Quantifying a free variable in an example from “How To Prove It†by Velleman
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This is an example from How To Prove It by Daniel J. Velleman (2nd Ed., p. 71):
Example 2.2.3. Analyze the logical forms of the following statements.
Statements about the natural numbers. The universe of discourse is $mathbb N$.
a. $x$ is a perfect square.
And here is the solution:
- a. This means that $x$ is the square of some natural number, or in other words $exists y(x = y^2)$.
My question is regarding the free variable $x$. Is it correct to leave it free, as the author has done? I was under the impression that, by convention, a free variable may be assumed to be universally quantified, which I think wouldn't make sense in this case.
If I wanted to fully formalize the statement, how would I do that? My best guess is as follows:
$$exists xexists y(x = y^2)$$
Is this correct?
logic proof-explanation first-order-logic predicate-logic quantifiers
edited Aug 10 at 16:44
Taroccoesbrocco
3,68451431
asked Aug 10 at 16:17
Calculemus
346117
add a comment |Â
up vote
1
down vote
favorite
This is an example from How To Prove It by Daniel J. Velleman (2nd Ed., p. 71):
Example 2.2.3. Analyze the logical forms of the following statements.
Statements about the natural numbers. The universe of discourse is $mathbb N$.
a. $x$ is a perfect square.
And here is the solution:
- a. This means that $x$ is the square of some natural number, or in other words $exists y(x = y^2)$.
My question is regarding the free variable $x$. Is it correct to leave it free, as the author has done? I was under the impression that, by convention, a free variable may be assumed to be universally quantified, which I think wouldn't make sense in this case.
If I wanted to fully formalize the statement, how would I do that? My best guess is as follows:
$$exists xexists y(x = y^2)$$
Is this correct?
logic proof-explanation first-order-logic predicate-logic quantifiers
edited Aug 10 at 16:44
Taroccoesbrocco
3,68451431
asked Aug 10 at 16:17
Calculemus
346117
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is an example from How To Prove It by Daniel J. Velleman (2nd Ed., p. 71):
Example 2.2.3. Analyze the logical forms of the following statements.
Statements about the natural numbers. The universe of discourse is $mathbb N$.
a. $x$ is a perfect square.
And here is the solution:
- a. This means that $x$ is the square of some natural number, or in other words $exists y(x = y^2)$.
My question is regarding the free variable $x$. Is it correct to leave it free, as the author has done? I was under the impression that, by convention, a free variable may be assumed to be universally quantified, which I think wouldn't make sense in this case.
If I wanted to fully formalize the statement, how would I do that? My best guess is as follows:
$$exists xexists y(x = y^2)$$
Is this correct?
logic proof-explanation first-order-logic predicate-logic quantifiers
edited Aug 10 at 16:44
Taroccoesbrocco
3,68451431
asked Aug 10 at 16:17
Calculemus
346117
This is an example from How To Prove It by Daniel J. Velleman (2nd Ed., p. 71):
Example 2.2.3. Analyze the logical forms of the following statements.
Statements about the natural numbers. The universe of discourse is $mathbb N$.
a. $x$ is a perfect square.
And here is the solution:
- a. This means that $x$ is the square of some natural number, or in other words $exists y(x = y^2)$.
My question is regarding the free variable $x$. Is it correct to leave it free, as the author has done? I was under the impression that, by convention, a free variable may be assumed to be universally quantified, which I think wouldn't make sense in this case.
If I wanted to fully formalize the statement, how would I do that? My best guess is as follows:
$$exists xexists y(x = y^2)$$
Is this correct?
logic proof-explanation first-order-logic predicate-logic quantifiers
edited Aug 10 at 16:44
Taroccoesbrocco
3,68451431
asked Aug 10 at 16:17
Calculemus
346117
edited Aug 10 at 16:44
Taroccoesbrocco
3,68451431
edited Aug 10 at 16:44
Taroccoesbrocco
3,68451431
edited Aug 10 at 16:44
Taroccoesbrocco
3,68451431
3,68451431
asked Aug 10 at 16:17
Calculemus
346117
asked Aug 10 at 16:17
Calculemus
346117
asked Aug 10 at 16:17
Calculemus
346117
346117
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You shouldn't confuse variables (the only ones that can be quantified) with individual constants.
The logical form of the sentence "$x$ is a perfect square" is indeed $exists y (x = y^2)$. In this formula, $x$ plays the role of an individual constant, not of a variable, because it refers to one specific individual in the universe of discourse $mathbbN$.
The meaning of $exists x exists y (x = y^2)$ is "there is a perfect square", which is completely different because it does not refer to one specific individual in the universe of discourse $mathbbN$.
