<p>How to compare 7/16 and 3/7 without a calculator?
Do we still have this kind of problem on the SAT</p>
<p>The trick I like to use is “cross-multiplying.” You set up the two fractions next to each other, multiply at diagonals. The top number represents that fraction. </p>
<p>ie you do 7 * 7 and 3 * 16. 49 > 48, so you take the one with 7 on top. 7/16 is greater.</p>
<p>I don’t think this comes up often though</p>
<p>It is easier with calculator ;} Devide 7 by 16 and 3 by 7 and see the result</p>
<p>ya math is with a calc, just use it.</p>
<p>Without a calculator, I specified it already
@steamed rice: I don’t understand much about your cross multiplying. Why do you take one with 7 on top?</p>
<p>Basically, you start from the bottom of the fraction and work diagonally upwards. If comparing (mind the poor division problem representation):
X@@@@Z
–@vs.@–
Y@@@@W
use:
WX vs. YZ</p>
<p>the product of WX corresponds to a the (X/Y) half of the ratio and the product of YZ corresponds to the (Z/W) half</p>
<p>so if WX > YZ, (X/Y) > (Z/W) and vice-versa</p>
<p>Why do we go by the head (numerator) but not by the tail (denominator)? And please prove it how if you can 'cause it doesn’t add up just saying multiply just that…</p>
<p>I’m not sure why a proof is necessary, but if you really need one…</p>
<p>Work backwards from WX > YZ. Divide through by Y and by X. You’ll get (W/Y) > (Z/X). Note how the numbers on top, W and Z, are on the same side of the inequality as they are in the first expression. If you divided by, say, Z and X instead. You’ll get (W/Z) > (Y/X). Again, the numerators are on the same side of the inequality as they are in the original expression.</p>
<p>Long story short: that’s how the algebra works out.</p>
<p>great post steamed rice</p>
<p>Another way to think about it: Take (W/X) > (Y/Z) and just multiply both sides by XZ. That’s really all you’re doing; the X cancels out on one side and the Z cancels on the other.</p>