2. Express the following as ‘for all’ assertions (using symbols and words):

(a) The equation $x^3 = 28$ does not have a natural number solution.
(b) $0$ is less than every natural number.
(c) The natural number n is a prime.

Question

2. Express the following as ‘for all’ assertions (using symbols and words):

(a) The equation  $x^3 = 28$  does not have a natural number solution. 
(b) $0$  is less than every natural number.
(c) The natural number n is a prime.

unklerhaukus · Answer

my answers so far

(a) $$(\forall x\in \mathbb N)(x^3\neq28)$$

(b) $$(\forall x\in\mathbb N)(0<x)$$

(c) $$(\forall n\in \mathbb N)\left((\neg\exists p,q \in \mathbb N)\wedge (pq= n)\right)$$

ganeshie8 · Answer

last one looks incomplete, i think its not covering (p,q) = (1,n) case

unklerhaukus · Answer

(c) $$(\forall n\in \mathbb N)\left[((\exists p,q \in \mathbb N)\wedge (pq= n ))\Rightarrow (p=n\vee q=n)\right]$$

ganeshie8 · Answer

$(\forall (n,p,q)\in \mathbb N)\left[(pq= n )\Rightarrow (p=n\vee q=n)\right] $

ganeshie8 · Answer

does it look correct

unklerhaukus · Answer

your statement looks stronger than mine

unklerhaukus · Answer

$$\large(\forall n,p,q\in \mathbb N)\left[pq= n \Rightarrow p=n\vee q=n\right]$$

ganeshie8 · Answer

lol i have no idea if its correct, i just manipulated ur symbols

ganeshie8 · Answer

i mean the usage of for all, can we pass multiple arguments... ?

unklerhaukus · Answer

i am not certain

ganeshie8 · Answer

$\large(\forall n,p\in \mathbb N)\left[p | n \Rightarrow p\in \{1,n\}\right] $

ganeshie8 · Answer

$\large(\forall n,p\in \mathbb N)\left[p | n  \Rightarrow p =1 \vee p=n\right] $

unklerhaukus · Answer

what happened to q /

experimentx · Answer

for all n belongs to natural number, there exist a natural number less than n and not equal to 1 such that n/p does not belong to the set of natural numbers.

unklerhaukus · Answer

?

experimentx · Answer

also try separating the latter part with | or :
$$ (\forall x\in \mathbb N)|(x^3\neq28) \
(\forall x\in \mathbb N):(x^3\neq28)$$

experimentx · Answer

$$ \exists n \in \mathbb N |(\exists p \in\mathbb N )\wedge ( 1 <p < n )\wedge (n/p \cancel{\in} \mathbb N)  $$

unklerhaukus · Answer

im looking for a  ‘for all’  assertion

experimentx · Answer

also if you are looking for just particular prime ... i think there shouldn't be "there exists" before N.

experimentx · Answer

*n

anonymous · Answer

for all in this case would mean $\forall  m\in \mathbb{N}, m\not| n$

anonymous · Answer

i guess you would have to say also $m\neq 1$