Mixing
Infinite non-periodic binary fraction
up vote
0
down vote
favorite
I have an infinite non-periodic binary fraction. For example:
$frac_1=0.101111011100110010001001010010000001001...$
Is it always true that $1-frac_1$ = non-periodic binary fraction?
fractions
asked 2 days ago
bvl
72
add a comment |
up vote
0
down vote
favorite
I have an infinite non-periodic binary fraction. For example:
$frac_1=0.101111011100110010001001010010000001001...$
Is it always true that $1-frac_1$ = non-periodic binary fraction?
fractions
asked 2 days ago
bvl
72
By non-periodic, I suppose you mean not "ultimately periodic". The answer is quite obviously yes. Notice that $1-frac_1$ has the same binary expansion where you switch $0$ and $1$.
– Jean-Claude Arbaut
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have an infinite non-periodic binary fraction. For example:
$frac_1=0.101111011100110010001001010010000001001...$
Is it always true that $1-frac_1$ = non-periodic binary fraction?
fractions
asked 2 days ago
bvl
72
I have an infinite non-periodic binary fraction. For example:
$frac_1=0.101111011100110010001001010010000001001...$
Is it always true that $1-frac_1$ = non-periodic binary fraction?
fractions
fractions
asked 2 days ago
bvl
72
asked 2 days ago
bvl
72
asked 2 days ago
bvl
72
asked 2 days ago
bvl
72
asked 2 days ago
bvl
72
72
By non-periodic, I suppose you mean not "ultimately periodic". The answer is quite obviously yes. Notice that $1-frac_1$ has the same binary expansion where you switch $0$ and $1$.
– Jean-Claude Arbaut
2 days ago
add a comment |
By non-periodic, I suppose you mean not "ultimately periodic". The answer is quite obviously yes. Notice that $1-frac_1$ has the same binary expansion where you switch $0$ and $1$.
– Jean-Claude Arbaut
2 days ago
By non-periodic, I suppose you mean not "ultimately periodic". The answer is quite obviously yes. Notice that $1-frac_1$ has the same binary expansion where you switch $0$ and $1$.
– Jean-Claude Arbaut
2 days ago
By non-periodic, I suppose you mean not "ultimately periodic". The answer is quite obviously yes. Notice that $1-frac_1$ has the same binary expansion where you switch $0$ and $1$.
– Jean-Claude Arbaut
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Well, yes. This is actually quite obvious.
If a fraction repeats (is periodic) then it is a fraction. If it's not, then it's not a fraction (that is, it's irrational).
Clearly, a number $q$ is a fraction if and only if $1-q$ is. Hence, if the digits of some number $q$ do not repeat, neither will the digits of $1-q$.
In fact, as pointed out in the comments, it's even easier to see it when you realize $1-q$ (if it's binary) is just $q$ with the digits flipped (changing $0$s into $1$s and vice versa). So, if $q$ doesn't repeat, neither will $1-q$.
answered 2 days ago
vrugtehagel
10.7k1549
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Well, yes. This is actually quite obvious.
If a fraction repeats (is periodic) then it is a fraction. If it's not, then it's not a fraction (that is, it's irrational).
Clearly, a number $q$ is a fraction if and only if $1-q$ is. Hence, if the digits of some number $q$ do not repeat, neither will the digits of $1-q$.
In fact, as pointed out in the comments, it's even easier to see it when you realize $1-q$ (if it's binary) is just $q$ with the digits flipped (changing $0$s into $1$s and vice versa). So, if $q$ doesn't repeat, neither will $1-q$.
answered 2 days ago
vrugtehagel
10.7k1549
add a comment |
up vote
0
down vote
accepted
Well, yes. This is actually quite obvious.
If a fraction repeats (is periodic) then it is a fraction. If it's not, then it's not a fraction (that is, it's irrational).
Clearly, a number $q$ is a fraction if and only if $1-q$ is. Hence, if the digits of some number $q$ do not repeat, neither will the digits of $1-q$.
In fact, as pointed out in the comments, it's even easier to see it when you realize $1-q$ (if it's binary) is just $q$ with the digits flipped (changing $0$s into $1$s and vice versa). So, if $q$ doesn't repeat, neither will $1-q$.
answered 2 days ago
vrugtehagel
10.7k1549
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Well, yes. This is actually quite obvious.
If a fraction repeats (is periodic) then it is a fraction. If it's not, then it's not a fraction (that is, it's irrational).
Clearly, a number $q$ is a fraction if and only if $1-q$ is. Hence, if the digits of some number $q$ do not repeat, neither will the digits of $1-q$.
In fact, as pointed out in the comments, it's even easier to see it when you realize $1-q$ (if it's binary) is just $q$ with the digits flipped (changing $0$s into $1$s and vice versa). So, if $q$ doesn't repeat, neither will $1-q$.
answered 2 days ago
vrugtehagel
10.7k1549
Well, yes. This is actually quite obvious.
If a fraction repeats (is periodic) then it is a fraction. If it's not, then it's not a fraction (that is, it's irrational).
Clearly, a number $q$ is a fraction if and only if $1-q$ is. Hence, if the digits of some number $q$ do not repeat, neither will the digits of $1-q$.
In fact, as pointed out in the comments, it's even easier to see it when you realize $1-q$ (if it's binary) is just $q$ with the digits flipped (changing $0$s into $1$s and vice versa). So, if $q$ doesn't repeat, neither will $1-q$.
answered 2 days ago
vrugtehagel
10.7k1549
answered 2 days ago
vrugtehagel
10.7k1549
answered 2 days ago
vrugtehagel
10.7k1549
answered 2 days ago
vrugtehagel
10.7k1549
10.7k1549
add a comment |
add a comment |
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%2f3005431%2finfinite-non-periodic-binary-fraction%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
By non-periodic, I suppose you mean not "ultimately periodic". The answer is quite obviously yes. Notice that $1-frac_1$ has the same binary expansion where you switch $0$ and $1$.
– Jean-Claude Arbaut
2 days ago