What is the fastest method to find which of 3 3 − 4 7 − 2 3 and 3 3 − 8 1 − 2 3 is bigger manually?

Question

What is the fastest method to find which of             3              3            −      4              7      −      2              3             and             3              3            −      8              1      −      2              3             is bigger manually?

elladanzaez · Accepted Answer

I would multiply by the conjugate of denominator  a  =            3              3            −      4              7      −      2              3              =      1    37        (    13          3        −    10    )    b  =            3              3            −      8              1      −      2              3              =      1    11        (    13          3        −    10    )  this makes   a  &amp;lt;  b

uznosititr · Accepted Answer

Using the rule   a      b    =            a      2        b   for   a  ,  b  &amp;gt;  0 one checks the signs of numerator and denominator to be both positive for the first fraction and both negative for the second fraction. This is therefore a comparison of two positive numbers (so no verdict yet), and by flipping the signs of numerator and denominator on the right we can ensure that all individual factors are positive: compare             3              3            −      4              7      −      2              3             and             8      −      3              3                    2              3            −      1      . Now multiply by the (positive!) product of the denominators to compare   (  3      3    −  4  )  (  2      3    −  1  ) and   (  8  −  3      3    )  (  7  −  2      3    ) which amounts to comparing   22  −  11      3   and   74  −  37      3   or finally   26      3   and   52. Since by happy coincidence   26  ×  2  =  52 and       3    &amp;lt;  2 the number on the left is smaller.I&#039;m not sure whether many people could do this easily by mental arithmetic (I certainly did not). Each of the terms in the final comparison are obtained by adding   4 products from the terms in the original expression. One product, namely   (  2      3    )  (  3      3    ), contributes twice with opposite signs and could therefore have been cancelled beforehand (leaving   7  ×  8  −  4  ×  1  =  52 for that term), but that&#039;s all I can see for simplifications. Personally my main problem with mental computation is that I very often get the sign of one term wrong in my head, but maybe with a lot of training one can limit this kind of errors.

What is the fastest method to find which of 3 <msqrt> 3 </m

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