multiply
sulekh...@gmail.com
suppose a is any non zero natural number ,
then a x 5 = a/2 x 10 is n't it ?
Richard Heathfield
It depends on the number representation you are using.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google Groups users, please read: <http://improve-usenet.org/>
"Usenet is a strange place" - dmr 29 July 1999
CBFalconer
>
> suppose a is any non zero natural number ,
> then a x 5 = a/2 x 10 is n't it ?
Only if a is even
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
Nishant Shukla
this statement is true:
'a' times 5 = ('a' divided by 2) times 10
Willem
) On Jul 2, 4:31 am, sulekhaswe...@gmail.com wrote:
)> Hi,
)>
)> suppose a is any non zero natural number ,
)> then a x 5 = a/2 x 10 is n't it ?
)
) this statement is true:
) 'a' times 5 = ('a' divided by 2) times 10
What does ('a' divided by 2) mean for natural numbers ?
If I ask someone: "What is 5 divided by 2", then
he could answer: "2, with a remainder of 1"
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
Richard Heathfield
> sulekh...@gmail.com wrote:
>>
>> suppose a is any non zero natural number ,
>> then a x 5 = a/2 x 10 is n't it ?
>
> Only if a is even
No, it's true for odd a (and even for a=0, despite the exception in the
OP).
Let a=7, for example. a * 5 = 35, and a/2 * 10 = 10a/2 = 5a = 35.
Mark Brader
>>> then a x 5 = a/2 x 10 is n't it ?
>> Only if a is even
> No, it's true for odd a ...
Depends on whether or not it's intended that the intermediate results
are also required to be natural numbers.
--
Mark Brader | "But [he] had already established his own reputation
Toronto | as someone who wrote poetry that mentioned the el."
m...@vex.net | --Al Kriman
dgates
wrote:
>Nishant Shukla wrote:
>) On Jul 2, 4:31 am, sulekhaswe...@gmail.com wrote:
>)> Hi,
>)>
>)> suppose a is any non zero natural number ,
>)> then a x 5 = a/2 x 10 is n't it ?
>)
>) this statement is true:
>) 'a' times 5 = ('a' divided by 2) times 10
>
>What does ('a' divided by 2) mean for natural numbers ?
>
>If I ask someone: "What is 5 divided by 2", then
>he could answer: "2, with a remainder of 1"
Hmm, and what would he say if you asked "What is '2, with a remainder
of 1' multiplied by 10?" :-)
Richard Heathfield
>>>> suppose a is any non zero natural number ,
>>>> then a x 5 = a/2 x 10 is n't it ?
>
>>> Only if a is even
>
>> No, it's true for odd a ...
>
> Depends on whether or not it's intended that the intermediate results
> are also required to be natural numbers.
And on whether we are allowed to rearrange the equation. In mathematics, it
is normal to be granted such licence, provided it doesn't break any rules.
(It is in quietly breaking such rules that paradoxists arrive at some of
their curios.)
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
Willem
) And on whether we are allowed to rearrange the equation. In mathematics, it
) is normal to be granted such licence, provided it doesn't break any rules.
) (It is in quietly breaking such rules that paradoxists arrive at some of
) their curios.)
In the natural numbers, the division operator is such that you would break
some rules when rearranging equations. I don't know offhand what they are
called, but I'm sure there are those more knowledgable who do.
Willem
) Hmm, and what would he say if you asked "What is '2, with a remainder
) of 1' multiplied by 10?" :-)
Err, he would probably say that that doesn't make sense.
What is '5, with a remainder of 3' multiplied by 10 ?
Simon Tatham
>> Depends on whether or not it's intended that the intermediate results
>> are also required to be natural numbers.
Richard Heathfield <r...@see.sig.invalid> wrote:
> And on whether we are allowed to rearrange the equation. In mathematics, it
> is normal to be granted such licence, provided it doesn't break any rules.
Yes, but that's precisely avoiding the question: Mark's point is
that he is stating what might be an implicit rule in this case,
which (if so) rearranging the equation would break.
The rationale behind rearranging and simplifying an equation is that
you apply transformations which do not affect the answer - or lack
of answer - given by the equation. That's the principle on which you
decide what rearrangements are valid in the first place. _If_ (as
Mark suggests is a possibility) we are constrained to work entirely
within the natural numbers and thus dividing an odd number by two is
forbidden, then we may not simplify "divide by two and then multiply
by ten" into "multiply by five", precisely _because_ the latter
would yield a result in cases where the former does not. You can't
get around such a restriction by citing rearrangement, because by
definition if the rearrangement gets you round a restriction then it
wasn't permitted anyway.
