Mixing
Comparing the magnitudes of expressions of surds
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up vote
6
down vote
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I recently tackled some questions on maths-challenge / maths-aptitude papers where the task was to order various expressions made up of surds (without a calculator, obviously).
I found myself wondering whether I was relying too much on knowing the numerical value of some common surds, when a more robust method was available (and would work in more difficult cases).
For example, one question asked which is the largest of:
(a) $sqrt10$
(b) $sqrt2+sqrt3$
(c) $5-sqrt3$
In this case, I relied on my knowledge that $sqrt10 approx 3.16$ and $sqrt2approx 1.41$ and $sqrt3 approx 1.73$ to find (a) ~3.16, (b) ~3.14 and (c) ~3.27 so that the required answer is (c).
But this seemed inelegant: I felt there might be some way to manipulate the surd expressions to make the ordering more explicit. I can't see what that might be, however (squaring all the expressions didn't really help).
I'd appreciate some views: am I missing a trick, or was this particular question simply testing knowledge of some common values?
arithmetic radicals
asked 3 hours ago
BBO555
939
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BBO555 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
6
down vote
favorite
I recently tackled some questions on maths-challenge / maths-aptitude papers where the task was to order various expressions made up of surds (without a calculator, obviously).
I found myself wondering whether I was relying too much on knowing the numerical value of some common surds, when a more robust method was available (and would work in more difficult cases).
For example, one question asked which is the largest of:
(a) $sqrt10$
(b) $sqrt2+sqrt3$
(c) $5-sqrt3$
In this case, I relied on my knowledge that $sqrt10 approx 3.16$ and $sqrt2approx 1.41$ and $sqrt3 approx 1.73$ to find (a) ~3.16, (b) ~3.14 and (c) ~3.27 so that the required answer is (c).
But this seemed inelegant: I felt there might be some way to manipulate the surd expressions to make the ordering more explicit. I can't see what that might be, however (squaring all the expressions didn't really help).
I'd appreciate some views: am I missing a trick, or was this particular question simply testing knowledge of some common values?
arithmetic radicals
asked 3 hours ago
BBO555
939
New contributor
BBO555 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
+1 for providing context (your first two sentences), something that nearly all questions at this level fail to do, and for providing a nice explanation of your concern. Incidentally, for math aptitude and other tests, it has always been my understanding that the questions are NOT testing whether you know the approximations, but whether you can perform the type of analysis in the answer by @Lord Shark the Unknown. Of course, unless the question writer puts some effort behind writing such questions, such questions can often be solved by your method.
– Dave L. Renfro
3 hours ago
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I recently tackled some questions on maths-challenge / maths-aptitude papers where the task was to order various expressions made up of surds (without a calculator, obviously).
I found myself wondering whether I was relying too much on knowing the numerical value of some common surds, when a more robust method was available (and would work in more difficult cases).
For example, one question asked which is the largest of:
(a) $sqrt10$
(b) $sqrt2+sqrt3$
(c) $5-sqrt3$
In this case, I relied on my knowledge that $sqrt10 approx 3.16$ and $sqrt2approx 1.41$ and $sqrt3 approx 1.73$ to find (a) ~3.16, (b) ~3.14 and (c) ~3.27 so that the required answer is (c).
But this seemed inelegant: I felt there might be some way to manipulate the surd expressions to make the ordering more explicit. I can't see what that might be, however (squaring all the expressions didn't really help).
I'd appreciate some views: am I missing a trick, or was this particular question simply testing knowledge of some common values?
arithmetic radicals
asked 3 hours ago
BBO555
939
New contributor
BBO555 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I recently tackled some questions on maths-challenge / maths-aptitude papers where the task was to order various expressions made up of surds (without a calculator, obviously).
I found myself wondering whether I was relying too much on knowing the numerical value of some common surds, when a more robust method was available (and would work in more difficult cases).
For example, one question asked which is the largest of:
(a) $sqrt10$
(b) $sqrt2+sqrt3$
(c) $5-sqrt3$
In this case, I relied on my knowledge that $sqrt10 approx 3.16$ and $sqrt2approx 1.41$ and $sqrt3 approx 1.73$ to find (a) ~3.16, (b) ~3.14 and (c) ~3.27 so that the required answer is (c).
