Mixing
When is this statement true? Never dealt with equations like these in logic before.
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When is this statement true? $$(frac{1}{x}<10) => (x>frac{1}{10})$$
I know that
for p=1 q=1 | p=>q=1
for p=1 q=0 | p=>q=0
for p=0 q=1 | p=>q=1
for p=0 q=0 | p=>q=1
But I don't know where to go from here
logic
asked Jan 16 at 18:32
B. CzostekB. Czostek
556
$endgroup$
add a comment |
$begingroup$
When is this statement true? $$(frac{1}{x}<10) => (x>frac{1}{10})$$
I know that
for p=1 q=1 | p=>q=1
for p=1 q=0 | p=>q=0
for p=0 q=1 | p=>q=1
for p=0 q=0 | p=>q=1
But I don't know where to go from here
logic
asked Jan 16 at 18:32
B. CzostekB. Czostek
556
$endgroup$
2
$begingroup$
It is a simple statement in arithmetic which is true if $x$ is a positive real number (or positive rational). So the value $10$ for example is simply a number and $x$ is not the kind of entity which takes the values true or false.
$endgroup$
– Mark Bennet
Jan 16 at 18:35
$begingroup$
So there is no catch? I thought I am definitely missing something
$endgroup$
– B. Czostek
Jan 16 at 18:44
$begingroup$
I think there is no catch
$endgroup$
– Mark Bennet
Jan 16 at 18:45
$begingroup$
But isn't the whole statement true if both of equations are false? You suggest that the answer is that statement is true if x is a positive real number but how for x=1/10 it's not false? I don't get it
$endgroup$
– B. Czostek
Jan 16 at 18:58
$begingroup$
You are right of course. I am thinking informally and wrongly. You might consider the $x=0$ case carefully.
$endgroup$
– Mark Bennet
Jan 16 at 19:07
add a comment |
$begingroup$
When is this statement true? $$(frac{1}{x}<10) => (x>frac{1}{10})$$
I know that
for p=1 q=1 | p=>q=1
for p=1 q=0 | p=>q=0
for p=0 q=1 | p=>q=1
for p=0 q=0 | p=>q=1
But I don't know where to go from here
logic
asked Jan 16 at 18:32
B. CzostekB. Czostek
556
$endgroup$
When is this statement true? $$(frac{1}{x}<10) => (x>frac{1}{10})$$
I know that
for p=1 q=1 | p=>q=1
for p=1 q=0 | p=>q=0
for p=0 q=1 | p=>q=1
for p=0 q=0 | p=>q=1
But I don't know where to go from here
logic
logic
asked Jan 16 at 18:32
B. CzostekB. Czostek
556
asked Jan 16 at 18:32
B. CzostekB. Czostek
556
asked Jan 16 at 18:32
B. CzostekB. Czostek
556
asked Jan 16 at 18:32
B. CzostekB. Czostek
556
asked Jan 16 at 18:32
B. CzostekB. Czostek
556
556
2
$begingroup$
It is a simple statement in arithmetic which is true if $x$ is a positive real number (or positive rational). So the value $10$ for example is simply a number and $x$ is not the kind of entity which takes the values true or false.
$endgroup$
– Mark Bennet
Jan 16 at 18:35
$begingroup$
So there is no catch? I thought I am definitely missing something
$endgroup$
– B. Czostek
Jan 16 at 18:44
$begingroup$
I think there is no catch
$endgroup$
– Mark Bennet
Jan 16 at 18:45
$begingroup$
But isn't the whole statement true if both of equations are false? You suggest that the answer is that statement is true if x is a positive real number but how for x=1/10 it's not false? I don't get it
$endgroup$
– B. Czostek
Jan 16 at 18:58
$begingroup$
You are right of course. I am thinking informally and wrongly. You might consider the $x=0$ case carefully.
$endgroup$
– Mark Bennet
Jan 16 at 19:07
add a comment |
2
$begingroup$
It is a simple statement in arithmetic which is true if $x$ is a positive real number (or positive rational). So the value $10$ for example is simply a number and $x$ is not the kind of entity which takes the values true or false.
$endgroup$
– Mark Bennet
Jan 16 at 18:35
$begingroup$
So there is no catch? I thought I am definitely missing something
$endgroup$
– B. Czostek
Jan 16 at 18:44
$begingroup$
I think there is no catch
$endgroup$
– Mark Bennet
Jan 16 at 18:45
$begingroup$
But isn't the whole statement true if both of equations are false? You suggest that the answer is that statement is true if x is a positive real number but how for x=1/10 it's not false? I don't get it
$endgroup$
– B. Czostek
Jan 16 at 18:58
$begingroup$
You are right of course. I am thinking informally and wrongly. You might consider the $x=0$ case carefully.
$endgroup$
– Mark Bennet
Jan 16 at 19:07
2
2
$begingroup$
It is a simple statement in arithmetic which is true if $x$ is a positive real number (or positive rational). So the value $10$ for example is simply a number and $x$ is not the kind of entity which takes the values true or false.
