Logic problem: Truth value of statement the product of $x^2$ and $x^3$ is $x^6$
1
$begingroup$
I want to understand why these statements below are false. I assumed that the statements are true because they are real numbers
- The product of $x^2$ and $x^3$ is $x^6$
- The $x^2>0$ for any real number $x$
discrete-mathematics
asked Jan 15 at 10:10
SamSam
44418
$endgroup$
|
show 1 more comment
1
$begingroup$
I want to understand why these statements below are false. I assumed that the statements are true because they are real numbers
- The product of $x^2$ and $x^3$ is $x^6$
- The $x^2>0$ for any real number $x$
discrete-mathematics
asked Jan 15 at 10:10
SamSam
44418
$endgroup$
1
$begingroup$
Without any quantifier, the first one is not even a statement, since it depends on $x$ (it doesn't matter that it's true for each real $x$), hence is neither true nor false. The second one is just false because it fails for $x=0$.
$endgroup$
– Wojowu
Jan 15 at 10:13
$begingroup$
@JoseArnaldoBebitaDris ...can we pretend I never said that? :P
$endgroup$
– Wojowu
Jan 15 at 10:19
$begingroup$
Was $x^6$ a typo for $x^5$?
$endgroup$
– bof
Jan 15 at 10:19
$begingroup$
@Wojowu: Sure thing! =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:19
$begingroup$
@bof probably not, as OP is trying to figure out why the given statements are false.
$endgroup$
– StackTD
Jan 15 at 10:20
|
show 1 more comment
1
1
1
$begingroup$
I want to understand why these statements below are false. I assumed that the statements are true because they are real numbers
- The product of $x^2$ and $x^3$ is $x^6$
- The $x^2>0$ for any real number $x$
discrete-mathematics
asked Jan 15 at 10:10
SamSam
44418
$endgroup$
I want to understand why these statements below are false. I assumed that the statements are true because they are real numbers
- The product of $x^2$ and $x^3$ is $x^6$
- The $x^2>0$ for any real number $x$
discrete-mathematics
discrete-mathematics
asked Jan 15 at 10:10
SamSam
44418
asked Jan 15 at 10:10
SamSam
44418
asked Jan 15 at 10:10
SamSam
44418
asked Jan 15 at 10:10
SamSam
44418
asked Jan 15 at 10:10
SamSam
44418
44418
1
$begingroup$
Without any quantifier, the first one is not even a statement, since it depends on $x$ (it doesn't matter that it's true for each real $x$), hence is neither true nor false. The second one is just false because it fails for $x=0$.
$endgroup$
– Wojowu
Jan 15 at 10:13
$begingroup$
@JoseArnaldoBebitaDris ...can we pretend I never said that? :P
$endgroup$
– Wojowu
Jan 15 at 10:19
$begingroup$
Was $x^6$ a typo for $x^5$?
$endgroup$
– bof
Jan 15 at 10:19
$begingroup$
@Wojowu: Sure thing! =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:19
$begingroup$
@bof probably not, as OP is trying to figure out why the given statements are false.
$endgroup$
– StackTD
Jan 15 at 10:20
|
show 1 more comment
1
$begingroup$
Without any quantifier, the first one is not even a statement, since it depends on $x$ (it doesn't matter that it's true for each real $x$), hence is neither true nor false. The second one is just false because it fails for $x=0$.
$endgroup$
– Wojowu
Jan 15 at 10:13
$begingroup$
@JoseArnaldoBebitaDris ...can we pretend I never said that? :P
$endgroup$
– Wojowu
Jan 15 at 10:19
$begingroup$
Was $x^6$ a typo for $x^5$?
$endgroup$
– bof
Jan 15 at 10:19
$begingroup$
@Wojowu: Sure thing! =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:19
$begingroup$
@bof probably not, as OP is trying to figure out why the given statements are false.
$endgroup$
– StackTD
Jan 15 at 10:20
1
1
$begingroup$
Without any quantifier, the first one is not even a statement, since it depends on $x$ (it doesn't matter that it's true for each real $x$), hence is neither true nor false. The second one is just false because it fails for $x=0$.
$endgroup$
– Wojowu
Jan 15 at 10:13
$begingroup$
Without any quantifier, the first one is not even a statement, since it depends on $x$ (it doesn't matter that it's true for each real $x$), hence is neither true nor false. The second one is just false because it fails for $x=0$.
$endgroup$
– Wojowu
Jan 15 at 10:13
$begingroup$
@JoseArnaldoBebitaDris ...can we pretend I never said that? :P
$endgroup$
– Wojowu
Jan 15 at 10:19
$begingroup$
@JoseArnaldoBebitaDris ...can we pretend I never said that? :P
$endgroup$
– Wojowu
Jan 15 at 10:19
$begingroup$
Was $x^6$ a typo for $x^5$?
$endgroup$
– bof
Jan 15 at 10:19
$begingroup$
Was $x^6$ a typo for $x^5$?
$endgroup$
– bof
Jan 15 at 10:19
$begingroup$
@Wojowu: Sure thing! =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:19
$begingroup$
@Wojowu: Sure thing! =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:19
$begingroup$
@bof probably not, as OP is trying to figure out why the given statements are false.
$endgroup$
– StackTD
Jan 15 at 10:20
$begingroup$
@bof probably not, as OP is trying to figure out why the given statements are false.
$endgroup$
– StackTD
Jan 15 at 10:20
|
show 1 more comment
1 Answer
1
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3
$begingroup$
- The product of $x^2$ and $x^3$ is $x^6$
Do you know a rule for $boxed{x^m x^n = ldots}$ ?