But this seemed inelegant: I felt there might be some way to manipulate the surd expressions to make the ordering more explicit. I can't see what that might be, however (squaring all the expressions didn't really help).
I'd appreciate some views: am I missing a trick, or was this particular question simply testing knowledge of some common values?
arithmetic radicals
arithmetic radicals
asked 3 hours ago
BBO555
939
New contributor
BBO555 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
BBO555
939
New contributor
BBO555 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
asked 3 hours ago
BBO555
939
New contributor
BBO555 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
BBO555
939
asked 3 hours ago
BBO555
939
939
New contributor
BBO555 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
BBO555 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
BBO555 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
+1 for providing context (your first two sentences), something that nearly all questions at this level fail to do, and for providing a nice explanation of your concern. Incidentally, for math aptitude and other tests, it has always been my understanding that the questions are NOT testing whether you know the approximations, but whether you can perform the type of analysis in the answer by @Lord Shark the Unknown. Of course, unless the question writer puts some effort behind writing such questions, such questions can often be solved by your method.
– Dave L. Renfro
3 hours ago
add a comment |Â
1
+1 for providing context (your first two sentences), something that nearly all questions at this level fail to do, and for providing a nice explanation of your concern. Incidentally, for math aptitude and other tests, it has always been my understanding that the questions are NOT testing whether you know the approximations, but whether you can perform the type of analysis in the answer by @Lord Shark the Unknown. Of course, unless the question writer puts some effort behind writing such questions, such questions can often be solved by your method.
– Dave L. Renfro
3 hours ago
1
1
+1 for providing context (your first two sentences), something that nearly all questions at this level fail to do, and for providing a nice explanation of your concern. Incidentally, for math aptitude and other tests, it has always been my understanding that the questions are NOT testing whether you know the approximations, but whether you can perform the type of analysis in the answer by @Lord Shark the Unknown. Of course, unless the question writer puts some effort behind writing such questions, such questions can often be solved by your method.
– Dave L. Renfro
3 hours ago
+1 for providing context (your first two sentences), something that nearly all questions at this level fail to do, and for providing a nice explanation of your concern. Incidentally, for math aptitude and other tests, it has always been my understanding that the questions are NOT testing whether you know the approximations, but whether you can perform the type of analysis in the answer by @Lord Shark the Unknown. Of course, unless the question writer puts some effort behind writing such questions, such questions can often be solved by your method.
– Dave L. Renfro
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
7
down vote
Comparing $sqrt10$ and $sqrt2+sqrt3$ is the same as comparing
$10$ and $(sqrt2+sqrt3)^2=5+2sqrt6$. That's the same as comparing
$5$ and $2sqrt6$. Which of these is bigger?
Likewise comparing $sqrt10$ and $5-sqrt3$ is the same as comparing
$10$ and $(5-sqrt3)^2=28-10sqrt3$. That's the same as comparing
$10sqrt3$ and $18$. Which of these is bigger?
answered 3 hours ago
Lord Shark the Unknown
89.8k955116
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
3 hours ago
Thank you for the hint!
– BBO555
3 hours ago
BBO555, regarding $5$ and $2sqrt6,$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be "seen" by considering the graph of $y = x^2$ for $xgeq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$
– Dave L. Renfro
3 hours ago
add a comment |Â
up vote
2
down vote
You can use:
(1) the fact that $f(x)=x^2$ is a monotonically increasing function when $xgeq0$
and
(2) the arithmetic-geometric mean inequality $sqrt(ab)leqfraca+b2$, when $a, bgeq0$.
Hence,
$$
(sqrt2+sqrt3)^2=5+2sqrt2cdot3leq5+2frac2+32=5+5=10=(sqrt10)^2
$$
Therefore, using (1), we obtain $sqrt2+sqrt3leq 10$.
answered 12 mins ago
minimax
1215
I would add that you can also "round down" at an intermediate stage of the computation. If you are trying to prove $a gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a gt c$, do some more manipulation, and show $c gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful.
