Pure Math / Logic / Quantifiers

dddx · **Joined:** Wed, 21 Sep 2011 18:36:35 UTC **Posts:** 29

Hi everyone,

i have an assignment and i am asked to write the following statements as English sentences and indicate whether they are true of false and to write the negation of each statement. Unfortunately i do not understand the statements here. Specially if someone could help me understand what does the commas (,) mean in these statements:

1) ∀x ∈ ℤ, ∃y ∈ ℚ, x < y.
2) ∃x ∈ ℤ, ∀y ∈ ℚ, x < y.
3) ∀x ∈ ℤ, x is odd ⇒ 4 | (x - X^3).

This is what i understand from them but i'm not sure if i'm reading it right:
1) For all integer x, there exists rational y, such that all x are greater than y ??
2) There exists an integer x, for all rational y such that x is greater than y ?? (--> I think this is false)
3) not sure

Any advise is welcome and appreciated.
Thanks in advance.
dddx

Shadow · **Posted:** Wed, 21 Sep 2011 19:48:32 UTC

dddx wrote:

Hi everyone,

i have an assignment and i am asked to write the following statements as English sentences and indicate whether they are true of false and to write the negation of each statement. Unfortunately i do not understand the statements here. Specially if someone could help me understand what does the commas (,) mean in these statements:

1) ∀x ∈ ℤ, ∃y ∈ ℚ, x < y.
2) ∃x ∈ ℤ, ∀y ∈ ℚ, x < y.
3) ∀x ∈ ℤ, x is odd ⇒ 4 | (x - x^3).

This is what i understand from them but i'm not sure if i'm reading it right:
1) For all integer x, there exists rational y, such that all x are greater than y ??
2) There exists an integer x, for all rational y such that x is greater than y ?? (--> I think this is false)
3) not sure

Any advise is welcome and appreciated.
Thanks in advance.
dddx

The commas are just like in normal English, they are there as a pause to separate different clauses.

For your first two check your last bit again, do they say "x is greater than y"?

For the second one you are right, it is false, as this would say that there is ONE integer which is SMALLER than ALL rationals, but the integers are rational, so this would say that there is a SMALLEST INTEGER, which is clearly false, you can always find a smaller integer than n by taking n-1.

Again, try reading the third one out, it says every odd integer has the property that four divides the difference of the integer and its cube.

dddx · **Joined:** Wed, 21 Sep 2011 18:36:35 UTC **Posts:** 29

Shadow wrote:

The commas are just like in normal English, they are there as a pause to separate different clauses.

For your first two check your last bit again, do they say "x is greater than y"?

For the second one you are right, it is false, as this would say that there is ONE integer which is SMALLER than ALL rationals, but the integers are rational, so this would say that there is a SMALLEST INTEGER, which is clearly false, you can always find a smaller integer than n by taking n-1.

Again, try reading the third one out, it says every odd integer has the property that four divides the difference of the integer and its cube.

Hi,

Thanks for your explanation. You are right. I misread the last part for the first two where the x is smaller than y

So just to confirm, the first one is true since the statement is possible. However, the second one is false since there is no "smallest" integer as what the statement says. We can always deduct 1 unit from it and get a smaller one.

As for the third one, i am still not clear about the truth of the statement, but i think this statement is false since not all odd numbers would fit in that property. for example if x is 5, then 4|(5-5^3)= -0.0333
does that make any sense?

For negation of the statements, if i am not mistaken ∀ becomes ∃ but would the comma or any other thing change?
for example:
Negation of #1 is:
1) ∃x ∈ ℤ, ∀y ∈ ℚ, x < y.

Negation of #2 is:
2) ∀x ∈ ℤ, ∃y ∈ ℚ, x < y.

I think that does not make any sense. does it?

Thanks

Shadow · **Posted:** Wed, 21 Sep 2011 21:13:28 UTC

dddx wrote:

Hi,

Thanks for your explanation. You are right. I misread the last part for the first two where the x is smaller than y

So just to confirm, the first one is true since the statement is possible. However, the second one is false since there is no "smallest" integer as what the statement says. We can always deduct 1 unit from it and get a smaller one.

As for the third one, i am still not clear about the truth of the statement, but i think this statement is false since not all odd numbers would fit in that property. for example if x is 5, then 4|(5-5^3)= -0.0333
does that make any sense?

For negation of the statements, if i am not mistaken ∀ becomes ∃ but would the comma or any other thing change?
for example:
Negation of #1 is:
1) ∃x ∈ ℤ, ∀y ∈ ℚ, x < y.

Negation of #2 is:
2) ∀x ∈ ℤ, ∃y ∈ ℚ, x < y.

I think that does not make any sense. does it?

Thanks

What?

is a logical statement, not a formula, you cannot have it equal something, in particular it's a statement about whether or not 4 divides a number, not the actual quotient of the two. What you want to say is that ${5-5^3\over 4}\not\in\mathbb{Z}$ , but that's false. . . (in fact the third statement IS true, and you can see that by taking an odd number, 2k+1

, and writing out (2k+1)-(2k+1)^3

, and factoring out a 4 from this.

