I don't have any experience with kids, so I have no idea if this would just make things more confusing. But you could try taking advantage of common denominators:
Assemble two piles that each contain 15 identical somethings (paper squares?). Now we can talk about coloring 2/3 the squares black in the first pile, and coloring 3/5 of the squares black in the second pile. To do this in a intuitive way, explain that this means, in the first pile, "two out of every three squares are shaded". So to demonstrate visually, count out three squares at a time from the first pile, and for each three counted out, color two of them. Do this until the entire pile has been accounted for.
Then move on to the second pile. Again, "three of every five squares are shaded", so count out five squares at a time, and for every five counted out, color 3 of them. As before, do this until the entire pile has been accounted for.
Now reiterate that 2/3 of the squares in the first pile were shaded black, and likewise for 3/5 of the squares in the second pile. Lastly, actually count the total number of black squares in each, and of course the 2/3 pile will have the most.
Somewhat unfortunately, the only reason this works so well is precisely because 15 is a number that is divisible by both 3 and 5. If she tries to investigate, say, 6/11 and 5/9 using a similar method (but an "incorrect" number of pieces), it won't work out as nicely. So you'll want to look at other answers or meditate further on how to convey the ultimate idea that x/y is answering, in some sense, "how many parts of a whole?", and why that makes "less than" and "greater than" comparisons trickier than with the integers.
One way of going about this would be to explain that the bottom number of a fraction isn't actually counting anything at all. Instead, it's indicating how many equal pieces the "whole" has been broken up into. Only the top number is counting something (how many pieces of that size). It's tougher comparing x/y and w/z given that the "whole" has been broken into pieces of different sizes depending on the denominator. Perhaps this can be demonstrated with a traditional "cut the apple" / "cut the pie" approach. Show, for example, that
manpreet
Best Answer
2 years ago
I was really upset while I was trying to explain for my daughter that 2323 is greater than 3535 and she always claimed that (3(3 is greater than 22 and 55 is greater than 3)3) then 3535 must be greater than 2323.
At this stage she can't calculate the decimal so that she can't realize that (23=0.66(23=0.66 and 35=0.6).35=0.6).
She is 88 years old.