Don't confuse it with the rule for $boxed{left(x^mright)^n = ldots}$ !
If not, look them up.
- The $x^2>0$ for any real number $x$
But also $0$ is a real number, so...
answered Jan 15 at 10:14
StackTDStackTD
24.2k2254
$endgroup$
$begingroup$
FWIW, the first statement is true if $x = 0$ or $x=1$, and false otherwise.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:21
$begingroup$
It's hard to tell if OP copied the problem literally or not; I'm trying to answer in the spirit of what I assume to be the underlying idea of the question. Of course the question would benefit from a more precise phrasing.
$endgroup$
– StackTD
Jan 15 at 10:27
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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oldest
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3
$begingroup$
- The product of $x^2$ and $x^3$ is $x^6$
Do you know a rule for $boxed{x^m x^n = ldots}$ ?
Don't confuse it with the rule for $boxed{left(x^mright)^n = ldots}$ !
If not, look them up.
- The $x^2>0$ for any real number $x$
But also $0$ is a real number, so...
answered Jan 15 at 10:14
StackTDStackTD
24.2k2254
$endgroup$
$begingroup$
FWIW, the first statement is true if $x = 0$ or $x=1$, and false otherwise.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:21
$begingroup$
It's hard to tell if OP copied the problem literally or not; I'm trying to answer in the spirit of what I assume to be the underlying idea of the question. Of course the question would benefit from a more precise phrasing.
$endgroup$
– StackTD
Jan 15 at 10:27
add a comment |
3
$begingroup$
- The product of $x^2$ and $x^3$ is $x^6$
Do you know a rule for $boxed{x^m x^n = ldots}$ ?
Don't confuse it with the rule for $boxed{left(x^mright)^n = ldots}$ !
If not, look them up.
- The $x^2>0$ for any real number $x$
But also $0$ is a real number, so...
answered Jan 15 at 10:14
StackTDStackTD
24.2k2254
$endgroup$
$begingroup$
FWIW, the first statement is true if $x = 0$ or $x=1$, and false otherwise.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:21
$begingroup$
It's hard to tell if OP copied the problem literally or not; I'm trying to answer in the spirit of what I assume to be the underlying idea of the question. Of course the question would benefit from a more precise phrasing.
$endgroup$
– StackTD
Jan 15 at 10:27
add a comment |
3
3
3
$begingroup$
- The product of $x^2$ and $x^3$ is $x^6$
Do you know a rule for $boxed{x^m x^n = ldots}$ ?
Don't confuse it with the rule for $boxed{left(x^mright)^n = ldots}$ !
If not, look them up.
- The $x^2>0$ for any real number $x$
But also $0$ is a real number, so...
answered Jan 15 at 10:14
StackTDStackTD
24.2k2254
$endgroup$
- The product of $x^2$ and $x^3$ is $x^6$
Do you know a rule for $boxed{x^m x^n = ldots}$ ?
Don't confuse it with the rule for $boxed{left(x^mright)^n = ldots}$ !
If not, look them up.
- The $x^2>0$ for any real number $x$
But also $0$ is a real number, so...
answered Jan 15 at 10:14
StackTDStackTD
24.2k2254
answered Jan 15 at 10:14
StackTDStackTD
24.2k2254
answered Jan 15 at 10:14
StackTDStackTD
24.2k2254
answered Jan 15 at 10:14
StackTDStackTD
24.2k2254
24.2k2254
$begingroup$
FWIW, the first statement is true if $x = 0$ or $x=1$, and false otherwise.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:21
$begingroup$
It's hard to tell if OP copied the problem literally or not; I'm trying to answer in the spirit of what I assume to be the underlying idea of the question. Of course the question would benefit from a more precise phrasing.
$endgroup$
– StackTD
Jan 15 at 10:27
add a comment |
$begingroup$
FWIW, the first statement is true if $x = 0$ or $x=1$, and false otherwise.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:21
$begingroup$
It's hard to tell if OP copied the problem literally or not; I'm trying to answer in the spirit of what I assume to be the underlying idea of the question. Of course the question would benefit from a more precise phrasing.
$endgroup$
– StackTD
Jan 15 at 10:27
$begingroup$
FWIW, the first statement is true if $x = 0$ or $x=1$, and false otherwise.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:21
$begingroup$
FWIW, the first statement is true if $x = 0$ or $x=1$, and false otherwise.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:21
$begingroup$
It's hard to tell if OP copied the problem literally or not; I'm trying to answer in the spirit of what I assume to be the underlying idea of the question. Of course the question would benefit from a more precise phrasing.
$endgroup$
– StackTD
Jan 15 at 10:27
$begingroup$
It's hard to tell if OP copied the problem literally or not; I'm trying to answer in the spirit of what I assume to be the underlying idea of the question. Of course the question would benefit from a more precise phrasing.
$endgroup$
– StackTD
Jan 15 at 10:27
add a comment |
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1
$begingroup$
Without any quantifier, the first one is not even a statement, since it depends on $x$ (it doesn't matter that it's true for each real $x$), hence is neither true nor false. The second one is just false because it fails for $x=0$.
$endgroup$
– Wojowu
Jan 15 at 10:13
$begingroup$
@JoseArnaldoBebitaDris ...can we pretend I never said that? :P
$endgroup$
– Wojowu
Jan 15 at 10:19
$begingroup$
Was $x^6$ a typo for $x^5$?
$endgroup$
– bof
Jan 15 at 10:19
$begingroup$
@Wojowu: Sure thing! =)
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 15 at 10:19
$begingroup$
@bof probably not, as OP is trying to figure out why the given statements are false.
$endgroup$
– StackTD
Jan 15 at 10:20