– Ross Millikan
5 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
Comparing $sqrt10$ and $sqrt2+sqrt3$ is the same as comparing
$10$ and $(sqrt2+sqrt3)^2=5+2sqrt6$. That's the same as comparing
$5$ and $2sqrt6$. Which of these is bigger?
Likewise comparing $sqrt10$ and $5-sqrt3$ is the same as comparing
$10$ and $(5-sqrt3)^2=28-10sqrt3$. That's the same as comparing
$10sqrt3$ and $18$. Which of these is bigger?
answered 3 hours ago
Lord Shark the Unknown
89.8k955116
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
3 hours ago
Thank you for the hint!
– BBO555
3 hours ago
BBO555, regarding $5$ and $2sqrt6,$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be "seen" by considering the graph of $y = x^2$ for $xgeq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$
– Dave L. Renfro
3 hours ago
add a comment |Â
up vote
7
down vote
Comparing $sqrt10$ and $sqrt2+sqrt3$ is the same as comparing
$10$ and $(sqrt2+sqrt3)^2=5+2sqrt6$. That's the same as comparing
$5$ and $2sqrt6$. Which of these is bigger?
Likewise comparing $sqrt10$ and $5-sqrt3$ is the same as comparing
$10$ and $(5-sqrt3)^2=28-10sqrt3$. That's the same as comparing
$10sqrt3$ and $18$. Which of these is bigger?
answered 3 hours ago
Lord Shark the Unknown
89.8k955116
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
3 hours ago
Thank you for the hint!
– BBO555
3 hours ago
BBO555, regarding $5$ and $2sqrt6,$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be "seen" by considering the graph of $y = x^2$ for $xgeq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$
– Dave L. Renfro
3 hours ago
add a comment |Â
up vote
7
down vote
up vote
7
down vote
Comparing $sqrt10$ and $sqrt2+sqrt3$ is the same as comparing
$10$ and $(sqrt2+sqrt3)^2=5+2sqrt6$. That's the same as comparing
$5$ and $2sqrt6$. Which of these is bigger?
Likewise comparing $sqrt10$ and $5-sqrt3$ is the same as comparing
$10$ and $(5-sqrt3)^2=28-10sqrt3$. That's the same as comparing
$10sqrt3$ and $18$. Which of these is bigger?
answered 3 hours ago
Lord Shark the Unknown
89.8k955116
Comparing $sqrt10$ and $sqrt2+sqrt3$ is the same as comparing
$10$ and $(sqrt2+sqrt3)^2=5+2sqrt6$. That's the same as comparing
$5$ and $2sqrt6$. Which of these is bigger?
Likewise comparing $sqrt10$ and $5-sqrt3$ is the same as comparing
$10$ and $(5-sqrt3)^2=28-10sqrt3$. That's the same as comparing
$10sqrt3$ and $18$. Which of these is bigger?
answered 3 hours ago
Lord Shark the Unknown
89.8k955116
answered 3 hours ago
Lord Shark the Unknown
89.8k955116
answered 3 hours ago
Lord Shark the Unknown
89.8k955116
answered 3 hours ago
Lord Shark the Unknown
89.8k955116
89.8k955116
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
3 hours ago
Thank you for the hint!
– BBO555
3 hours ago
BBO555, regarding $5$ and $2sqrt6,$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be "seen" by considering the graph of $y = x^2$ for $xgeq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$
– Dave L. Renfro
3 hours ago
add a comment |Â
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
3 hours ago
Thank you for the hint!
– BBO555
3 hours ago
BBO555, regarding $5$ and $2sqrt6,$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be "seen" by considering the graph of $y = x^2$ for $xgeq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$
– Dave L. Renfro
3 hours ago
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
3 hours ago
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
3 hours ago
Thank you for the hint!
– BBO555
3 hours ago
Thank you for the hint!
– BBO555
3 hours ago
BBO555, regarding $5$ and $2sqrt6,$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be "seen" by considering the graph of $y = x^2$ for $xgeq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$
– Dave L. Renfro
3 hours ago
BBO555, regarding $5$ and $2sqrt6,$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be "seen" by considering the graph of $y = x^2$ for $xgeq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$
– Dave L. Renfro
3 hours ago
add a comment |Â
up vote
2
down vote
You can use:
(1) the fact that $f(x)=x^2$ is a monotonically increasing function when $xgeq0$
and
(2) the arithmetic-geometric mean inequality $sqrt(ab)leqfraca+b2$, when $a, bgeq0$.