It is certainly true that naively rearranging this particular
equation can turn it into one which yields the same answer in all
cases where the original _did_ yield an answer and which also
produces answers with an internal consistency in some additional
cases, and this sort of possibility has historically tended to
inspire mathematicians to search for an expanded algebraic structure
which contains the previous one and in which more rearrangements
become valid and more equations soluble. Such processes have been
responsible for extending the natural numbers into integers,
rationals, reals, complex numbers and no end of rings, fields,
groups and more exotic stuff still. But none of that excuses the
breaking of the rules that bind you here and now in any given
situation; they merely suggest that _after_ you've finished solving
this particular problem and are no longer bound by its rules, then
there's an interesting area in which you might like to do further
research.
--
Simon Tatham "A cynic is a person who smells flowers and
<ana...@pobox.com> immediately looks around for a coffin."
Mike Williams
>Richard Heathfield wrote:
>) And on whether we are allowed to rearrange the equation. In mathematics, it
>) is normal to be granted such licence, provided it doesn't break any rules.
>) (It is in quietly breaking such rules that paradoxists arrive at some of
>) their curios.)
>
>In the natural numbers, the division operator is such that you would break
>some rules when rearranging equations. I don't know offhand what they are
>called, but I'm sure there are those more knowledgable who do.
I think you're looking for the word "associative".
If a set is associative under multiplication, it means that
(a * b) * c = a * (b * c)
for all a, b and c in the set.
Thus: (a/2 * 2) * 5 = a/2 * (2 * 5)
The natural numbers are associative under multiplication.
So if "a/2" is allowed to have any meaning for odd naturals, and
conforms to the expected behaviour (a/2 * 2) = a then we're OK.
--
Mike Williams
Gentleman of Leisure
dgates
wrote:
>dgates wrote:
>) Hmm, and what would he say if you asked "What is '2, with a remainder
>) of 1' multiplied by 10?" :-)
>
>Err, he would probably say that that doesn't make sense.
>
>What is '5, with a remainder of 3' multiplied by 10 ?
Oh, you're right...
Once you ask the second question, it forgets that the original
"remainder of 1" was tied to having divided by 2, and thus represents
1/2.
Not that it made much sense anyway :-)
Jeffrey Turner
> CBFalconer said:
>
>
>>sulekh...@gmail.com wrote:
>>
>>>suppose a is any non zero natural number ,
>>>then a x 5 = a/2 x 10 is n't it ?
>>
>>Only if a is even
>
>
> No, it's true for odd a
Not if a is an int.
--Jeff
--
The struggle with evil by means of violence
is the same as an attempt to stop a cloud,
in order that there may be no rain. -Leo Tolstoy
Jeffrey Turner
> dgates wrote:
> ) Hmm, and what would he say if you asked "What is '2, with a remainder
> ) of 1' multiplied by 10?" :-)
>
> Err, he would probably say that that doesn't make sense.
>
> What is '5, with a remainder of 3' multiplied by 10 ?
50 or 51 to 57 or 58, depending on rounding.
Greg Neill
news:tfOdnbl_IfG1yPPV...@posted.localnet
>
>> CBFalconer said:
>>
>>
>>> sulekh...@gmail.com wrote:
>>>
>>>> suppose a is any non zero natural number ,
>>>> then a x 5 = a/2 x 10 is n't it ?
>>>
>>> Only if a is even
>>
>>
>> No, it's true for odd a
>
> Not if a is an int.
Why not? The specification was that a is a natural
number. It doesn't say that a/2 must be also.
Richard Heathfield
> Richard Heathfield wrote:
>
>> CBFalconer said:
>>
>>
>>>sulekh...@gmail.com wrote:
>>>
>>>>suppose a is any non zero natural number ,
>>>>then a x 5 = a/2 x 10 is n't it ?
>>>
>>>Only if a is even
>>
>>
>> No, it's true for odd a
>
> Not if a is an int.
Firstly, it isn't an int. It's a non-zero natural number.
Secondly, on my system the following program (in which a is indeed an int
and which faithfully reproduces the problem) produced no output:
#include <stdio.h>
#include <limits.h>
int main(void)
{
int a;
for(a = INT_MIN; a < INT_MAX; a++)
{
double two = 2;
double five = 5;
double ten = 10;
double first = a * five;
double second = (a / two) * ten;
if(first != second)
{
printf("%d: %f, %f\n", a, first, second);
}
}
return 0;
}
You can test INT_MAX for yourself if you like.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
Mark Brader
>>>> then a x 5 = a/2 x 10 is n't it ?
>>> Only if a is even
>> No, it's true for odd a
> Not if a is an int.
If a is an int, a x 5 = a/2 x 10 is a syntax error.
--
Mark Brader At any rate, C++ != C. Actually, the value of
Toronto the expression "C++ != C" is [undefined].
m...@vex.net -- Peter da Silva