MHB Some help with some problems about proofing quantifiers.

[zenakent]

Hello, Just want some help if my answers are correct.

1.) Determine the truth value for the following statement if the domain consist of the set of negative integers.

∀x(-2x>x)

Answer is True
Let x = -1, since the domain is in the negative integer and the product of two negative integers is positive. so
(-2(-1)>(-1)) = 2 > -1.
This is true for all value of x in the negative domain.

2.) Let Q(x,y) be the statement x - y = 1. Find the truth value of the statement where x and y are integers.

∃x∀y Q(x,y)
The answer is True
when solving for x we get x = 1 + y.
(1 + y) - y = 1.
1 + y - y = 1.
1 = 1.
This is true for every value of integer y.

 

[Evgeny.Makarov]

1.) Determine the truth value for the following statement if the domain consist of the set of negative integers.

∀x(-2x>x)

Answer is True
Let x = -1, since the domain is in the negative integer and the product of two negative integers is positive. so
(-2(-1)>(-1)) = 2 > -1.
This is true for all value of x in the negative domain.

Starting with "Let $x=-1$" when you are supposed to prove a universal statement (the one that starts with ∀) is only acceptable when you are helping a person who has difficulty grasping a general proof and needs an example. Starting this way is not possible in a homework or an exam answer. A proof must cover all cases: "Let $x$ be an arbitrary negative integer. Then $-2x$ is positive and is therefore greater than $x$".

2.) Let Q(x,y) be the statement x - y = 1. Find the truth value of the statement where x and y are integers.

∃x∀y Q(x,y)
The answer is True
when solving for x we get x = 1 + y.
(1 + y) - y = 1.
1 + y - y = 1.
1 = 1.
This is true for every value of integer y.

You took the definition of $Q(x,y)$ and then substituted it in the same definition again. No wonder you got something that is always true.

The question is, does there exist an $x$ which is greater by 1 than every integer $y$? Can you name such an $x$? Note the order of quantifiers: there supposed to be a single $x$ that works for all $y$. If the formula started with ∀y∃x, then it asks to find its own $x$ for every $y$.

 

[zenakent]

1.)

since integer x is in the negative domain, every time x is multiplied to -2x the result will always be greater than x and -2x will always be positive.

that should be it yes?
no need to give an example since every x is a negative and -2x will always be positive.2.)

I don't think I could name such an x for every y. so therefore, the truth value is false.

 

[Evgeny.Makarov]

Yes, this is correct.

 

[zenakent]

Thank you so much for your help. (Smile)