How do you order the following from least to greatest #sqrt14, sqrt(49/36), 3 3/5, 3#?

2 Answers
Aug 4, 2016

Answer:

#sqrt(49/36), 3, 3 3/5, sqrt14#

Explanation:

The first problem is that the values are all in different formats.
Comparing decimals is the easiest to do. Change them all to decimals.

The obvious and easiest method is to just use a calculator, but that's a cop-out and only involves pushing buttons. Let's think through the question instead....

#sqrt 14 " lies between " sqrt 9 and sqrt 16, and "is 3....."#
We need to calculate this one. #sqrt14 = 3.742#

#sqrt(49/36) = 1 13/36#

#3 3/5 = 3.6#

Now we can arrange them in order using the original formats.

#sqrt(49/36), 3, 3 3/5, sqrt14#

Aug 4, 2016

Answer:

#color(red)("Using a 'trick'")#
The order from smallest to largest is: #sqrt(49/36);" " 3;" "3 3/5;" "sqrt(14)#

Explanation:

#color(blue)("A bit of a cheat method")#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(magenta)("I have just realised that there is a slight flaw in my logic")#
As it happens the values are such that the answer is till correct but this is more by luck!

The problem is that #sqrt(49/36)>49/36#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Original values:#" "sqrt(14),sqrt(49/36),3 3/5,3#
#color(white)(.)#

#color(green)("There is a trap in the logic I previously applied. Which was:")#
#color(brown)("If you magnify the values by some consistent operation you")# #color(brown)("increase the difference but do not change the order.")#

#color(magenta)("I squared everything and then compared. Wrong!!!!")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Solution re-worked")#

Consider #sqrt(49/36) = (sqrt(49))/(sqrt(36)) = 7/6# giving

#sqrt(14)," "7/6," "3 3/5," "3#

#color(magenta)("This is a much better method!")#

Magnify the difference by multiplying by, say 100, and round the numbers to the nearest whole number for comparison.

Giving:

#sqrt(14),color(white)(.)" "7/6,color(white)(.)" "3 3/5,color(white)(...)" "3#

#color(white)(.)374,color(white)(.)" "116,color(white)(.)" "360,color(white)(.)" "300 color(brown)(larr" magnified approximations")#

#color(white)(.)uarr" "color(white)(.) uarr" " color(white)(.) uarr" "color(white)(.) uarr#
#color(white)(..)4" "1" "3" "2 color(brown)(" "larr" ranking: lowest rank is 1")#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Remember that #7/6-=sqrt(49/36)#
So the order lowest to highest is:

#sqrt(49/36);" " 3;" "3 3/5;" "sqrt(14)#