The commas do not change, remember to negate a string of quantifiers you switch the foralls and exists, and negate the statement that follows them, for example $\forall x,\exists y : P$ has negation $\exists x,\forall y:(\lnot P)$ , and so in your case, you're almost right, but you need to negate x<y

by replacing it with $x\ge y$ .

dddx · **Joined:** Wed, 21 Sep 2011 18:36:35 UTC **Posts:** 29

Shadow wrote:

What?

is a logical statement, not a formula, you cannot have it equal something, in particular it's a statement about whether or not 4 divides a number, not the actual quotient of the two. What you want to say is that ${5-5^3\over 4}\not\in\mathbb{Z}$ , but that's false. . . (in fact the third statement IS true, and you can see that by taking an odd number, 2k+1

, and writing out (2k+1)-(2k+1)^3

, and factoring out a 4 from this.

The commas do not change, remember to negate a string of quantifiers you switch the foralls and exists, and negate the statement that follows them, for example $\forall x,\exists y : P$ has negation $\exists x,\forall y:(\lnot P)$ , and so in your case, you're almost right, but you need to negate x<y

by replacing it with $x\ge y$ .

Hi Shadow,

Thanks a lot for all the explanation. you have been a great help

For the third one, i did as you instructed by taking an odd integer k such that x=2k+1 but it turned out that the result is even:
4|(x-x^3) = 4|((2k+1)-(2k+1)^3)
=2(-4k^3 -6k^2 -2k)

which is even right?

Does this means that this statement is false or it means that mine is false

?

Also as for negation to this statement, will this be true:

∀x ∈ ℤ, x is odd ⇒ 4 | (x - x^3)

negation: ∃x ∈ ℤ, x is odd and not (4 | (x - x^3))

Many Thanks

Shadow · **Posted:** Thu, 22 Sep 2011 08:37:43 UTC

dddx wrote:

Shadow wrote:

What?

is a logical statement, not a formula, you cannot have it equal something, in particular it's a statement about whether or not 4 divides a number, not the actual quotient of the two. What you want to say is that ${5-5^3\over 4}\not\in\mathbb{Z}$ , but that's false. . . (in fact the third statement IS true, and you can see that by taking an odd number, 2k+1

, and writing out (2k+1)-(2k+1)^3

, and factoring out a 4 from this.

The commas do not change, remember to negate a string of quantifiers you switch the foralls and exists, and negate the statement that follows them, for example $\forall x,\exists y : P$ has negation $\exists x,\forall y:(\lnot P)$ , and so in your case, you're almost right, but you need to negate x<y

by replacing it with $x\ge y$ .

Hi Shadow,

Thanks a lot for all the explanation. you have been a great help

For the third one, i did as you instructed by taking an odd integer k such that x=2k+1 but it turned out that the result is even:
4|(x-x^3) = 4|((2k+1)-(2k+1)^3)
=2(-4k^3 -6k^2 -2k)

which is even right?

Does this means that this statement is false or it means that mine is false

?

Also as for negation to this statement, will this be true:

∀x ∈ ℤ, x is odd ⇒ 4 | (x - x^3)

negation: ∃x ∈ ℤ, x is odd and not (4 | (x - x^3))

Many Thanks

Almost there, but you only showed it was EVEN, you need it is divisible by 4, so factor out another 2 from the expression to get x-x^3=4(-2k^3-3k^2-k)

and that shows divisibility by 4.

Shadow · **Posted:** Thu, 22 Sep 2011 08:49:49 UTC

On another note, depending on what machine I access with I sometimes cannot see your quantifiers, which leads me to believe they're part of some font package. As a result, I recommend you learn how to use $\LaTeX$ when you ask for hep on this site, to ensure everyone can ready your questions so that they can help you.

For help on how to use this typesetting, visit our immensely handy message board on the topic where there are some primers, at this link.

dddx · **Joined:** Wed, 21 Sep 2011 18:36:35 UTC **Posts:** 29

Shadow wrote:

Almost there, but you only showed it was EVEN, you need it is divisible by 4, so factor out another 2 from the expression to get x-x^3=4(-2k^3-3k^2-k)

and that shows divisibility by 4.

You are right, i can factor another 2 to show the divisibility by 4

Shadow wrote:

On another note, depending on what machine I access with I sometimes cannot see your quantifiers, which leads me to believe they're part of some font package. As a result, I recommend you learn how to use $\LaTeX$ when you ask for hep on this site, to ensure everyone can ready your questions so that they can help you.

For help on how to use this typesetting, visit our immensely handy message board on the topic where there are some primers, at this link.

Thanks Shadow, yes i was using special fonts. I will read the $\LaTeX$ guides and make sure that next time i do it correctly.

By the way, thanks a lot for helping me out. You were really helpful. I will most definitely seek your assistance for my future assignments (if i can't do them myself).

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