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Mathematics 7 Online
OpenStudy (swissgirl):

For all x and Y in Q: if x

OpenStudy (swissgirl):

i would say this is true but i bet i am wrong lol

OpenStudy (swissgirl):

ull find some fraction that will make it false somehow

OpenStudy (anonymous):

right

OpenStudy (swissgirl):

Its correct???? YAYYYYYYY

OpenStudy (anonymous):

yes

OpenStudy (swissgirl):

Awesome :D

OpenStudy (kinggeorge):

If you want to prove it, you would select two reduced fractions \(x=a/b\) and \(y=c/d\) such that \(x<y\). Then, look at the numbers \[\frac{ad}{bd}\]\[\frac{bc}{bd}\]We know that \(ad<bc\) and they are integers by our assumption. If \(bc\neq ad+1\) then \[\frac{ad}{bd}<\frac{ad+1}{bd}<\frac{bc}{bd}\]Otherwise multiply x and y by \(\frac{2}{2}\), and choose \[\frac{2ad+1}{2bd}\]

OpenStudy (swissgirl):

thaankkksss :DDDDDDDDDDDDDDDDD

OpenStudy (anonymous):

@swissgirl u have a good sense of logic and reasoning;)

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