Remark. The starting point of my answer above is the fact that the exercise claims that "$x$ is a perfect square" is a statement, that is a a meaningful declarative sentence that is true or false because every sign in it is interpreted, hence $x$ should refer to a specific individual in the universe of discourse.
If you see "$x$ is a perfect square" just as a phrase which is not a sentence (and so it is not true or false because not all its signs are interpreted), then you can see $x$ as a free variable (waiting for an interpretation to assign a truth value to the phrase). Even in this case the logical (but meaningless) form of the phrase is $exists y (x = y^2)$.
answered Aug 10 at 16:31
Taroccoesbrocco
3,68451431
Thank you. Is there an easy way to distinguish between individual constants and free variables? Or do I need to figure it out from context?
– Calculemus
Aug 10 at 16:36
1
@Calculemus - In a formal language, variables and individual constants belong to two disjoint sets, so it is immediate to distinguish each other (but in this setting a formula is meaningless without a formal interpretation). In an informal setting, such as the one in your exercise, you need to figure it out from the context.
– Taroccoesbrocco
Aug 10 at 16:41
Thanks again :)
– Calculemus
Aug 10 at 16:43
1
I don't quite agree with this answer. The English language sentence "$x$ is a perfect square" and the mathematical notation sentence "$exists y(x=y^2)$" can be regarded as equivalent sentences with $x$ as a free variable in each of those sentences.
– Lee Mosher
Aug 10 at 16:55
Also, regarding the OP's statement that "a free variable may be assumed to be universally quantified", that's not always true. Open mathematical sentences, meaning mathematical sentences with unquantified free variables, are very common and useful things. For one, they can be used to define sets. For example, the subset $[0,infty)$ is the solution set of the open sentence "$x$ is a perfect square" when variables are restricted to the real numbers.
– Lee Mosher
Aug 10 at 16:59
 |Â
show 7 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You shouldn't confuse variables (the only ones that can be quantified) with individual constants.
The logical form of the sentence "$x$ is a perfect square" is indeed $exists y (x = y^2)$. In this formula, $x$ plays the role of an individual constant, not of a variable, because it refers to one specific individual in the universe of discourse $mathbbN$.
The meaning of $exists x exists y (x = y^2)$ is "there is a perfect square", which is completely different because it does not refer to one specific individual in the universe of discourse $mathbbN$.
Remark. The starting point of my answer above is the fact that the exercise claims that "$x$ is a perfect square" is a statement, that is a a meaningful declarative sentence that is true or false because every sign in it is interpreted, hence $x$ should refer to a specific individual in the universe of discourse.
If you see "$x$ is a perfect square" just as a phrase which is not a sentence (and so it is not true or false because not all its signs are interpreted), then you can see $x$ as a free variable (waiting for an interpretation to assign a truth value to the phrase). Even in this case the logical (but meaningless) form of the phrase is $exists y (x = y^2)$.
answered Aug 10 at 16:31
Taroccoesbrocco
3,68451431
Thank you. Is there an easy way to distinguish between individual constants and free variables? Or do I need to figure it out from context?
– Calculemus
Aug 10 at 16:36
1
@Calculemus - In a formal language, variables and individual constants belong to two disjoint sets, so it is immediate to distinguish each other (but in this setting a formula is meaningless without a formal interpretation). In an informal setting, such as the one in your exercise, you need to figure it out from the context.
– Taroccoesbrocco
Aug 10 at 16:41
Thanks again :)
– Calculemus
Aug 10 at 16:43
1
I don't quite agree with this answer. The English language sentence "$x$ is a perfect square" and the mathematical notation sentence "$exists y(x=y^2)$" can be regarded as equivalent sentences with $x$ as a free variable in each of those sentences.
– Lee Mosher
Aug 10 at 16:55
Also, regarding the OP's statement that "a free variable may be assumed to be universally quantified", that's not always true. Open mathematical sentences, meaning mathematical sentences with unquantified free variables, are very common and useful things. For one, they can be used to define sets. For example, the subset $[0,infty)$ is the solution set of the open sentence "$x$ is a perfect square" when variables are restricted to the real numbers.
– Lee Mosher
Aug 10 at 16:59
 |Â
show 7 more comments
up vote
1
down vote
accepted
You shouldn't confuse variables (the only ones that can be quantified) with individual constants.
The logical form of the sentence "$x$ is a perfect square" is indeed $exists y (x = y^2)$. In this formula, $x$ plays the role of an individual constant, not of a variable, because it refers to one specific individual in the universe of discourse $mathbbN$.
The meaning of $exists x exists y (x = y^2)$ is "there is a perfect square", which is completely different because it does not refer to one specific individual in the universe of discourse $mathbbN$.
