Is $0.sqrt9$ a valid number?
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
add a comment |
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
1
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
– Math_QED
Jan 4 at 18:39
3
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 $0-9$?
– Eleven-Eleven
Jan 4 at 18:40
add a comment |
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
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
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
– Math_QED
Jan 4 at 18:39
3
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 $0-9$?
– Eleven-Eleven
Jan 4 at 18:40
add a comment |
1
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
– Math_QED
Jan 4 at 18:39
3
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 $0-9$?
– Eleven-Eleven
Jan 4 at 18:40
1
1
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
– Math_QED
Jan 4 at 18:39
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
– Math_QED
Jan 4 at 18:39
3
3
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 $0-9$?
– Eleven-Eleven
Jan 4 at 18:40
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 $0-9$?
– Eleven-Eleven
Jan 4 at 18:40
add a comment |
2 Answers
2
active
oldest
votes
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.
23.4k22137
add a comment |
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
1,262214
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
23.4k22137
add a comment |
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.
23.4k22137
add a comment |
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.
23.4k22137
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.
23.4k22137
answered Jan 4 at 18:41
J.G.J.G.
23.4k22137
answered Jan 4 at 18:41
J.G.J.G.
23.4k22137
answered Jan 4 at 18:41
J.G.J.G.
23.4k22137
23.4k22137
add a comment |
add a comment |
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
1,262214
add a comment |
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
1,262214
add a comment |
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
1,262214
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
1,262214
answered Jan 4 at 19:46
SkipSkip
1,262214
answered Jan 4 at 19:46
SkipSkip
1,262214
answered Jan 4 at 19:46
SkipSkip
1,262214
1,262214
add a comment |
add a comment |
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1
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
– Math_QED
Jan 4 at 18:39
3
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 $0-9$?
– Eleven-Eleven
Jan 4 at 18:40