Digit Sum Inequality Equation
up vote
0
down vote
favorite
Let $Q (n)$ be the digit sum of the integers $n$. Now I want to prove that
$Q (m + n) ≤ Q (m) + Q (n)$
is valid for all positive integers $m$ and $n$.
So we can write:
$n = a_n * 10^n + a_{n−1} * 10^{n−1} + · · · + a_1 * 10^1 + a_0 *10^0$
Thus we have:
$Q(n) = a_n + a_{n−1} + · · · + a_1 + a_0$
and of course:
$Q(m) = a_m + a_{m−1} + · · · + a_1 + a_0$.
Now:
$Q(m+n)= a_n+m + a_{n+m−1} + · · · + a_1 + a_0$
This equation seems to be smaller or equal to:
$Q(n) + Q(m) = a_n + a_{n−1} + · · · + a_1 + a_0 + a_m + a_{m−1} + · · · + a_1 + a_0$.
I'll try to do an induction on n and a fixed m.
number-theory inequality
asked Nov 19 at 19:55
calculatormathematical
389
add a comment |
up vote
0
down vote
favorite
Let $Q (n)$ be the digit sum of the integers $n$. Now I want to prove that
$Q (m + n) ≤ Q (m) + Q (n)$
is valid for all positive integers $m$ and $n$.
So we can write:
$n = a_n * 10^n + a_{n−1} * 10^{n−1} + · · · + a_1 * 10^1 + a_0 *10^0$
Thus we have:
$Q(n) = a_n + a_{n−1} + · · · + a_1 + a_0$
and of course:
$Q(m) = a_m + a_{m−1} + · · · + a_1 + a_0$.
Now:
$Q(m+n)= a_n+m + a_{n+m−1} + · · · + a_1 + a_0$
This equation seems to be smaller or equal to:
$Q(n) + Q(m) = a_n + a_{n−1} + · · · + a_1 + a_0 + a_m + a_{m−1} + · · · + a_1 + a_0$.
I'll try to do an induction on n and a fixed m.
number-theory inequality
asked Nov 19 at 19:55
calculatormathematical
389
1
If there are no carries, then you have an equality. If there is a carry, you've added two digits and necessarily produced a smaller sum (of the $1$ carry digit and the remaining sum). QED.
– David G. Stork
Nov 19 at 19:57
Why do you think $Q(m+n)=Q(Q(n)+Q(m))$? Try $m=n=99$
– Macavity
Nov 20 at 1:22
Try induction on $n$ for any fixed $m$.
– Macavity
Nov 20 at 1:23
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $Q (n)$ be the digit sum of the integers $n$. Now I want to prove that
$Q (m + n) ≤ Q (m) + Q (n)$
is valid for all positive integers $m$ and $n$.
So we can write:
$n = a_n * 10^n + a_{n−1} * 10^{n−1} + · · · + a_1 * 10^1 + a_0 *10^0$
Thus we have:
$Q(n) = a_n + a_{n−1} + · · · + a_1 + a_0$
and of course:
$Q(m) = a_m + a_{m−1} + · · · + a_1 + a_0$.
Now:
$Q(m+n)= a_n+m + a_{n+m−1} + · · · + a_1 + a_0$
This equation seems to be smaller or equal to:
$Q(n) + Q(m) = a_n + a_{n−1} + · · · + a_1 + a_0 + a_m + a_{m−1} + · · · + a_1 + a_0$.
I'll try to do an induction on n and a fixed m.
number-theory inequality
asked Nov 19 at 19:55
calculatormathematical
389
Let $Q (n)$ be the digit sum of the integers $n$. Now I want to prove that
$Q (m + n) ≤ Q (m) + Q (n)$
is valid for all positive integers $m$ and $n$.
So we can write:
$n = a_n * 10^n + a_{n−1} * 10^{n−1} + · · · + a_1 * 10^1 + a_0 *10^0$
Thus we have:
$Q(n) = a_n + a_{n−1} + · · · + a_1 + a_0$
and of course:
$Q(m) = a_m + a_{m−1} + · · · + a_1 + a_0$.
Now:
$Q(m+n)= a_n+m + a_{n+m−1} + · · · + a_1 + a_0$
This equation seems to be smaller or equal to:
$Q(n) + Q(m) = a_n + a_{n−1} + · · · + a_1 + a_0 + a_m + a_{m−1} + · · · + a_1 + a_0$.
I'll try to do an induction on n and a fixed m.
number-theory inequality
number-theory inequality
asked Nov 19 at 19:55
calculatormathematical
389
asked Nov 19 at 19:55
calculatormathematical
389
edited Nov 20 at 18:21
asked Nov 19 at 19:55
calculatormathematical
389
asked Nov 19 at 19:55
calculatormathematical
389
asked Nov 19 at 19:55
calculatormathematical
389
389
1
If there are no carries, then you have an equality. If there is a carry, you've added two digits and necessarily produced a smaller sum (of the $1$ carry digit and the remaining sum). QED.
– David G. Stork
Nov 19 at 19:57
Why do you think $Q(m+n)=Q(Q(n)+Q(m))$? Try $m=n=99$
– Macavity
Nov 20 at 1:22
Try induction on $n$ for any fixed $m$.
– Macavity
Nov 20 at 1:23
add a comment |
1
If there are no carries, then you have an equality. If there is a carry, you've added two digits and necessarily produced a smaller sum (of the $1$ carry digit and the remaining sum). QED.
– David G. Stork
Nov 19 at 19:57
Why do you think $Q(m+n)=Q(Q(n)+Q(m))$? Try $m=n=99$
– Macavity
Nov 20 at 1:22
Try induction on $n$ for any fixed $m$.
– Macavity
Nov 20 at 1:23
1
1
If there are no carries, then you have an equality. If there is a carry, you've added two digits and necessarily produced a smaller sum (of the $1$ carry digit and the remaining sum). QED.
– David G. Stork
Nov 19 at 19:57
If there are no carries, then you have an equality. If there is a carry, you've added two digits and necessarily produced a smaller sum (of the $1$ carry digit and the remaining sum). QED.
– David G. Stork
Nov 19 at 19:57
Why do you think $Q(m+n)=Q(Q(n)+Q(m))$? Try $m=n=99$
– Macavity
Nov 20 at 1:22
Why do you think $Q(m+n)=Q(Q(n)+Q(m))$? Try $m=n=99$
– Macavity
Nov 20 at 1:22
Try induction on $n$ for any fixed $m$.
– Macavity
Nov 20 at 1:23
Try induction on $n$ for any fixed $m$.
– Macavity
Nov 20 at 1:23
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005437%2fdigit-sum-inequality-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
If there are no carries, then you have an equality. If there is a carry, you've added two digits and necessarily produced a smaller sum (of the $1$ carry digit and the remaining sum). QED.
– David G. Stork
Nov 19 at 19:57
Why do you think $Q(m+n)=Q(Q(n)+Q(m))$? Try $m=n=99$
– Macavity
Nov 20 at 1:22
Try induction on $n$ for any fixed $m$.
– Macavity
Nov 20 at 1:23