My Math Forum > Math Which side is greater? my challenge #1
 User Name Remember Me? Password

 Math General Math Forum - For general math related discussion and news

 June 19th, 2014, 10:02 PM #1 Banned Camp   Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191 Which side is greater? my challenge #1 Determine which side has the larger value: $\displaystyle \sqrt{2 \ } \ + \ \sqrt{7 \ } \ + \ \sqrt{287 \ } \ \ \ versus \ \ \ 21$ You are not allowed to do it by estimating roots. You may square each side when appropriate. You can subtract or add radicals and/or non-radicals from each side as the case may be. When you are done with the process, you should have only an integer on each side of the word "versus" so you can answer the question.
 June 20th, 2014, 03:26 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2201 21 = 1.414 + 2.645 + 16.941 = √1.999396 + √6.996025 + √286.997481 < √2 + √7 + √287
June 20th, 2014, 07:08 AM   #3
Banned Camp

Joined: Jun 2014
From: Earth

Posts: 945
Thanks: 191

Quote:
 Originally Posted by skipjack 21 = 1.414 + 2.645 + 16.941 = √1.999396 + √6.996025 + √286.997481 < √2 + √7 + √287
skipjack, your solution does not count (in my case), because I stated
that I won't allow it to be done by estimating roots.

This includes the estimates of the roots. My solution does not use any decimals.

The problem is still open.

Here is a simpler example of how it works:

(√3 + √5) versus 4

(√3 + √5)^2 versus (4)^2

3 + 2√(15) + 5 versus 16

8 + 2√(15) versus 16

2√(15) versus 8

√(15) versus 4

[√(15)]^2 versus (4)^2

15 < 16

So √3 + √5 < 4.

Or, to answer the question, the right side is greater.

Last edited by Math Message Board tutor; June 20th, 2014 at 07:35 AM.

 June 20th, 2014, 07:20 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms The minimal polynomial of $\sqrt2+\sqrt7+\sqrt{287}$ is $P(x)=x^8 - 1184x^6 + 504920x^4 - 91695104x^2 + 5964163984$ so presumably we can compare $P(\sqrt2+\sqrt7+\sqrt{287})=0$ to $P(21)=-258263$ to conclude that $\sqrt2+\sqrt7+\sqrt{287}>21.$ With more patience you could do this in the step-by-step way you ask for, I think.
 June 20th, 2014, 07:55 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra \begin{align*} \sqrt{2} + \sqrt{7} + \sqrt{287} \qquad &\text{versus} \qquad 21 \\ 2 + 2\sqrt{14} + 2\sqrt{574} + 7 + 2\sqrt{2009} + 287 \qquad &\text{versus} \qquad 441 \\ 2 + 2\sqrt{14} + 2\sqrt{574} + 7 + 2\sqrt{2009} + 287 &= 2 + \sqrt{56} + \sqrt{2296} + 7 + \sqrt{8036} + 287 \\ &\gt 2+\sqrt{49}+\sqrt{2209}+7+\sqrt{7921}+287 = 504 \\ &\gt 504 \end{align*} Thanks from CRGreathouse
 June 20th, 2014, 08:19 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Of course, I had to estimate the roots to get to 2209 and 7921. The only difference was that I didn't include any decimal places. Last edited by skipjack; June 20th, 2014 at 11:58 PM.
June 20th, 2014, 11:07 PM   #7
Global Moderator

Joined: Dec 2006

Posts: 20,919
Thanks: 2201

That's incorrect, as $2+\sqrt{49}+\sqrt{2209}+7+\sqrt{7921}+287 = 2 + 7 + 47 + 7 + 89 + 287 = 439$, not $504$.

Quote:
 Originally Posted by Math Message Board tutor This includes the estimates of the roots. My solution does not use any decimals.
You hadn't prohibited decimals and they can trivially be avoided by writing
21000 = 1414 + 2645 + 16941 = √1999396 + √6996025 + √286997481 < 1000√2 + 1000√7 + 1000√287,
which implies the desired inequality.

That uses knowledge of some large perfect squares, but doesn't explicitly use estimation of square roots.

The kind of approach you wanted is quite easy, but eventually involves large squares.

One compares (√7 + √287)² to (21 - √2)², i.e., 7(42 + 2√41) to 443 - 42√2.
It suffices now to compare 14√41 to 149 - 42√2, and squaring again means
it suffices to compare 8036 to 25729 - 12516√2.
After rearranging and squaring again, the comparison becomes
2(12516²) versus 17693², i.e., 313300512 versus 313042249,
and the left-hand side is greater, as required.

Quote:
 Originally Posted by CRGreathouse . . . presumably we can compare $P(\sqrt2+\sqrt7+\sqrt{287})=0$ to $P(21)=-258263$ to conclude that $\sqrt2+\sqrt7+\sqrt{287}>21.$
Not quite. You need to know a little more about the polynomial's zeros to justify that argument.

Last edited by skipjack; June 21st, 2014 at 12:14 AM.

June 22nd, 2014, 08:01 AM   #8
Global Moderator

Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Quote:
 Originally Posted by skipjack Not quite. You need to know a little more about the polynomial's zeros to justify that argument.
Agreed. I just didn't feel like working through the details that there were no zeros between the two points. You can do it by studying the derivative (and its derivative?).

 Tags challenge, greater, side

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post noman.rasheed Algebra 2 February 25th, 2014 05:31 AM tellios Algebra 2 January 21st, 2014 07:55 PM eugen123 Algebra 3 January 31st, 2012 06:23 PM sivela Calculus 1 January 18th, 2011 08:19 PM arun Algebra 9 December 24th, 2010 02:11 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top