Reflexive relation, Symmetric relation and Transitive relation
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Let X be defined between [0, 1] defined functions range. Describe what properties (Reflexive relation, Symmetric relation and Transitive relation) has relation R in group X, if $$fRgLeftrightarrowforall xin[0,1],f(x)neq g(x)$$
I have tried to solve it on my own, is it right to say that this isn't reflexive, because f(x)!=g(x)
discrete-mathematics
asked Dec 12 '18 at 6:55
Fighter334Fighter334
61
$endgroup$
add a comment |
$begingroup$
Let X be defined between [0, 1] defined functions range. Describe what properties (Reflexive relation, Symmetric relation and Transitive relation) has relation R in group X, if $$fRgLeftrightarrowforall xin[0,1],f(x)neq g(x)$$
I have tried to solve it on my own, is it right to say that this isn't reflexive, because f(x)!=g(x)
discrete-mathematics
asked Dec 12 '18 at 6:55
Fighter334Fighter334
61
$endgroup$
4
$begingroup$
Correct, your logic shows non-reflexivity. What about the rest?
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 6:56
add a comment |
$begingroup$
Let X be defined between [0, 1] defined functions range. Describe what properties (Reflexive relation, Symmetric relation and Transitive relation) has relation R in group X, if $$fRgLeftrightarrowforall xin[0,1],f(x)neq g(x)$$
I have tried to solve it on my own, is it right to say that this isn't reflexive, because f(x)!=g(x)
discrete-mathematics
asked Dec 12 '18 at 6:55
Fighter334Fighter334
61
$endgroup$
Let X be defined between [0, 1] defined functions range. Describe what properties (Reflexive relation, Symmetric relation and Transitive relation) has relation R in group X, if $$fRgLeftrightarrowforall xin[0,1],f(x)neq g(x)$$
I have tried to solve it on my own, is it right to say that this isn't reflexive, because f(x)!=g(x)
discrete-mathematics
discrete-mathematics
asked Dec 12 '18 at 6:55
Fighter334Fighter334
61
asked Dec 12 '18 at 6:55
Fighter334Fighter334
61
asked Dec 12 '18 at 6:55
Fighter334Fighter334
61
asked Dec 12 '18 at 6:55
Fighter334Fighter334
61
asked Dec 12 '18 at 6:55
Fighter334Fighter334
61
61
4
$begingroup$
Correct, your logic shows non-reflexivity. What about the rest?
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 6:56
add a comment |
4
$begingroup$
Correct, your logic shows non-reflexivity. What about the rest?
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 6:56
4
4
$begingroup$
Correct, your logic shows non-reflexivity. What about the rest?
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 6:56
$begingroup$
Correct, your logic shows non-reflexivity. What about the rest?
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 6:56
add a comment |
1 Answer
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$begingroup$
It isn't reflexive by your reasoning. It is symmetric (as $f(x)neq g(x)Longleftrightarrow g(x)neq f(x)$) and non-transitive ($f(x) neq g(x) and g(x)neq f(x)$ but $f(x)=f(x)$).
Hope it is helpful
answered Dec 12 '18 at 7:05
MartundMartund
1,623213
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$begingroup$
It isn't reflexive by your reasoning. It is symmetric (as $f(x)neq g(x)Longleftrightarrow g(x)neq f(x)$) and non-transitive ($f(x) neq g(x) and g(x)neq f(x)$ but $f(x)=f(x)$).
Hope it is helpful
answered Dec 12 '18 at 7:05
MartundMartund
1,623213
$endgroup$
add a comment |
$begingroup$
It isn't reflexive by your reasoning. It is symmetric (as $f(x)neq g(x)Longleftrightarrow g(x)neq f(x)$) and non-transitive ($f(x) neq g(x) and g(x)neq f(x)$ but $f(x)=f(x)$).
Hope it is helpful
answered Dec 12 '18 at 7:05
MartundMartund
1,623213
$endgroup$
add a comment |
$begingroup$
It isn't reflexive by your reasoning. It is symmetric (as $f(x)neq g(x)Longleftrightarrow g(x)neq f(x)$) and non-transitive ($f(x) neq g(x) and g(x)neq f(x)$ but $f(x)=f(x)$).
Hope it is helpful
answered Dec 12 '18 at 7:05
MartundMartund
1,623213
$endgroup$
It isn't reflexive by your reasoning. It is symmetric (as $f(x)neq g(x)Longleftrightarrow g(x)neq f(x)$) and non-transitive ($f(x) neq g(x) and g(x)neq f(x)$ but $f(x)=f(x)$).
Hope it is helpful
answered Dec 12 '18 at 7:05
MartundMartund
1,623213
answered Dec 12 '18 at 7:05
MartundMartund
1,623213
answered Dec 12 '18 at 7:05
MartundMartund
1,623213
answered Dec 12 '18 at 7:05
MartundMartund
1,623213
1,623213
add a comment |
add a comment |
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$begingroup$
Correct, your logic shows non-reflexivity. What about the rest?
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 6:56