Example of a proposition that is true when its quantifier ranges over integers, but false over rationals
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I was asked to give an example of a proposition with a quantifier which is true if the quantifier ranges over the integers, but false if it ranges over the rational numbers.
My attempt:
$$(exists n in mathbb{Z}, 2n =n^2)$$
Is this the correct approach?
calculus proof-verification proof-explanation
edited Jan 10 at 1:08
Blue
48.6k870156
asked Jan 9 at 19:11
ForextraderForextrader
808
$endgroup$
|
show 1 more comment
$begingroup$
I was asked to give an example of a proposition with a quantifier which is true if the quantifier ranges over the integers, but false if it ranges over the rational numbers.
My attempt:
$$(exists n in mathbb{Z}, 2n =n^2)$$
Is this the correct approach?
calculus proof-verification proof-explanation
edited Jan 10 at 1:08
Blue
48.6k870156
asked Jan 9 at 19:11
ForextraderForextrader
808
$endgroup$
2
$begingroup$
If $(exists x in mathbb{Z}) P(x)$ is true, then so is $(exists x in mathbb{Q}) P(x)$, as the same $x$ works for both.
$endgroup$
– Connor Harris
Jan 9 at 19:13
1
$begingroup$
In general, you'll want to look for propositions with universal rather than existential quantifiers.
$endgroup$
– Connor Harris
Jan 9 at 19:16
$begingroup$
This is my first mathematical reasoning class. Totally new to me. I'll try and work off of what you said to understand it a little more.
$endgroup$
– Forextrader
Jan 9 at 19:18
$begingroup$
Ok what if I wrote then $(forall n in mathbb{Z}, n^2geq n)$ is that an appropriate answer to the question?
$endgroup$
– Forextrader
Jan 9 at 20:29
$begingroup$
That works................
$endgroup$
– DanielWainfleet
Jan 10 at 3:32
|
show 1 more comment
$begingroup$
I was asked to give an example of a proposition with a quantifier which is true if the quantifier ranges over the integers, but false if it ranges over the rational numbers.
My attempt:
$$(exists n in mathbb{Z}, 2n =n^2)$$
Is this the correct approach?
calculus proof-verification proof-explanation
edited Jan 10 at 1:08
Blue
48.6k870156
asked Jan 9 at 19:11
ForextraderForextrader
808
$endgroup$
I was asked to give an example of a proposition with a quantifier which is true if the quantifier ranges over the integers, but false if it ranges over the rational numbers.
My attempt:
$$(exists n in mathbb{Z}, 2n =n^2)$$
Is this the correct approach?
calculus proof-verification proof-explanation
calculus proof-verification proof-explanation
edited Jan 10 at 1:08
Blue
48.6k870156
asked Jan 9 at 19:11
ForextraderForextrader
808
edited Jan 10 at 1:08
Blue
48.6k870156
asked Jan 9 at 19:11
ForextraderForextrader
808
edited Jan 10 at 1:08
Blue
48.6k870156
edited Jan 10 at 1:08
Blue
48.6k870156
edited Jan 10 at 1:08
Blue
48.6k870156
48.6k870156
asked Jan 9 at 19:11
ForextraderForextrader
808
asked Jan 9 at 19:11
ForextraderForextrader
808
asked Jan 9 at 19:11
ForextraderForextrader
808
808
2
$begingroup$
If $(exists x in mathbb{Z}) P(x)$ is true, then so is $(exists x in mathbb{Q}) P(x)$, as the same $x$ works for both.
$endgroup$
– Connor Harris
Jan 9 at 19:13
1
$begingroup$
In general, you'll want to look for propositions with universal rather than existential quantifiers.
$endgroup$
– Connor Harris
Jan 9 at 19:16
$begingroup$
This is my first mathematical reasoning class. Totally new to me. I'll try and work off of what you said to understand it a little more.
$endgroup$
– Forextrader
Jan 9 at 19:18
$begingroup$
Ok what if I wrote then $(forall n in mathbb{Z}, n^2geq n)$ is that an appropriate answer to the question?
$endgroup$
– Forextrader
Jan 9 at 20:29
$begingroup$
That works................
$endgroup$
– DanielWainfleet
Jan 10 at 3:32
|
show 1 more comment
2
$begingroup$
If $(exists x in mathbb{Z}) P(x)$ is true, then so is $(exists x in mathbb{Q}) P(x)$, as the same $x$ works for both.
$endgroup$
– Connor Harris
Jan 9 at 19:13
1
$begingroup$
In general, you'll want to look for propositions with universal rather than existential quantifiers.
$endgroup$
– Connor Harris
Jan 9 at 19:16
$begingroup$
This is my first mathematical reasoning class. Totally new to me. I'll try and work off of what you said to understand it a little more.
$endgroup$
– Forextrader
Jan 9 at 19:18
$begingroup$
Ok what if I wrote then $(forall n in mathbb{Z}, n^2geq n)$ is that an appropriate answer to the question?
$endgroup$
– Forextrader
Jan 9 at 20:29
$begingroup$
That works................