Remark. The starting point of my answer above is the fact that the exercise claims that "$x$ is a perfect square" is a statement, that is a a meaningful declarative sentence that is true or false because every sign in it is interpreted, hence $x$ should refer to a specific individual in the universe of discourse.
If you see "$x$ is a perfect square" just as a phrase which is not a sentence (and so it is not true or false because not all its signs are interpreted), then you can see $x$ as a free variable (waiting for an interpretation to assign a truth value to the phrase). Even in this case the logical (but meaningless) form of the phrase is $exists y (x = y^2)$.
answered Aug 10 at 16:31
Taroccoesbrocco
3,68451431
Thank you. Is there an easy way to distinguish between individual constants and free variables? Or do I need to figure it out from context?
– Calculemus
Aug 10 at 16:36
1
@Calculemus - In a formal language, variables and individual constants belong to two disjoint sets, so it is immediate to distinguish each other (but in this setting a formula is meaningless without a formal interpretation). In an informal setting, such as the one in your exercise, you need to figure it out from the context.
– Taroccoesbrocco
Aug 10 at 16:41
Thanks again :)
– Calculemus
Aug 10 at 16:43
1
I don't quite agree with this answer. The English language sentence "$x$ is a perfect square" and the mathematical notation sentence "$exists y(x=y^2)$" can be regarded as equivalent sentences with $x$ as a free variable in each of those sentences.
– Lee Mosher
Aug 10 at 16:55
Also, regarding the OP's statement that "a free variable may be assumed to be universally quantified", that's not always true. Open mathematical sentences, meaning mathematical sentences with unquantified free variables, are very common and useful things. For one, they can be used to define sets. For example, the subset $[0,infty)$ is the solution set of the open sentence "$x$ is a perfect square" when variables are restricted to the real numbers.
– Lee Mosher
Aug 10 at 16:59
 |Â
show 7 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You shouldn't confuse variables (the only ones that can be quantified) with individual constants.
The logical form of the sentence "$x$ is a perfect square" is indeed $exists y (x = y^2)$. In this formula, $x$ plays the role of an individual constant, not of a variable, because it refers to one specific individual in the universe of discourse $mathbbN$.
The meaning of $exists x exists y (x = y^2)$ is "there is a perfect square", which is completely different because it does not refer to one specific individual in the universe of discourse $mathbbN$.
Remark. The starting point of my answer above is the fact that the exercise claims that "$x$ is a perfect square" is a statement, that is a a meaningful declarative sentence that is true or false because every sign in it is interpreted, hence $x$ should refer to a specific individual in the universe of discourse.
If you see "$x$ is a perfect square" just as a phrase which is not a sentence (and so it is not true or false because not all its signs are interpreted), then you can see $x$ as a free variable (waiting for an interpretation to assign a truth value to the phrase). Even in this case the logical (but meaningless) form of the phrase is $exists y (x = y^2)$.
answered Aug 10 at 16:31
Taroccoesbrocco
3,68451431
You shouldn't confuse variables (the only ones that can be quantified) with individual constants.
The logical form of the sentence "$x$ is a perfect square" is indeed $exists y (x = y^2)$. In this formula, $x$ plays the role of an individual constant, not of a variable, because it refers to one specific individual in the universe of discourse $mathbbN$.
The meaning of $exists x exists y (x = y^2)$ is "there is a perfect square", which is completely different because it does not refer to one specific individual in the universe of discourse $mathbbN$.
Remark. The starting point of my answer above is the fact that the exercise claims that "$x$ is a perfect square" is a statement, that is a a meaningful declarative sentence that is true or false because every sign in it is interpreted, hence $x$ should refer to a specific individual in the universe of discourse.
If you see "$x$ is a perfect square" just as a phrase which is not a sentence (and so it is not true or false because not all its signs are interpreted), then you can see $x$ as a free variable (waiting for an interpretation to assign a truth value to the phrase). Even in this case the logical (but meaningless) form of the phrase is $exists y (x = y^2)$.
answered Aug 10 at 16:31
Taroccoesbrocco
3,68451431
edited Aug 10 at 18:04
answered Aug 10 at 16:31
Taroccoesbrocco
3,68451431
answered Aug 10 at 16:31
Taroccoesbrocco
3,68451431
answered Aug 10 at 16:31
Taroccoesbrocco
3,68451431
3,68451431
Thank you. Is there an easy way to distinguish between individual constants and free variables? Or do I need to figure it out from context?
