If $q neq 5$ or $k neq 1$, do the following conditions follow?
$begingroup$
This is a follow-up to MSE question #2998091.
From that question, we were able to get the bounds
$$frac{q-1}{q(q+1)} = f(1) leq f(k) < frac{q-2}{q(q-1)}$$
and
$$0 < f(q) leq f(5) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)},$$
where $f(q,k)$ is the function
$$f(q,k) = frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)},$$
defined for $q geq 5$ and $k geq 1$.
Now assume that $q neq 5$ and $k neq 1$. Solving the (resulting?) inequality
$$frac{q-1}{q(q+1)} < f(q,k) < frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
for $q$ and $k$ I get (with some help from WolframAlpha):
$$q > 4 text{ and } k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)}.$$
Since $q geq 5$ and by assumption, $q neq 5$, the first inequality could be strengthened to $q > 5$.
How about the second inequality?
$$k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)}$$
When $q = 5$, all it says is that $k > 1$. When $q > 5$, all we could deduce is that
$$frac{4}{q} < frac{4}{5} implies 1 - frac{4}{q} > frac{1}{5} implies frac{q}{q-4} = frac{1}{1-frac{4}{q}} < 5,$$
so that we could only obtain an upper bound of $1$ for
$$frac{logbigg(frac{q}{q-4}bigg)}{log(5)}$$
but not a lower bound.
Indeed, it appears that the inequality
$$k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)} = log_{5}bigg(frac{q}{q-4}bigg)$$
is trivial, because we have the limit
$$lim_{q to infty}{log_{5}bigg(frac{q}{q-4}bigg)} = 0$$
whereas we already have $k > 1$, by assumption.
Alternatively, one can perhaps get by
$$log_{5}bigg(frac{q}{q-4}bigg) < k$$
through solving for $q$:
$$q > frac{4cdot{5^k}}{5^k - 1},$$
from which we get that $q > 5$ if $k = 1$.
So perhaps the correct assumption would have been
$$lnot(q = 5 land k = 1)?$$
elementary-number-theory multivariable-calculus proof-verification inequality wolfram-alpha
asked Dec 19 '18 at 9:58
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
$endgroup$
add a comment |
$begingroup$
This is a follow-up to MSE question #2998091.
From that question, we were able to get the bounds
$$frac{q-1}{q(q+1)} = f(1) leq f(k) < frac{q-2}{q(q-1)}$$
and
$$0 < f(q) leq f(5) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)},$$
where $f(q,k)$ is the function
$$f(q,k) = frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)},$$
defined for $q geq 5$ and $k geq 1$.
Now assume that $q neq 5$ and $k neq 1$. Solving the (resulting?) inequality
$$frac{q-1}{q(q+1)} < f(q,k) < frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
for $q$ and $k$ I get (with some help from WolframAlpha):
$$q > 4 text{ and } k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)}.$$
Since $q geq 5$ and by assumption, $q neq 5$, the first inequality could be strengthened to $q > 5$.
How about the second inequality?
$$k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)}$$
When $q = 5$, all it says is that $k > 1$. When $q > 5$, all we could deduce is that
$$frac{4}{q} < frac{4}{5} implies 1 - frac{4}{q} > frac{1}{5} implies frac{q}{q-4} = frac{1}{1-frac{4}{q}} < 5,$$
so that we could only obtain an upper bound of $1$ for
$$frac{logbigg(frac{q}{q-4}bigg)}{log(5)}$$
but not a lower bound.
Indeed, it appears that the inequality
$$k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)} = log_{5}bigg(frac{q}{q-4}bigg)$$
is trivial, because we have the limit
$$lim_{q to infty}{log_{5}bigg(frac{q}{q-4}bigg)} = 0$$
whereas we already have $k > 1$, by assumption.
Alternatively, one can perhaps get by
$$log_{5}bigg(frac{q}{q-4}bigg) < k$$
through solving for $q$:
$$q > frac{4cdot{5^k}}{5^k - 1},$$
from which we get that $q > 5$ if $k = 1$.
