quantifier problem

**armeros** · June 13th 2009, 02:29 AM

$\exists t \geq 0 \forall s \geq 0,$ we have $s \geq t$

$t = 0$ works.

$\exists s \geq 0 \forall t \geq 0,$ we have $s \geq t$

Why does $s = t+1$ not work? The book said
$\exists s \geq 0 \forall t \geq 0,$ we have $s \geq t$ is false since its negation is true. But I would like to know why could s = t+1 not be used?

Thanks

**Plato** · June 13th 2009, 04:00 AM

Originally Posted by armeros

[tex]
$\exists s \geq 0 \forall t \geq 0,$ we have $s \geq t$
Why does $s = t+1$ not work? The book said
$\exists s \geq 0 \forall t \geq 0,$ we have $s \geq t$ is false since its negation is true. But I would like to know why could s = t+1 not be used?

The answer may appear to be too subtle: it is a matter of order.
Let’s translate the sentence into standard English.
It would be: “Some non-negative number is greater than or equal to every non-negative number.”
Surely put that way the expression is clearly false.

Do you see why order makes a different?
In a word, you are saying that ‘s is fixed’, it comes first, and it has a certain relation to every other t.

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