Hence,
$$
(sqrt2+sqrt3)^2=5+2sqrt2cdot3leq5+2frac2+32=5+5=10=(sqrt10)^2
$$
Therefore, using (1), we obtain $sqrt2+sqrt3leq 10$.
answered 12 mins ago
minimax
1215
I would add that you can also "round down" at an intermediate stage of the computation. If you are trying to prove $a gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a gt c$, do some more manipulation, and show $c gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful.
– Ross Millikan
5 mins ago
add a comment |Â
up vote
2
down vote
You can use:
(1) the fact that $f(x)=x^2$ is a monotonically increasing function when $xgeq0$
and
(2) the arithmetic-geometric mean inequality $sqrt(ab)leqfraca+b2$, when $a, bgeq0$.
Hence,
$$
(sqrt2+sqrt3)^2=5+2sqrt2cdot3leq5+2frac2+32=5+5=10=(sqrt10)^2
$$
Therefore, using (1), we obtain $sqrt2+sqrt3leq 10$.
answered 12 mins ago
minimax
1215
I would add that you can also "round down" at an intermediate stage of the computation. If you are trying to prove $a gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a gt c$, do some more manipulation, and show $c gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful.
– Ross Millikan
5 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You can use:
(1) the fact that $f(x)=x^2$ is a monotonically increasing function when $xgeq0$
and
(2) the arithmetic-geometric mean inequality $sqrt(ab)leqfraca+b2$, when $a, bgeq0$.
Hence,
$$
(sqrt2+sqrt3)^2=5+2sqrt2cdot3leq5+2frac2+32=5+5=10=(sqrt10)^2
$$
Therefore, using (1), we obtain $sqrt2+sqrt3leq 10$.
answered 12 mins ago
minimax
1215
You can use:
(1) the fact that $f(x)=x^2$ is a monotonically increasing function when $xgeq0$
and
(2) the arithmetic-geometric mean inequality $sqrt(ab)leqfraca+b2$, when $a, bgeq0$.
Hence,
$$
(sqrt2+sqrt3)^2=5+2sqrt2cdot3leq5+2frac2+32=5+5=10=(sqrt10)^2
$$
Therefore, using (1), we obtain $sqrt2+sqrt3leq 10$.
answered 12 mins ago
minimax
1215
answered 12 mins ago
minimax
1215
answered 12 mins ago
minimax
1215
answered 12 mins ago
minimax
1215
1215
I would add that you can also "round down" at an intermediate stage of the computation. If you are trying to prove $a gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a gt c$, do some more manipulation, and show $c gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful.
– Ross Millikan
5 mins ago
add a comment |Â
I would add that you can also "round down" at an intermediate stage of the computation. If you are trying to prove $a gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a gt c$, do some more manipulation, and show $c gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful.
– Ross Millikan
5 mins ago
I would add that you can also "round down" at an intermediate stage of the computation. If you are trying to prove $a gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a gt c$, do some more manipulation, and show $c gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful.
– Ross Millikan
5 mins ago
I would add that you can also "round down" at an intermediate stage of the computation. If you are trying to prove $a gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a gt c$, do some more manipulation, and show $c gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful.
– Ross Millikan
5 mins ago
add a comment |Â
BBO555 is a new contributor. Be nice, and check out our Code of Conduct.
BBO555 is a new contributor. Be nice, and check out our Code of Conduct.
BBO555 is a new contributor. Be nice, and check out our Code of Conduct.
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1
+1 for providing context (your first two sentences), something that nearly all questions at this level fail to do, and for providing a nice explanation of your concern. Incidentally, for math aptitude and other tests, it has always been my understanding that the questions are NOT testing whether you know the approximations, but whether you can perform the type of analysis in the answer by @Lord Shark the Unknown. Of course, unless the question writer puts some effort behind writing such questions, such questions can often be solved by your method.
– Dave L. Renfro
3 hours ago