Comparing the magnitudes of expressions of surds
Clash Royale CLAN TAG#URR8PPP
up vote
8
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) $approx 3.16$, (b) $approx ~3.14$ and (c) $approx ~3.27$ 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?
EDIT: after the very helpful answers, which certainly showed that there was a much satisfying and general way of approaching the original question, can I also ask about another version of the question which included (d) $sqrt[4]101$.
When approaching the question by approximation, I simply observed that $sqrt[4]101$ is only a tiny bit greater than $sqrt10$, and hence it still was clear to choose (c) as the answer. Is there any elegant way to extend the more robust methods to handle this case?
arithmetic radicals
asked Sep 19 at 10:20
BBO555
10710
add a comment |Â
up vote
8
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) $approx 3.16$, (b) $approx ~3.14$ and (c) $approx ~3.27$ 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?
EDIT: after the very helpful answers, which certainly showed that there was a much satisfying and general way of approaching the original question, can I also ask about another version of the question which included (d) $sqrt[4]101$.
When approaching the question by approximation, I simply observed that $sqrt[4]101$ is only a tiny bit greater than $sqrt10$, and hence it still was clear to choose (c) as the answer. Is there any elegant way to extend the more robust methods to handle this case?
arithmetic radicals
asked Sep 19 at 10:20
BBO555
10710
2
+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
Sep 19 at 10:41
Thank you for all the comments and answers. I am pleased I chose to ask the question at MSE -- there was indeed something to learn here!
– BBO555
Sep 20 at 10:05
add a comment |Â
up vote
8
down vote
favorite
up vote
8
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) $approx 3.16$, (b) $approx ~3.14$ and (c) $approx ~3.27$ 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?
EDIT: after the very helpful answers, which certainly showed that there was a much satisfying and general way of approaching the original question, can I also ask about another version of the question which included (d) $sqrt[4]101$.
When approaching the question by approximation, I simply observed that $sqrt[4]101$ is only a tiny bit greater than $sqrt10$, and hence it still was clear to choose (c) as the answer. Is there any elegant way to extend the more robust methods to handle this case?
arithmetic radicals
asked Sep 19 at 10:20
BBO555
10710
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) $approx 3.16$, (b) $approx ~3.14$ and (c) $approx ~3.27$ 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?
EDIT: after the very helpful answers, which certainly showed that there was a much satisfying and general way of approaching the original question, can I also ask about another version of the question which included (d) $sqrt[4]101$.
When approaching the question by approximation, I simply observed that $sqrt[4]101$ is only a tiny bit greater than $sqrt10$, and hence it still was clear to choose (c) as the answer. Is there any elegant way to extend the more robust methods to handle this case?
arithmetic radicals
arithmetic radicals
asked Sep 19 at 10:20
BBO555
10710
asked Sep 19 at 10:20
BBO555
10710
edited Sep 19 at 14:09
asked Sep 19 at 10:20
BBO555
10710
asked Sep 19 at 10:20
BBO555
10710
asked Sep 19 at 10:20
BBO555
10710
10710
2
+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
Sep 19 at 10:41
Thank you for all the comments and answers. I am pleased I chose to ask the question at MSE -- there was indeed something to learn here!
– BBO555
Sep 20 at 10:05
add a comment |Â
2
+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
Sep 19 at 10:41
Thank you for all the comments and answers. I am pleased I chose to ask the question at MSE -- there was indeed something to learn here!
– BBO555
Sep 20 at 10:05
2
2
+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
Sep 19 at 10:41
+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
Sep 19 at 10:41
Thank you for all the comments and answers. I am pleased I chose to ask the question at MSE -- there was indeed something to learn here!
– BBO555
Sep 20 at 10:05
Thank you for all the comments and answers. I am pleased I chose to ask the question at MSE -- there was indeed something to learn here!
– BBO555
Sep 20 at 10:05
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
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 $sqrtableqfraca+b2$, when $a, bgeq0$.
Hence,
$$
(sqrt2+sqrt3)^2=5+2sqrt2cdot3leq5+2frac2+32=5+5=10=(sqrt10)^2
$$
Therefore, using (1), we obtain $sqrt2+sqrt3leq 10$.
I forgot about this:
$$
5-sqrt3=3+2-sqrt3=3+frac12+sqrt3geq3+frac12+2=3.25
$$
One can easily verify that
$(3+1/4)^2>10.5>10$. One also finds that $10.5^2>110>101$.
Then, performing argument (1) twice, one finds that $5-sqrt3>(101)^1/4$.
Consequently, $5-sqrt3$ is the bigger number.
edited Sep 20 at 10:28
BBO555
10710
answered Sep 19 at 13:54
minimax
38817
3
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
Sep 19 at 14:01
Thanks for including a route to handling case (d) !
– BBO555
Sep 20 at 8:07
add a comment |Â
up vote
14
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 Sep 19 at 10:25
Lord Shark the Unknown
91.1k955117
2
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
Sep 19 at 10:30
Thank you for the hint!
– BBO555
Sep 19 at 10:32
2
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
Sep 19 at 10:47
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
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 $sqrtableqfraca+b2$, when $a, bgeq0$.
Hence,
$$
(sqrt2+sqrt3)^2=5+2sqrt2cdot3leq5+2frac2+32=5+5=10=(sqrt10)^2
$$
Therefore, using (1), we obtain $sqrt2+sqrt3leq 10$.
I forgot about this:
$$
5-sqrt3=3+2-sqrt3=3+frac12+sqrt3geq3+frac12+2=3.25
$$
One can easily verify that
$(3+1/4)^2>10.5>10$. One also finds that $10.5^2>110>101$.
Then, performing argument (1) twice, one finds that $5-sqrt3>(101)^1/4$.
