Question

Prove:
For every positive real number x, there is a positive real number y less than x, with the property that for all positive real numbers z, yz is greater or equal to z

Answer 1

there is no \(x\) in your conclusion

Answer 2

There doesnt have to be

Answer 3

ooh i see

Answer 4

It doesnt have to be true. you can provide a counterexample

Answer 5

it looks false to me

Answer 6

take the case of x=1

Answer 7

then y<1

Answer 8

and yz < z

Answer 9

ok i see

Answer 10

i think this is not correct

Answer 11

\[yz\geq z\] and \(y, z>0\implies y\geq 1\)

Answer 12

and it is certainly not the case that given ANY \(x\) there is a \(y<x\) where \(y\geq 1\)

Answer 13

for example \(x=\frac{1}{2}\)

Answer 14

ok if i was paying attention i would have seen @asnaseer wrote this already

Answer 15

:) thats alright @satellite73 - your proof is much neater! :)

Answer 16

must be miller time

Answer 17

hahahah THANKS GUYS :))))))

Answer 18

yw :)