Doubt interpreting meaning of $\forall$ and number belonging to set

Question

$\forall b \space (b>1\implies b>5)$

is false since $b$ can be $2$ and $2 < 5$

It is correct?

Now,

if $b \in (1, \infty) \space$ $\implies$ $\space b > 5$

Is the second statement true or false or neither?

“If $\forall b$“ is the wrong order. It should be “$\forall b, $ if…” But, as a rule, you shouldn’t mix language with the symbols $\forall$ or $\exists.$ Here, you’d just write:$$\forall b(b>1\implies b>5)$$ Or write it in English: “For all $b,$ if $b>1$ then $b>5.$” — Thomas Andrews, Jan 20 '22 at 05:02
$\forall b \space (b>1\implies b>5)\tag1$
Statement $(1)$ is read as “Every $b$ exceeding $1$ exceeds $5$.”

$\text{if};b \in (1, \infty) \implies b > 5\tag2$

The word “if” needs to be omitted from the open formula $(2),$ after which it will in most contexts be understood as if it's statement $(1),$ even though it is missing the quantifier $∀b.$ — ryang, Jan 20 '22 at 14:34

Vince Vickler · Accepted Answer · 2022-01-22T18:56:32.583

1

Let $P_1$ and $P_2$ be two properties.

A statement of the form :

$$\forall (b) \Big[ P_1 (b) \implies P_2(b)\Big]$$

is equivalent to : there is no $b$ such that $b$ has the property $P_1$ and $b$ does not have the property $P_2$.

Note.- The literal meaning of an expression of this form is : for all $b$, if $b$ has property $P_1$, then, $b$ ( also) has property $P_2$.

Do you think there is no $b$ that

In case you can exhibit any number that has property $1$ but does not have property $2$, you've proved that the sentence is false.

edited Jan 22 '22 at 18:56

answered Jan 22 '22 at 11:55

Vince Vickler

1

2 has both properties, hence, the sentence is false. – Carlitos_30 Jan 22 '22 at 15:32
1

Or rather $ 2$ has the first property but does not have the second. Hence , the sentence is false. – Vince Vickler Jan 22 '22 at 18:17

1 Answers1