So perhaps the correct assumption would have been
$$lnot(q = 5 land k = 1)?$$
elementary-number-theory multivariable-calculus proof-verification inequality wolfram-alpha
asked Dec 19 '18 at 9:58
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
$endgroup$
add a comment |
$begingroup$
This is a follow-up to MSE question #2998091.
From that question, we were able to get the bounds
$$frac{q-1}{q(q+1)} = f(1) leq f(k) < frac{q-2}{q(q-1)}$$
and
$$0 < f(q) leq f(5) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)},$$
where $f(q,k)$ is the function
$$f(q,k) = frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)},$$
defined for $q geq 5$ and $k geq 1$.
Now assume that $q neq 5$ and $k neq 1$. Solving the (resulting?) inequality
$$frac{q-1}{q(q+1)} < f(q,k) < frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
for $q$ and $k$ I get (with some help from WolframAlpha):
$$q > 4 text{ and } k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)}.$$
Since $q geq 5$ and by assumption, $q neq 5$, the first inequality could be strengthened to $q > 5$.
How about the second inequality?
$$k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)}$$
When $q = 5$, all it says is that $k > 1$. When $q > 5$, all we could deduce is that
$$frac{4}{q} < frac{4}{5} implies 1 - frac{4}{q} > frac{1}{5} implies frac{q}{q-4} = frac{1}{1-frac{4}{q}} < 5,$$
so that we could only obtain an upper bound of $1$ for
$$frac{logbigg(frac{q}{q-4}bigg)}{log(5)}$$
but not a lower bound.
Indeed, it appears that the inequality
$$k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)} = log_{5}bigg(frac{q}{q-4}bigg)$$
is trivial, because we have the limit
$$lim_{q to infty}{log_{5}bigg(frac{q}{q-4}bigg)} = 0$$
whereas we already have $k > 1$, by assumption.
Alternatively, one can perhaps get by
$$log_{5}bigg(frac{q}{q-4}bigg) < k$$
through solving for $q$:
$$q > frac{4cdot{5^k}}{5^k - 1},$$
from which we get that $q > 5$ if $k = 1$.
So perhaps the correct assumption would have been
$$lnot(q = 5 land k = 1)?$$
elementary-number-theory multivariable-calculus proof-verification inequality wolfram-alpha
asked Dec 19 '18 at 9:58
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
$endgroup$
This is a follow-up to MSE question #2998091.
From that question, we were able to get the bounds
$$frac{q-1}{q(q+1)} = f(1) leq f(k) < frac{q-2}{q(q-1)}$$
and
$$0 < f(q) leq f(5) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)},$$
where $f(q,k)$ is the function
$$f(q,k) = frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)},$$
defined for $q geq 5$ and $k geq 1$.
Now assume that $q neq 5$ and $k neq 1$. Solving the (resulting?) inequality
$$frac{q-1}{q(q+1)} < f(q,k) < frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
for $q$ and $k$ I get (with some help from WolframAlpha):
$$q > 4 text{ and } k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)}.$$
Since $q geq 5$ and by assumption, $q neq 5$, the first inequality could be strengthened to $q > 5$.
How about the second inequality?
$$k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)}$$
When $q = 5$, all it says is that $k > 1$. When $q > 5$, all we could deduce is that
$$frac{4}{q} < frac{4}{5} implies 1 - frac{4}{q} > frac{1}{5} implies frac{q}{q-4} = frac{1}{1-frac{4}{q}} < 5,$$
so that we could only obtain an upper bound of $1$ for
$$frac{logbigg(frac{q}{q-4}bigg)}{log(5)}$$
but not a lower bound.
Indeed, it appears that the inequality
$$k > frac{logbigg(frac{q}{q-4}bigg)}{log(5)} = log_{5}bigg(frac{q}{q-4}bigg)$$
is trivial, because we have the limit
$$lim_{q to infty}{log_{5}bigg(frac{q}{q-4}bigg)} = 0$$
whereas we already have $k > 1$, by assumption.
Alternatively, one can perhaps get by
$$log_{5}bigg(frac{q}{q-4}bigg) < k$$
through solving for $q$:
$$q > frac{4cdot{5^k}}{5^k - 1},$$
from which we get that $q > 5$ if $k = 1$.
