Question

when disproving universal quantifiers, a counterexample is stated. When proving existential quantifier, one example that suits the expression is all that's needed. What about in proving universal quantifiers?

Answer 1

\[\forall x f(x)\]
Where f(x) is a truth function (could either be true or false)
To prove it, get an arbitrary x, and show that f(x) is true.

Answer 2

but even if you get arbitrary x, it does not mean it is applicable for all x

Answer 3

Yes it does...

Answer 4

for axample what's given is \[Q(x) = x < 6\]

Answer 5

When, proving a universal quantifier A is true for all B, you could disprove the converse statement A is untrue for some B.

Answer 6

even if you say \[\text{At} \; Q(4): \\ 4 < 6 \\ \therefore \forall x Q(x) \equiv T\]

Answer 7

this cannot be said...

Answer 8

because at Q(7)

7 < 4

\(\forall\)x Q(x) \(\equiv\) F

Answer 9

The statement is false. You can produce a counterexample.

Answer 10

that's what i'm sayin

Answer 11

saying*

Answer 12

And you did not take an arbitrary value for x, you took a specific value.

Answer 13

by arbitrary you mean..?

Answer 14

I think it best to illustrate via an example:
To prove that
For all integers, if it is odd, then its square is odd.
\[\forall x \in Z, x \ is \ odd \rightarrow x^{2} \ is \ odd\]
To prove it
Let x be an arbitrary integer (THIS IS WHAT I MEAN)
Suppose x is odd.
Then x = 2k + 1 for some integer k.
Then x^2 = (2k + 1)^2
= 4k^2 + 4k + 1
= 2(2k^2 + 2k) + 1
But 2k^2 + 2k is an integer, therefore
x^2 is odd.
QED

Answer 15

that's already proving....

Answer 16

i mean "proving" proving

Answer 17

I know, and what did, that's what I mean by taking an arbitrary value, you're not supposed to assign a value to it, as you did by assigning x = 4 and x = 7

Answer 18

but what i did is what i meant

Answer 19

that was what i was asking

Answer 20

I thought you were asking how to prove universal quantifiers...

Answer 21

i was asking how to prove universal quantifiers using *numbers*

Answer 22

Well, that, I can't say for sure, unless you can show me an example of what to prove...

Answer 23

Let P(x) = "x+ 1 > x". Find the truth value of \(\forall\)x P(x) if the domain consists of all real numbers

Answer 24

Hmm... I'm afraid I can only prove this using the method shown above... As you said, examples ain't gonna do it for proving universal quantifiers...

Answer 25

i supose there's no other way then

Answer 26

Let x be a real number
1 > 0
Since x is a real number, by the addition property of inequality,
x + 1 > 0 + x
x + 1 > x

It's possible to prove 1 > 0, but I take it your instructor is not quite that sadistic...

Answer 27

instructor? i have no instructor

Answer 28

Then replace "your instructor is not quite that sadistic..." with "you're not quite that masochistic"