Mixing
Is $0.sqrt9$ a valid number?
$begingroup$
Is $0.sqrt9$ valid number? Are such numbers allowed?
First, I thought the value of the above number can be 0.3 but then it occurred how I would interpret this number: $0.65sqrt2$ or $0.65sqrt229$
Are such numbers valid?
real-numbers
asked Jan 4 at 18:36
user33786user33786
116129
$endgroup$
add a comment |
$begingroup$
Is $0.sqrt9$ valid number? Are such numbers allowed?
First, I thought the value of the above number can be 0.3 but then it occurred how I would interpret this number: $0.65sqrt2$ or $0.65sqrt229$
Are such numbers valid?
real-numbers
asked Jan 4 at 18:36
user33786user33786
116129
$endgroup$
1
$begingroup$
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
$endgroup$
– Math_QED
Jan 4 at 18:39
3
$begingroup$
I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
$endgroup$
– Eleven-Eleven
Jan 4 at 18:40
add a comment |
$begingroup$
Is $0.sqrt9$ valid number? Are such numbers allowed?
First, I thought the value of the above number can be 0.3 but then it occurred how I would interpret this number: $0.65sqrt2$ or $0.65sqrt229$
Are such numbers valid?
real-numbers
asked Jan 4 at 18:36
user33786user33786
116129
$endgroup$
Is $0.sqrt9$ valid number? Are such numbers allowed?
First, I thought the value of the above number can be 0.3 but then it occurred how I would interpret this number: $0.65sqrt2$ or $0.65sqrt229$
Are such numbers valid?
real-numbers
real-numbers
asked Jan 4 at 18:36
user33786user33786
116129
asked Jan 4 at 18:36
user33786user33786
116129
asked Jan 4 at 18:36
user33786user33786
116129
asked Jan 4 at 18:36
user33786user33786
116129
asked Jan 4 at 18:36
user33786user33786
116129
116129
1
$begingroup$
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
$endgroup$
– Math_QED
Jan 4 at 18:39
3
$begingroup$
I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
$endgroup$
– Eleven-Eleven
Jan 4 at 18:40
add a comment |
1
$begingroup$
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
$endgroup$
– Math_QED
Jan 4 at 18:39
3
$begingroup$
I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
$endgroup$
– Eleven-Eleven
Jan 4 at 18:40
1
1
$begingroup$
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
$endgroup$
– Math_QED
Jan 4 at 18:39
$begingroup$
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
$endgroup$
– Math_QED
Jan 4 at 18:39
3
3
$begingroup$
I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
$endgroup$
– Eleven-Eleven
Jan 4 at 18:40
$begingroup$
I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
$endgroup$
– Eleven-Eleven
Jan 4 at 18:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+bsqrt{c}$, so you might as well just write that.
answered Jan 4 at 18:41
J.G.J.G.
24.5k22539
$endgroup$
add a comment |
$begingroup$
I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as
$$f(x) = 10^{-lfloor log_{10}(x)+1 rfloor}x.$$
Using this definition, then indeed it turns out
$$f(sqrt{9}) = 0.3.$$
$65 sqrt 2$ is commonly interpreted as $65 cdot sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || sqrt 2 = 651.41421...$. In either case that you mean, we have
$$f(65 cdot sqrt 2) = frac{13}{10sqrt 2}, f(65||sqrt2) = frac{650+sqrt 2}{1000}$$
and so on.
answered Jan 4 at 19:46
SkipSkip
1,387214
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3061949%2fis-0-sqrt9-a-valid-number%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+bsqrt{c}$, so you might as well just write that.
answered Jan 4 at 18:41
J.G.J.G.
24.5k22539
$endgroup$
add a comment |
$begingroup$
They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+bsqrt{c}$, so you might as well just write that.
answered Jan 4 at 18:41
J.G.J.G.
24.5k22539
$endgroup$
add a comment |
$begingroup$
They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+bsqrt{c}$, so you might as well just write that.
answered Jan 4 at 18:41
J.G.J.G.
24.5k22539
$endgroup$
They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+bsqrt{c}$, so you might as well just write that.
answered Jan 4 at 18:41
J.G.J.G.
24.5k22539
answered Jan 4 at 18:41
J.G.J.G.
24.5k22539
answered Jan 4 at 18:41
J.G.J.G.
24.5k22539
answered Jan 4 at 18:41
J.G.J.G.
24.5k22539
24.5k22539
add a comment |
add a comment |
$begingroup$
I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as
$$f(x) = 10^{-lfloor log_{10}(x)+1 rfloor}x.$$
Using this definition, then indeed it turns out
$$f(sqrt{9}) = 0.3.$$
$65 sqrt 2$ is commonly interpreted as $65 cdot sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || sqrt 2 = 651.41421...$. In either case that you mean, we have
$$f(65 cdot sqrt 2) = frac{13}{10sqrt 2}, f(65||sqrt2) = frac{650+sqrt 2}{1000}$$
and so on.
answered Jan 4 at 19:46
SkipSkip
1,387214
$endgroup$
add a comment |
$begingroup$
I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as
$$f(x) = 10^{-lfloor log_{10}(x)+1 rfloor}x.$$
Using this definition, then indeed it turns out
$$f(sqrt{9}) = 0.3.$$
$65 sqrt 2$ is commonly interpreted as $65 cdot sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || sqrt 2 = 651.41421...$. In either case that you mean, we have
$$f(65 cdot sqrt 2) = frac{13}{10sqrt 2}, f(65||sqrt2) = frac{650+sqrt 2}{1000}$$
and so on.
answered Jan 4 at 19:46
SkipSkip
1,387214
$endgroup$
add a comment |
$begingroup$
I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as
$$f(x) = 10^{-lfloor log_{10}(x)+1 rfloor}x.$$
Using this definition, then indeed it turns out
$$f(sqrt{9}) = 0.3.$$
$65 sqrt 2$ is commonly interpreted as $65 cdot sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || sqrt 2 = 651.41421...$. In either case that you mean, we have
$$f(65 cdot sqrt 2) = frac{13}{10sqrt 2}, f(65||sqrt2) = frac{650+sqrt 2}{1000}$$
and so on.
answered Jan 4 at 19:46
SkipSkip
1,387214
$endgroup$
I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as
$$f(x) = 10^{-lfloor log_{10}(x)+1 rfloor}x.$$
Using this definition, then indeed it turns out
$$f(sqrt{9}) = 0.3.$$
$65 sqrt 2$ is commonly interpreted as $65 cdot sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || sqrt 2 = 651.41421...$. In either case that you mean, we have
$$f(65 cdot sqrt 2) = frac{13}{10sqrt 2}, f(65||sqrt2) = frac{650+sqrt 2}{1000}$$
and so on.
answered Jan 4 at 19:46
SkipSkip
1,387214
answered Jan 4 at 19:46
SkipSkip
1,387214
answered Jan 4 at 19:46
SkipSkip
1,387214
answered Jan 4 at 19:46
SkipSkip
1,387214
1,387214
add a comment |
add a comment |
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.
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%2f3061949%2fis-0-sqrt9-a-valid-number%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
$begingroup$
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
$endgroup$
– Math_QED
Jan 4 at 18:39
3
$begingroup$
I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
$endgroup$
– Eleven-Eleven
Jan 4 at 18:40