So perhaps the correct assumption would have been
$$lnot(q = 5 land k = 1)?$$
elementary-number-theory multivariable-calculus proof-verification inequality wolfram-alpha
elementary-number-theory multivariable-calculus proof-verification inequality wolfram-alpha
asked Dec 19 '18 at 9:58
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
asked Dec 19 '18 at 9:58
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
asked Dec 19 '18 at 9:58
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
asked Dec 19 '18 at 9:58
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
asked Dec 19 '18 at 9:58
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
5,48541945
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that: $$ neg (p wedge q ) Leftrightarrow neg p vee neg q$$
$endgroup$
1
$begingroup$
Yes, of course I know that @someone. But +1 nonetheless for pointing it out.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:03
$begingroup$
Sorry, this doesn't answer your question completely. I actually wanted to post this as a comment, but I guess I don't have that privilege yet, as I'm new here.
$endgroup$
– user626177
Dec 19 '18 at 11:06
$begingroup$
No worries. I perfectly understood. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:07
add a comment |
$begingroup$
Blimey, how could I have missed it!
When $f$ is considered as a function of $k$,
$$f(q,1) = frac{q-1}{q(q+1)}$$
which when evaluated at $q=5$ equals $2/15$.
When $f$ is considered as a function of $q$,
$$f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
which when evaluated at $k=1$ equals $2/15$.
So indeed, in order to have an inequality in-between
$$frac{q-1}{q(q+1)} = f(q,1) neq f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
it suffices to consider
$$frac{q-1}{q(q+1)} = f(q,1) < f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)},$$
which is then equivalent to the condition
$$lnot(q = 5 land k = 1) iff lnot(q=5) lor lnot(k=1) iff (q > 5) lor (k > 1).$$
answered Dec 19 '18 at 11:25
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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$begingroup$
Note that: $$ neg (p wedge q ) Leftrightarrow neg p vee neg q$$
$endgroup$
1
$begingroup$
Yes, of course I know that @someone. But +1 nonetheless for pointing it out.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:03
$begingroup$
Sorry, this doesn't answer your question completely. I actually wanted to post this as a comment, but I guess I don't have that privilege yet, as I'm new here.
$endgroup$
– user626177
Dec 19 '18 at 11:06
$begingroup$
No worries. I perfectly understood. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:07
add a comment |
$begingroup$
Note that: $$ neg (p wedge q ) Leftrightarrow neg p vee neg q$$
$endgroup$
1
$begingroup$
Yes, of course I know that @someone. But +1 nonetheless for pointing it out.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:03
$begingroup$
Sorry, this doesn't answer your question completely. I actually wanted to post this as a comment, but I guess I don't have that privilege yet, as I'm new here.
$endgroup$
– user626177
Dec 19 '18 at 11:06
$begingroup$
No worries. I perfectly understood. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:07
add a comment |
$begingroup$
Note that: $$ neg (p wedge q ) Leftrightarrow neg p vee neg q$$
$endgroup$
Note that: $$ neg (p wedge q ) Leftrightarrow neg p vee neg q$$
answered Dec 19 '18 at 11:01
user626177
1
$begingroup$
Yes, of course I know that @someone. But +1 nonetheless for pointing it out.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:03
$begingroup$
Sorry, this doesn't answer your question completely. I actually wanted to post this as a comment, but I guess I don't have that privilege yet, as I'm new here.
$endgroup$
– user626177
Dec 19 '18 at 11:06
$begingroup$
No worries. I perfectly understood. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:07
add a comment |
1
$begingroup$
Yes, of course I know that @someone. But +1 nonetheless for pointing it out.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:03
$begingroup$
Sorry, this doesn't answer your question completely. I actually wanted to post this as a comment, but I guess I don't have that privilege yet, as I'm new here.
$endgroup$
– user626177
Dec 19 '18 at 11:06
$begingroup$
No worries. I perfectly understood. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:07
1
1
$begingroup$
Yes, of course I know that @someone. But +1 nonetheless for pointing it out.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:03
$begingroup$
Yes, of course I know that @someone. But +1 nonetheless for pointing it out.
