Comparing the magnitudes of expressions of surds

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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?










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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















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?










share|cite|improve this question









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













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?










share|cite|improve this question









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






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edited 2 hours ago





















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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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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













  • 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











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?






share|cite|improve this answer




















  • 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


















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 $sqrt2+sqrt3leq 10$.








share|cite|improve this answer




















  • 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










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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?






share|cite|improve this answer




















  • 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















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?






share|cite|improve this answer




















  • 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













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?






share|cite|improve this answer












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?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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

















  • 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











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 $sqrt2+sqrt3leq 10$.








share|cite|improve this answer




















  • 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














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 $sqrt2+sqrt3leq 10$.








share|cite|improve this answer




















  • 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












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 $sqrt2+sqrt3leq 10$.








share|cite|improve this answer












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 $sqrt2+sqrt3leq 10$.









share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • 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










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