$endgroup$
– DanielWainfleet
Jan 10 at 3:32
2
2
$begingroup$
If $(exists x in mathbb{Z}) P(x)$ is true, then so is $(exists x in mathbb{Q}) P(x)$, as the same $x$ works for both.
$endgroup$
– Connor Harris
Jan 9 at 19:13
$begingroup$
If $(exists x in mathbb{Z}) P(x)$ is true, then so is $(exists x in mathbb{Q}) P(x)$, as the same $x$ works for both.
$endgroup$
– Connor Harris
Jan 9 at 19:13
1
1
$begingroup$
In general, you'll want to look for propositions with universal rather than existential quantifiers.
$endgroup$
– Connor Harris
Jan 9 at 19:16
$begingroup$
In general, you'll want to look for propositions with universal rather than existential quantifiers.
$endgroup$
– Connor Harris
Jan 9 at 19:16
$begingroup$
This is my first mathematical reasoning class. Totally new to me. I'll try and work off of what you said to understand it a little more.
$endgroup$
– Forextrader
Jan 9 at 19:18
$begingroup$
This is my first mathematical reasoning class. Totally new to me. I'll try and work off of what you said to understand it a little more.
$endgroup$
– Forextrader
Jan 9 at 19:18
$begingroup$
Ok what if I wrote then $(forall n in mathbb{Z}, n^2geq n)$ is that an appropriate answer to the question?
$endgroup$
– Forextrader
Jan 9 at 20:29
$begingroup$
Ok what if I wrote then $(forall n in mathbb{Z}, n^2geq n)$ is that an appropriate answer to the question?
$endgroup$
– Forextrader
Jan 9 at 20:29
$begingroup$
That works................
$endgroup$
– DanielWainfleet
Jan 10 at 3:32
$begingroup$
That works................
$endgroup$
– DanielWainfleet
Jan 10 at 3:32
|
show 1 more comment
1 Answer
1
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$begingroup$
One way is to find an expression
that gives different values for integers
than non-integers.
An obvious candidate is
$lfloor x rfloor$,
the integer part of $x$.
This is $x$ when $x$ is an integer
and less than $x$ otherwise.
Therefore one proposition that works is
$P(x) equiv (x = lfloor x rfloor)$.
answered Jan 10 at 0:57
marty cohenmarty cohen
74k549128
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One way is to find an expression
that gives different values for integers
than non-integers.
An obvious candidate is
$lfloor x rfloor$,
the integer part of $x$.
This is $x$ when $x$ is an integer
and less than $x$ otherwise.
Therefore one proposition that works is
$P(x) equiv (x = lfloor x rfloor)$.
answered Jan 10 at 0:57
marty cohenmarty cohen
74k549128
$endgroup$
add a comment |
$begingroup$
One way is to find an expression
that gives different values for integers
than non-integers.
An obvious candidate is
$lfloor x rfloor$,
the integer part of $x$.
This is $x$ when $x$ is an integer
and less than $x$ otherwise.
Therefore one proposition that works is
$P(x) equiv (x = lfloor x rfloor)$.
answered Jan 10 at 0:57
marty cohenmarty cohen
74k549128
$endgroup$
add a comment |
$begingroup$
One way is to find an expression
that gives different values for integers
than non-integers.
An obvious candidate is
$lfloor x rfloor$,
the integer part of $x$.
This is $x$ when $x$ is an integer
and less than $x$ otherwise.
Therefore one proposition that works is
$P(x) equiv (x = lfloor x rfloor)$.
answered Jan 10 at 0:57
marty cohenmarty cohen
74k549128
$endgroup$
One way is to find an expression
that gives different values for integers
than non-integers.
An obvious candidate is
$lfloor x rfloor$,
the integer part of $x$.
This is $x$ when $x$ is an integer
and less than $x$ otherwise.
Therefore one proposition that works is
$P(x) equiv (x = lfloor x rfloor)$.
answered Jan 10 at 0:57
marty cohenmarty cohen
74k549128
answered Jan 10 at 0:57
marty cohenmarty cohen
74k549128
answered Jan 10 at 0:57
marty cohenmarty cohen
74k549128
answered Jan 10 at 0:57
marty cohenmarty cohen
74k549128
74k549128
add a comment |
add a comment |
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$begingroup$
If $(exists x in mathbb{Z}) P(x)$ is true, then so is $(exists x in mathbb{Q}) P(x)$, as the same $x$ works for both.
$endgroup$
– Connor Harris
Jan 9 at 19:13
1
$begingroup$
In general, you'll want to look for propositions with universal rather than existential quantifiers.
$endgroup$
– Connor Harris
Jan 9 at 19:16
$begingroup$
This is my first mathematical reasoning class. Totally new to me. I'll try and work off of what you said to understand it a little more.
$endgroup$
– Forextrader
Jan 9 at 19:18
$begingroup$
Ok what if I wrote then $(forall n in mathbb{Z}, n^2geq n)$ is that an appropriate answer to the question?
$endgroup$
– Forextrader
Jan 9 at 20:29
$begingroup$
That works................
$endgroup$
– DanielWainfleet
Jan 10 at 3:32