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:03
$begingroup$
Sorry, this doesn't answer your question completely. I actually wanted to post this as a comment, but I guess I don't have that privilege yet, as I'm new here.
$endgroup$
– user626177
Dec 19 '18 at 11:06
$begingroup$
Sorry, this doesn't answer your question completely. I actually wanted to post this as a comment, but I guess I don't have that privilege yet, as I'm new here.
$endgroup$
– user626177
Dec 19 '18 at 11:06
$begingroup$
No worries. I perfectly understood. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:07
$begingroup$
No worries. I perfectly understood. =)
$endgroup$
– Jose Arnaldo Bebita Dris
Dec 19 '18 at 11:07
add a comment |
$begingroup$
Blimey, how could I have missed it!
When $f$ is considered as a function of $k$,
$$f(q,1) = frac{q-1}{q(q+1)}$$
which when evaluated at $q=5$ equals $2/15$.
When $f$ is considered as a function of $q$,
$$f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
which when evaluated at $k=1$ equals $2/15$.
So indeed, in order to have an inequality in-between
$$frac{q-1}{q(q+1)} = f(q,1) neq f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
it suffices to consider
$$frac{q-1}{q(q+1)} = f(q,1) < f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)},$$
which is then equivalent to the condition
$$lnot(q = 5 land k = 1) iff lnot(q=5) lor lnot(k=1) iff (q > 5) lor (k > 1).$$
answered Dec 19 '18 at 11:25
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
$endgroup$
add a comment |
$begingroup$
Blimey, how could I have missed it!
When $f$ is considered as a function of $k$,
$$f(q,1) = frac{q-1}{q(q+1)}$$
which when evaluated at $q=5$ equals $2/15$.
When $f$ is considered as a function of $q$,
$$f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
which when evaluated at $k=1$ equals $2/15$.
So indeed, in order to have an inequality in-between
$$frac{q-1}{q(q+1)} = f(q,1) neq f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
it suffices to consider
$$frac{q-1}{q(q+1)} = f(q,1) < f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)},$$
which is then equivalent to the condition
$$lnot(q = 5 land k = 1) iff lnot(q=5) lor lnot(k=1) iff (q > 5) lor (k > 1).$$
answered Dec 19 '18 at 11:25
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
$endgroup$
add a comment |
$begingroup$
Blimey, how could I have missed it!
When $f$ is considered as a function of $k$,
$$f(q,1) = frac{q-1}{q(q+1)}$$
which when evaluated at $q=5$ equals $2/15$.
When $f$ is considered as a function of $q$,
$$f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
which when evaluated at $k=1$ equals $2/15$.
So indeed, in order to have an inequality in-between
$$frac{q-1}{q(q+1)} = f(q,1) neq f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
it suffices to consider
$$frac{q-1}{q(q+1)} = f(q,1) < f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)},$$
which is then equivalent to the condition
$$lnot(q = 5 land k = 1) iff lnot(q=5) lor lnot(k=1) iff (q > 5) lor (k > 1).$$
answered Dec 19 '18 at 11:25
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
$endgroup$
Blimey, how could I have missed it!
When $f$ is considered as a function of $k$,
$$f(q,1) = frac{q-1}{q(q+1)}$$
which when evaluated at $q=5$ equals $2/15$.
When $f$ is considered as a function of $q$,
$$f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
which when evaluated at $k=1$ equals $2/15$.
So indeed, in order to have an inequality in-between
$$frac{q-1}{q(q+1)} = f(q,1) neq f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)}$$
it suffices to consider
$$frac{q-1}{q(q+1)} = f(q,1) < f(5,k) = frac{(5^k - 1)(5^{k+1} - 2cdot{5^k} + 1)}{4cdot{5^k}(5^{k+1} - 1)},$$
which is then equivalent to the condition
$$lnot(q = 5 land k = 1) iff lnot(q=5) lor lnot(k=1) iff (q > 5) lor (k > 1).$$
answered Dec 19 '18 at 11:25
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
answered Dec 19 '18 at 11:25
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
answered Dec 19 '18 at 11:25
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
answered Dec 19 '18 at 11:25
Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris
5,48541945
5,48541945
add a comment |
add a comment |
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