Prove that $lfloor sqrt{(p-1)p} rfloor = p - 1$ and likewise $lceil sqrt{(p-1)p} rceil = p$.
$begingroup$
Here $p$ is prime but is not necessary for the problem just that $p ge 0$. I suspect that a statement like $p-1 le sqrt{(p-1)p} le p$ would be the case but I am not certain how to establish this condition.
elementary-number-theory maxima-minima floor-function ceiling-function
edited Dec 12 '18 at 17:53
Bernard
121k740116
asked Dec 12 '18 at 17:24
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
add a comment |
$begingroup$
Here $p$ is prime but is not necessary for the problem just that $p ge 0$. I suspect that a statement like $p-1 le sqrt{(p-1)p} le p$ would be the case but I am not certain how to establish this condition.
elementary-number-theory maxima-minima floor-function ceiling-function
edited Dec 12 '18 at 17:53
Bernard
121k740116
asked Dec 12 '18 at 17:24
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
$begingroup$
We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
$endgroup$
– I like Serena
Dec 12 '18 at 17:48
add a comment |
$begingroup$
Here $p$ is prime but is not necessary for the problem just that $p ge 0$. I suspect that a statement like $p-1 le sqrt{(p-1)p} le p$ would be the case but I am not certain how to establish this condition.
elementary-number-theory maxima-minima floor-function ceiling-function
edited Dec 12 '18 at 17:53
Bernard
121k740116
asked Dec 12 '18 at 17:24
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
Here $p$ is prime but is not necessary for the problem just that $p ge 0$. I suspect that a statement like $p-1 le sqrt{(p-1)p} le p$ would be the case but I am not certain how to establish this condition.
elementary-number-theory maxima-minima floor-function ceiling-function
elementary-number-theory maxima-minima floor-function ceiling-function
edited Dec 12 '18 at 17:53
Bernard
121k740116
asked Dec 12 '18 at 17:24
Lorenz H MenkeLorenz H Menke
10411
edited Dec 12 '18 at 17:53
Bernard
121k740116
asked Dec 12 '18 at 17:24
Lorenz H MenkeLorenz H Menke
10411
edited Dec 12 '18 at 17:53
Bernard
121k740116
edited Dec 12 '18 at 17:53
Bernard
121k740116
edited Dec 12 '18 at 17:53
Bernard
121k740116
121k740116
asked Dec 12 '18 at 17:24
Lorenz H MenkeLorenz H Menke
10411
asked Dec 12 '18 at 17:24
Lorenz H MenkeLorenz H Menke
10411
asked Dec 12 '18 at 17:24
Lorenz H MenkeLorenz H Menke
10411
10411
$begingroup$
We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
$endgroup$
– I like Serena
Dec 12 '18 at 17:48
add a comment |
$begingroup$
We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
$endgroup$
– I like Serena
Dec 12 '18 at 17:48
$begingroup$
We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
$endgroup$
– I like Serena
Dec 12 '18 at 17:48
$begingroup$
We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
$endgroup$
– I like Serena
Dec 12 '18 at 17:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: $(p-1)^2 le (p-1)p le p^2$ for $p ge 0$
answered Dec 12 '18 at 17:29
karakfakarakfa
2,015811
$endgroup$
add a comment |
$begingroup$
For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$
and
$$p-1<sqrt{(p-1)p}<p.$$
As the extreme members are integer,
$$p-1leleftlfloorsqrt{(p-1)p}rightrfloorleleftlceilsqrt{(p-1)p}rightrceille p.$$
answered Dec 12 '18 at 17:31
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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$begingroup$
Hint: $(p-1)^2 le (p-1)p le p^2$ for $p ge 0$
answered Dec 12 '18 at 17:29
karakfakarakfa
2,015811
$endgroup$
add a comment |
$begingroup$
Hint: $(p-1)^2 le (p-1)p le p^2$ for $p ge 0$
answered Dec 12 '18 at 17:29
karakfakarakfa
2,015811
$endgroup$
add a comment |
$begingroup$
Hint: $(p-1)^2 le (p-1)p le p^2$ for $p ge 0$
answered Dec 12 '18 at 17:29
karakfakarakfa
2,015811
$endgroup$
Hint: $(p-1)^2 le (p-1)p le p^2$ for $p ge 0$
answered Dec 12 '18 at 17:29
karakfakarakfa
2,015811
answered Dec 12 '18 at 17:29
karakfakarakfa
2,015811
answered Dec 12 '18 at 17:29
karakfakarakfa
2,015811
answered Dec 12 '18 at 17:29
karakfakarakfa
2,015811
2,015811
add a comment |
add a comment |
$begingroup$
For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$
and
$$p-1<sqrt{(p-1)p}<p.$$
As the extreme members are integer,
$$p-1leleftlfloorsqrt{(p-1)p}rightrfloorleleftlceilsqrt{(p-1)p}rightrceille p.$$
answered Dec 12 '18 at 17:31
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$
and
$$p-1<sqrt{(p-1)p}<p.$$
As the extreme members are integer,
$$p-1leleftlfloorsqrt{(p-1)p}rightrfloorleleftlceilsqrt{(p-1)p}rightrceille p.$$
answered Dec 12 '18 at 17:31
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$
and
$$p-1<sqrt{(p-1)p}<p.$$
As the extreme members are integer,
$$p-1leleftlfloorsqrt{(p-1)p}rightrfloorleleftlceilsqrt{(p-1)p}rightrceille p.$$
answered Dec 12 '18 at 17:31
Yves DaoustYves Daoust
128k675227
$endgroup$
For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$
and
$$p-1<sqrt{(p-1)p}<p.$$
As the extreme members are integer,
$$p-1leleftlfloorsqrt{(p-1)p}rightrfloorleleftlceilsqrt{(p-1)p}rightrceille p.$$
answered Dec 12 '18 at 17:31
Yves DaoustYves Daoust
128k675227
answered Dec 12 '18 at 17:31
Yves DaoustYves Daoust
128k675227
answered Dec 12 '18 at 17:31
Yves DaoustYves Daoust
128k675227
answered Dec 12 '18 at 17:31
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
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$begingroup$
We need that $pge 1$, since otherwise the square root is not defined. That is also a sufficient condition.
$endgroup$
– I like Serena
Dec 12 '18 at 17:48