:
19 * 12 = 30
this equation is not true but calculate once more the cross sums
10 * 3 = 3
this equation is still not true, so calculate once more the cross sums
1 * 3 = 3
Now the equation is true and if you calculate the mod 9 values of
487 * 327 = 159249 you get the last cross sum equation

In former centuries the cross sum method was a nice trick to

I hope I remembered all those things correctly

Regards,
Peter

5. Re: HP Req: Question? Why is Mod = 13 ?

On Sep 7, 6:14*pm, PeterW wrote:
> On 7 Sep., 19:25, John Henry wrote:
>
> > I really don't understand the significance
> > of the number 13 as a default modulo setting.
> > I see it's a prime...
> > Why not 12, or 10, or 8 ( or 3, or 23, or 1999..) ?

>
> Example: if you calculate modulo 9 (which is not prime)
> you get 3 * 3 mod 9 = 0. This means in the integers modulo 9,
> the rule "if a product is zero, then one of the factors must be zero"
> is not true.
>
> If you calculate modulo a prime, this rule is true and you have a
> so called finite field.
>

Here's my guess:

There has to be an initial default value to be used for the ARITH/
MODUL functions so that they will work (even if not to the user's
expectation) instead of giving an error. A prime modulus is inherently
more interesting than a composite one (see the above post by PeterW),
so 13 is as good as any other. Plus, it's not "too big" or "too small"
to be handled by the mind.

Also, INVMOD is only guaranteed to work for all arguments if the
modulus is prime (see above by PeterW). This is because all cyclic
groups Z/pZ (p prime) will have no zero divisors. This means that all
elements of Z/pZ are units and are therefore, by definition,
invertible.

Other than that, I don't see any particular reason why they chose 13
as the default modulus.

S.C.

6. Re: HP Req: Question? Why is Mod = 13 ?

sc_usenet@hotmail.com wrote in
m:

> On Sep 7, 6:14*pm, PeterW wrote:
>> On 7 Sep., 19:25, John Henry wrote:
>>
>> > I really don't understand the significance
>> > of the number 13 as a default modulo setting.
>> > I see it's a prime...
>> > Why not 12, or 10, or 8 ( or 3, or 23, or 1999..) ?

>>
>> Example: if you calculate modulo 9 (which is not prime)
>> you get 3 * 3 mod 9 = 0. This means in the integers modulo 9,
>> the rule "if a product is zero, then one of the factors must be
>> zero" is not true.
>>
>> If you calculate modulo a prime, this rule is true and you have a
>> so called finite field.
>>

>
>
> Here's my guess:
>
> There has to be an initial default value to be used for the ARITH/
> MODUL functions so that they will work (even if not to the user's
> expectation) instead of giving an error. A prime modulus is
> inherently more interesting than a composite one (see the above
> post by PeterW), so 13 is as good as any other. Plus, it's not
> "too big" or "too small" to be handled by the mind.
>
> Also, INVMOD is only guaranteed to work for all arguments if the
> modulus is prime (see above by PeterW). This is because all cyclic
> groups Z/pZ (p prime) will have no zero divisors. This means that
> all elements of Z/pZ are units and are therefore, by definition,
> invertible.
>
> Other than that, I don't see any particular reason why they chose
> 13 as the default modulus.
>
> S.C.
>

Hi again-

Thank you both very much for the response. I think I will have to
chase around the W^3 for additional information. Interesting..

JH

7. Re: HP Req: Question? Why is Mod = 13 ?

Hi

On 2008-09-08 08:14:14 +1000, PeterW said:
>
> In former centuries the cross sum method was a nice trick to