When is it true that $(y_1-x_1)^p + (y_2 -x_2)^p leq (y_2 - x_1)^p + (y_1 - x_2)^p$?
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Consider four integers $x_1, x_2, y_1, y_2$ with $x_2 > x_1$ and $y_2 > y_1$. Is it true that the only integer values of $p$ so that:
$$(y_1-x_1)^p + (y_2 -x_2)^p leq (y_2 - x_1)^p + (y_1 - x_2)^p$$
are $p=0, 1$ and all even $p geq 2$?
algebra-precalculus inequality
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Consider four integers $x_1, x_2, y_1, y_2$ with $x_2 > x_1$ and $y_2 > y_1$. Is it true that the only integer values of $p$ so that:
$$(y_1-x_1)^p + (y_2 -x_2)^p leq (y_2 - x_1)^p + (y_1 - x_2)^p$$
are $p=0, 1$ and all even $p geq 2$?
algebra-precalculus inequality
how do you define $(-1)^1.5$?
– Calvin Khor
Jul 22 at 9:56
@CalvinKhor Thanks. I have fixed the question.
– Anush
Jul 22 at 10:47
It’s true for $p=2$ as well.
– Michael Hoppe
Jul 22 at 16:41
@MichaelHoppe Thank you. Is it true for any other integers $p$?
– Anush
Jul 22 at 16:42
Just play a around with some intergers and make a guess.
– Michael Hoppe
Jul 22 at 16:44
 |Â
show 1 more comment
up vote
0
down vote
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up vote
0
down vote
favorite
Consider four integers $x_1, x_2, y_1, y_2$ with $x_2 > x_1$ and $y_2 > y_1$. Is it true that the only integer values of $p$ so that:
$$(y_1-x_1)^p + (y_2 -x_2)^p leq (y_2 - x_1)^p + (y_1 - x_2)^p$$
are $p=0, 1$ and all even $p geq 2$?
algebra-precalculus inequality
Consider four integers $x_1, x_2, y_1, y_2$ with $x_2 > x_1$ and $y_2 > y_1$. Is it true that the only integer values of $p$ so that:
$$(y_1-x_1)^p + (y_2 -x_2)^p leq (y_2 - x_1)^p + (y_1 - x_2)^p$$
are $p=0, 1$ and all even $p geq 2$?
algebra-precalculus inequality
edited Jul 22 at 16:57
asked Jul 22 at 9:44
Anush
190224
190224
how do you define $(-1)^1.5$?
– Calvin Khor
Jul 22 at 9:56
@CalvinKhor Thanks. I have fixed the question.
– Anush
Jul 22 at 10:47
It’s true for $p=2$ as well.
– Michael Hoppe
Jul 22 at 16:41
@MichaelHoppe Thank you. Is it true for any other integers $p$?
– Anush
Jul 22 at 16:42
Just play a around with some intergers and make a guess.
– Michael Hoppe
Jul 22 at 16:44
 |Â
show 1 more comment
how do you define $(-1)^1.5$?
– Calvin Khor
Jul 22 at 9:56
@CalvinKhor Thanks. I have fixed the question.
– Anush
Jul 22 at 10:47
It’s true for $p=2$ as well.
– Michael Hoppe
Jul 22 at 16:41
@MichaelHoppe Thank you. Is it true for any other integers $p$?
– Anush
Jul 22 at 16:42
Just play a around with some intergers and make a guess.
– Michael Hoppe
Jul 22 at 16:44
how do you define $(-1)^1.5$?
– Calvin Khor
Jul 22 at 9:56
how do you define $(-1)^1.5$?
– Calvin Khor
Jul 22 at 9:56
@CalvinKhor Thanks. I have fixed the question.
– Anush
Jul 22 at 10:47
@CalvinKhor Thanks. I have fixed the question.
– Anush
Jul 22 at 10:47
It’s true for $p=2$ as well.
– Michael Hoppe
Jul 22 at 16:41
It’s true for $p=2$ as well.
– Michael Hoppe
Jul 22 at 16:41
@MichaelHoppe Thank you. Is it true for any other integers $p$?
– Anush
Jul 22 at 16:42
@MichaelHoppe Thank you. Is it true for any other integers $p$?
– Anush
Jul 22 at 16:42
Just play a around with some intergers and make a guess.
– Michael Hoppe
Jul 22 at 16:44
Just play a around with some intergers and make a guess.
– Michael Hoppe
Jul 22 at 16:44
 |Â
show 1 more comment
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how do you define $(-1)^1.5$?
– Calvin Khor
Jul 22 at 9:56
@CalvinKhor Thanks. I have fixed the question.
– Anush
Jul 22 at 10:47
It’s true for $p=2$ as well.
– Michael Hoppe
Jul 22 at 16:41
@MichaelHoppe Thank you. Is it true for any other integers $p$?
– Anush
Jul 22 at 16:42
Just play a around with some intergers and make a guess.
– Michael Hoppe
Jul 22 at 16:44