– Calculemus
Aug 10 at 16:36
1
@Calculemus - In a formal language, variables and individual constants belong to two disjoint sets, so it is immediate to distinguish each other (but in this setting a formula is meaningless without a formal interpretation). In an informal setting, such as the one in your exercise, you need to figure it out from the context.
– Taroccoesbrocco
Aug 10 at 16:41
Thanks again :)
– Calculemus
Aug 10 at 16:43
1
I don't quite agree with this answer. The English language sentence "$x$ is a perfect square" and the mathematical notation sentence "$exists y(x=y^2)$" can be regarded as equivalent sentences with $x$ as a free variable in each of those sentences.
– Lee Mosher
Aug 10 at 16:55
Also, regarding the OP's statement that "a free variable may be assumed to be universally quantified", that's not always true. Open mathematical sentences, meaning mathematical sentences with unquantified free variables, are very common and useful things. For one, they can be used to define sets. For example, the subset $[0,infty)$ is the solution set of the open sentence "$x$ is a perfect square" when variables are restricted to the real numbers.
– Lee Mosher
Aug 10 at 16:59
 |Â
show 7 more comments
Thank you. Is there an easy way to distinguish between individual constants and free variables? Or do I need to figure it out from context?
– Calculemus
Aug 10 at 16:36
1
@Calculemus - In a formal language, variables and individual constants belong to two disjoint sets, so it is immediate to distinguish each other (but in this setting a formula is meaningless without a formal interpretation). In an informal setting, such as the one in your exercise, you need to figure it out from the context.
– Taroccoesbrocco
Aug 10 at 16:41
Thanks again :)
– Calculemus
Aug 10 at 16:43
1
I don't quite agree with this answer. The English language sentence "$x$ is a perfect square" and the mathematical notation sentence "$exists y(x=y^2)$" can be regarded as equivalent sentences with $x$ as a free variable in each of those sentences.
– Lee Mosher
Aug 10 at 16:55
Also, regarding the OP's statement that "a free variable may be assumed to be universally quantified", that's not always true. Open mathematical sentences, meaning mathematical sentences with unquantified free variables, are very common and useful things. For one, they can be used to define sets. For example, the subset $[0,infty)$ is the solution set of the open sentence "$x$ is a perfect square" when variables are restricted to the real numbers.
– Lee Mosher
Aug 10 at 16:59
Thank you. Is there an easy way to distinguish between individual constants and free variables? Or do I need to figure it out from context?
– Calculemus
Aug 10 at 16:36
Thank you. Is there an easy way to distinguish between individual constants and free variables? Or do I need to figure it out from context?
– Calculemus
Aug 10 at 16:36
1
1
@Calculemus - In a formal language, variables and individual constants belong to two disjoint sets, so it is immediate to distinguish each other (but in this setting a formula is meaningless without a formal interpretation). In an informal setting, such as the one in your exercise, you need to figure it out from the context.
– Taroccoesbrocco
Aug 10 at 16:41
@Calculemus - In a formal language, variables and individual constants belong to two disjoint sets, so it is immediate to distinguish each other (but in this setting a formula is meaningless without a formal interpretation). In an informal setting, such as the one in your exercise, you need to figure it out from the context.
– Taroccoesbrocco
Aug 10 at 16:41
Thanks again :)
– Calculemus
Aug 10 at 16:43
Thanks again :)
– Calculemus
Aug 10 at 16:43
1
1
I don't quite agree with this answer. The English language sentence "$x$ is a perfect square" and the mathematical notation sentence "$exists y(x=y^2)$" can be regarded as equivalent sentences with $x$ as a free variable in each of those sentences.
– Lee Mosher
Aug 10 at 16:55
I don't quite agree with this answer. The English language sentence "$x$ is a perfect square" and the mathematical notation sentence "$exists y(x=y^2)$" can be regarded as equivalent sentences with $x$ as a free variable in each of those sentences.
– Lee Mosher
Aug 10 at 16:55
Also, regarding the OP's statement that "a free variable may be assumed to be universally quantified", that's not always true. Open mathematical sentences, meaning mathematical sentences with unquantified free variables, are very common and useful things. For one, they can be used to define sets. For example, the subset $[0,infty)$ is the solution set of the open sentence "$x$ is a perfect square" when variables are restricted to the real numbers.
– Lee Mosher
Aug 10 at 16:59
Also, regarding the OP's statement that "a free variable may be assumed to be universally quantified", that's not always true. Open mathematical sentences, meaning mathematical sentences with unquantified free variables, are very common and useful things. For one, they can be used to define sets. For example, the subset $[0,infty)$ is the solution set of the open sentence "$x$ is a perfect square" when variables are restricted to the real numbers.
– Lee Mosher
Aug 10 at 16:59
 |Â
show 7 more comments
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