$endgroup$
– Mark Bennet
Jan 16 at 18:35
$begingroup$
It is a simple statement in arithmetic which is true if $x$ is a positive real number (or positive rational). So the value $10$ for example is simply a number and $x$ is not the kind of entity which takes the values true or false.
$endgroup$
– Mark Bennet
Jan 16 at 18:35
$begingroup$
So there is no catch? I thought I am definitely missing something
$endgroup$
– B. Czostek
Jan 16 at 18:44
$begingroup$
So there is no catch? I thought I am definitely missing something
$endgroup$
– B. Czostek
Jan 16 at 18:44
$begingroup$
I think there is no catch
$endgroup$
– Mark Bennet
Jan 16 at 18:45
$begingroup$
I think there is no catch
$endgroup$
– Mark Bennet
Jan 16 at 18:45
$begingroup$
But isn't the whole statement true if both of equations are false? You suggest that the answer is that statement is true if x is a positive real number but how for x=1/10 it's not false? I don't get it
$endgroup$
– B. Czostek
Jan 16 at 18:58
$begingroup$
But isn't the whole statement true if both of equations are false? You suggest that the answer is that statement is true if x is a positive real number but how for x=1/10 it's not false? I don't get it
$endgroup$
– B. Czostek
Jan 16 at 18:58
$begingroup$
You are right of course. I am thinking informally and wrongly. You might consider the $x=0$ case carefully.
$endgroup$
– Mark Bennet
Jan 16 at 19:07
$begingroup$
You are right of course. I am thinking informally and wrongly. You might consider the $x=0$ case carefully.
$endgroup$
– Mark Bennet
Jan 16 at 19:07
add a comment |
1 Answer
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Well, I think Mark Bennet was right the first time. According to text-books in logic, an implication is true iff the LHS is false and/or the RHS is true. So the implication in the question is true iff $1/x ge 10$ and/or $x > 1/10$. Since $1/x ge 10$ iff $0 < x le 1/10$, the implication is true iff $x > 0$.
answered Jan 16 at 22:41
Michael BehrendMichael Behrend
1,22746
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1 Answer
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$begingroup$
Well, I think Mark Bennet was right the first time. According to text-books in logic, an implication is true iff the LHS is false and/or the RHS is true. So the implication in the question is true iff $1/x ge 10$ and/or $x > 1/10$. Since $1/x ge 10$ iff $0 < x le 1/10$, the implication is true iff $x > 0$.
answered Jan 16 at 22:41
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
Well, I think Mark Bennet was right the first time. According to text-books in logic, an implication is true iff the LHS is false and/or the RHS is true. So the implication in the question is true iff $1/x ge 10$ and/or $x > 1/10$. Since $1/x ge 10$ iff $0 < x le 1/10$, the implication is true iff $x > 0$.
answered Jan 16 at 22:41
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
Well, I think Mark Bennet was right the first time. According to text-books in logic, an implication is true iff the LHS is false and/or the RHS is true. So the implication in the question is true iff $1/x ge 10$ and/or $x > 1/10$. Since $1/x ge 10$ iff $0 < x le 1/10$, the implication is true iff $x > 0$.
answered Jan 16 at 22:41
Michael BehrendMichael Behrend
1,22746
$endgroup$
Well, I think Mark Bennet was right the first time. According to text-books in logic, an implication is true iff the LHS is false and/or the RHS is true. So the implication in the question is true iff $1/x ge 10$ and/or $x > 1/10$. Since $1/x ge 10$ iff $0 < x le 1/10$, the implication is true iff $x > 0$.
answered Jan 16 at 22:41
Michael BehrendMichael Behrend
1,22746
answered Jan 16 at 22:41
Michael BehrendMichael Behrend
1,22746
answered Jan 16 at 22:41
Michael BehrendMichael Behrend
1,22746
answered Jan 16 at 22:41
Michael BehrendMichael Behrend
1,22746
1,22746
add a comment |
add a comment |
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$begingroup$
It is a simple statement in arithmetic which is true if $x$ is a positive real number (or positive rational). So the value $10$ for example is simply a number and $x$ is not the kind of entity which takes the values true or false.
$endgroup$
– Mark Bennet
Jan 16 at 18:35
$begingroup$
So there is no catch? I thought I am definitely missing something
$endgroup$
– B. Czostek
Jan 16 at 18:44
$begingroup$
I think there is no catch
$endgroup$
– Mark Bennet
Jan 16 at 18:45
$begingroup$
But isn't the whole statement true if both of equations are false? You suggest that the answer is that statement is true if x is a positive real number but how for x=1/10 it's not false? I don't get it
$endgroup$
– B. Czostek
Jan 16 at 18:58
$begingroup$
You are right of course. I am thinking informally and wrongly. You might consider the $x=0$ case carefully.
$endgroup$
– Mark Bennet
Jan 16 at 19:07