Consequently, $5-sqrt3$ is the bigger number.
edited Sep 20 at 10:28
BBO555
10710
answered Sep 19 at 13:54
minimax
38817
3
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
Sep 19 at 14:01
Thanks for including a route to handling case (d) !
– BBO555
Sep 20 at 8:07
add a comment |Â
up vote
5
down vote
accepted
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 $sqrtableqfraca+b2$, when $a, bgeq0$.
Hence,
$$
(sqrt2+sqrt3)^2=5+2sqrt2cdot3leq5+2frac2+32=5+5=10=(sqrt10)^2
$$
Therefore, using (1), we obtain $sqrt2+sqrt3leq 10$.
I forgot about this:
$$
5-sqrt3=3+2-sqrt3=3+frac12+sqrt3geq3+frac12+2=3.25
$$
One can easily verify that
$(3+1/4)^2>10.5>10$. One also finds that $10.5^2>110>101$.
Then, performing argument (1) twice, one finds that $5-sqrt3>(101)^1/4$.
Consequently, $5-sqrt3$ is the bigger number.
edited Sep 20 at 10:28
BBO555
10710
answered Sep 19 at 13:54
minimax
38817
3
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
Sep 19 at 14:01
Thanks for including a route to handling case (d) !
– BBO555
Sep 20 at 8:07
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
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 $sqrtableqfraca+b2$, when $a, bgeq0$.
Hence,
$$
(sqrt2+sqrt3)^2=5+2sqrt2cdot3leq5+2frac2+32=5+5=10=(sqrt10)^2
$$
Therefore, using (1), we obtain $sqrt2+sqrt3leq 10$.
I forgot about this:
$$
5-sqrt3=3+2-sqrt3=3+frac12+sqrt3geq3+frac12+2=3.25
$$
One can easily verify that
$(3+1/4)^2>10.5>10$. One also finds that $10.5^2>110>101$.
Then, performing argument (1) twice, one finds that $5-sqrt3>(101)^1/4$.
Consequently, $5-sqrt3$ is the bigger number.
edited Sep 20 at 10:28
BBO555
10710
answered Sep 19 at 13:54
minimax
38817
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 $sqrtableqfraca+b2$, when $a, bgeq0$.
Hence,
$$
(sqrt2+sqrt3)^2=5+2sqrt2cdot3leq5+2frac2+32=5+5=10=(sqrt10)^2
$$
Therefore, using (1), we obtain $sqrt2+sqrt3leq 10$.
I forgot about this:
$$
5-sqrt3=3+2-sqrt3=3+frac12+sqrt3geq3+frac12+2=3.25
$$
One can easily verify that
$(3+1/4)^2>10.5>10$. One also finds that $10.5^2>110>101$.
Then, performing argument (1) twice, one finds that $5-sqrt3>(101)^1/4$.
Consequently, $5-sqrt3$ is the bigger number.
edited Sep 20 at 10:28
BBO555
10710
answered Sep 19 at 13:54
minimax
38817
edited Sep 20 at 10:28
BBO555
10710
edited Sep 20 at 10:28
BBO555
10710
edited Sep 20 at 10:28
BBO555
10710
10710
answered Sep 19 at 13:54
minimax
38817
answered Sep 19 at 13:54
minimax
38817
answered Sep 19 at 13:54
minimax
38817
38817
3
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
Sep 19 at 14:01
Thanks for including a route to handling case (d) !
– BBO555
Sep 20 at 8:07
add a comment |Â
3
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
Sep 19 at 14:01
Thanks for including a route to handling case (d) !
– BBO555
Sep 20 at 8:07
3
3
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
Sep 19 at 14:01
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
Sep 19 at 14:01
Thanks for including a route to handling case (d) !
– BBO555
Sep 20 at 8:07
Thanks for including a route to handling case (d) !
– BBO555
Sep 20 at 8:07
add a comment |Â
up vote
14
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 Sep 19 at 10:25
Lord Shark the Unknown
91.1k955117
2
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
Sep 19 at 10:30
Thank you for the hint!
– BBO555
Sep 19 at 10:32
2
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
Sep 19 at 10:47
add a comment |Â
up vote
14
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 Sep 19 at 10:25
Lord Shark the Unknown
91.1k955117
2
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
Sep 19 at 10:30
Thank you for the hint!
– BBO555
Sep 19 at 10:32
2
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
Sep 19 at 10:47
add a comment |Â
up vote
14
down vote
up vote
14
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 Sep 19 at 10:25
Lord Shark the Unknown
91.1k955117
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 Sep 19 at 10:25
Lord Shark the Unknown
91.1k955117
answered Sep 19 at 10:25
Lord Shark the Unknown
91.1k955117
answered Sep 19 at 10:25
Lord Shark the Unknown
91.1k955117
answered Sep 19 at 10:25
Lord Shark the Unknown
91.1k955117
91.1k955117
2
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
Sep 19 at 10:30
Thank you for the hint!
– BBO555
Sep 19 at 10:32
2
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
Sep 19 at 10:47
add a comment |Â
2
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
Sep 19 at 10:30
Thank you for the hint!
– BBO555
Sep 19 at 10:32
2
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
Sep 19 at 10:47
2
2
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
Sep 19 at 10:30
Ah .... yes of course ... $5=sqrt25>sqrt24=2sqrt6$
– BBO555
Sep 19 at 10:30
Thank you for the hint!
– BBO555
Sep 19 at 10:32
Thank you for the hint!
– BBO555
Sep 19 at 10:32
2
2
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
Sep 19 at 10:47
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
Sep 19 at 10:47
add a comment |Â
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2
+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
Sep 19 at 10:41
Thank you for all the comments and answers. I am pleased I chose to ask the question at MSE -- there was indeed something to learn here!
– BBO555
Sep 20 at 10:05