Question

What is the difference?

For every positive real number x there is a positive real number y with the property that if y 13 years ago

Answer 1

but it's not true.....

Answer 2

The first option is true and the second one is false

Answer 3

So there must be some difference that i am missing

Answer 4

The first can be written \[\forall x\in\mathbb{R}^+ \;\;\exists y\in\mathbb{R}^+ \;\;s.t. \;\;y<x\implies (\forall z\in\mathbb{R}^+ \;\; yz \geq z)\]The second is\[\forall x\in\mathbb{R}^+ \;\;\exists y\in\mathbb{R}^+ \;\;s.t. \;\;y<x \;\text{and}\; (\forall z\in\mathbb{R}^+ \;\; yz \geq z)\]The difference is in that implication.

Answer 5

so how wld that difference make one true and one false?

Answer 6

maybe if i use it

Answer 7

its contrapositive?

Answer 8

I'm not quite sure why that's different at first glance. Give me a couple more minutes to think about this.

Answer 9

okkk

Answer 10

I believe the key is that y=1 is the only option for y.

Answer 11

In the first way of writing it, there are two cases
Case 1: x>1. Here, we choose y=1. We have that \(y<x\), and \(yz\geq z\)So that's taken care of.
Case 2: \(x\leq1\). Here, we choose \(y\geq1\). Since this makes \(y<x\) false, it doesn't matter if \(yz\geq z\) or not, since the implication will always be true.

Answer 12

YAYYYYYYYYY

Answer 13

In the second way of writing it, we have the same cases. The only difference is in case 2.
Case 2: \(x\leq 1\). Here, we are forced to choose \(y<1\), and then we can choose a \(z\) such that \(yz>z\) which makes the statement false.

Answer 14

yup I forgot abt that one

Answer 15

THANNNKSSSSS :))))))))))

Answer 16

You're welcome.