Triangle Inequality help
$begingroup$
Wondering where my logic is going wrong in this assignment:
Show that $||x|-|y|| leq |x-y|$
Using the fact $||x|-|y||, |x-y| geq 0$
It follows $(|x|-|y|)^2 leq (x-y)^2$
Using the fact $|x|^2 = x^2$
$x^2 -2|x||y| +y^2 leq x^2 -2xy +y^2$
Cancelling down:
$|xy| leq xy$
Which I know is not true. Thanks for any input.
inequality triangle
asked Dec 17 '18 at 20:30
PolynomialCPolynomialC
876
$endgroup$
add a comment |
$begingroup$
Wondering where my logic is going wrong in this assignment:
Show that $||x|-|y|| leq |x-y|$
Using the fact $||x|-|y||, |x-y| geq 0$
It follows $(|x|-|y|)^2 leq (x-y)^2$
Using the fact $|x|^2 = x^2$
$x^2 -2|x||y| +y^2 leq x^2 -2xy +y^2$
Cancelling down:
$|xy| leq xy$
Which I know is not true. Thanks for any input.
inequality triangle
asked Dec 17 '18 at 20:30
PolynomialCPolynomialC
876
$endgroup$
2
$begingroup$
You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
$endgroup$
– Crostul
Dec 17 '18 at 20:32
1
$begingroup$
You divided by $-2$ but didn't change the inequality direction.
$endgroup$
– orange
Dec 17 '18 at 20:33
add a comment |
$begingroup$
Wondering where my logic is going wrong in this assignment:
Show that $||x|-|y|| leq |x-y|$
Using the fact $||x|-|y||, |x-y| geq 0$
It follows $(|x|-|y|)^2 leq (x-y)^2$
Using the fact $|x|^2 = x^2$
$x^2 -2|x||y| +y^2 leq x^2 -2xy +y^2$
Cancelling down:
$|xy| leq xy$
Which I know is not true. Thanks for any input.
inequality triangle
asked Dec 17 '18 at 20:30
PolynomialCPolynomialC
876
$endgroup$
Wondering where my logic is going wrong in this assignment:
Show that $||x|-|y|| leq |x-y|$
Using the fact $||x|-|y||, |x-y| geq 0$
It follows $(|x|-|y|)^2 leq (x-y)^2$
Using the fact $|x|^2 = x^2$
$x^2 -2|x||y| +y^2 leq x^2 -2xy +y^2$
Cancelling down:
$|xy| leq xy$
Which I know is not true. Thanks for any input.
inequality triangle
inequality triangle
asked Dec 17 '18 at 20:30
PolynomialCPolynomialC
876
asked Dec 17 '18 at 20:30
PolynomialCPolynomialC
876
asked Dec 17 '18 at 20:30
PolynomialCPolynomialC
876
asked Dec 17 '18 at 20:30
PolynomialCPolynomialC
876
asked Dec 17 '18 at 20:30
PolynomialCPolynomialC
876
876
2
$begingroup$
You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
$endgroup$
– Crostul
Dec 17 '18 at 20:32
1
$begingroup$
You divided by $-2$ but didn't change the inequality direction.
$endgroup$
– orange
Dec 17 '18 at 20:33
add a comment |
2
$begingroup$
You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
$endgroup$
– Crostul
Dec 17 '18 at 20:32
1
$begingroup$
You divided by $-2$ but didn't change the inequality direction.
$endgroup$
– orange
Dec 17 '18 at 20:33
2
2
$begingroup$
You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
$endgroup$
– Crostul
Dec 17 '18 at 20:32
$begingroup$
You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
$endgroup$
– Crostul
Dec 17 '18 at 20:32
1
1
$begingroup$
You divided by $-2$ but didn't change the inequality direction.
$endgroup$
– orange
Dec 17 '18 at 20:33
$begingroup$
You divided by $-2$ but didn't change the inequality direction.
$endgroup$
– orange
Dec 17 '18 at 20:33
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since
$$ xyle|x||y|$$
then you have
$$ -2|x||y|le -2xy. $$
answered Dec 17 '18 at 20:32
xpaulxpaul
23.3k24655
$endgroup$
$begingroup$
Thankyou so much!
$endgroup$
– PolynomialC
Dec 17 '18 at 20:33
add a comment |
$begingroup$
You divided by $-2$ but did not change the sign. Another approach.
Observe $$|x|=|x-y+y|leq|x-y|+|y|implies |x|-|y|leq|x-y|$$
Simillarly we have $|y|-|x|leq|x-y|$.
Hence $$-|x-y|leq|x|-|y|leq|x-y|implies big||x|-|y|big|leq|x-y|$$
answered Dec 17 '18 at 20:34
Yadati KiranYadati Kiran
2,0331621
$endgroup$
add a comment |
$begingroup$
First, in line 3 you are already using the fact that $||x|-|y||leq|x-y|$.
However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows:
$-2|x||y|leq-2xy$ or equivalently $-|x||y|leq -xy$.
There, you multiply both sides of the inequality by $-1$, inverting the inequality $bigl((aleq b)rightarrow (-bleq -a)bigr)$ ending up with $xyleq |x||y|$.
answered Dec 17 '18 at 20:50
Jesús Miguel Martínez CamarenaJesús Miguel Martínez Camarena
363
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add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Since
$$ xyle|x||y|$$
then you have
$$ -2|x||y|le -2xy. $$
answered Dec 17 '18 at 20:32
xpaulxpaul
23.3k24655
$endgroup$
$begingroup$
Thankyou so much!
$endgroup$
– PolynomialC
Dec 17 '18 at 20:33
add a comment |
$begingroup$
Since
$$ xyle|x||y|$$
then you have
$$ -2|x||y|le -2xy. $$
answered Dec 17 '18 at 20:32
xpaulxpaul
23.3k24655
$endgroup$
$begingroup$
Thankyou so much!
$endgroup$
– PolynomialC
Dec 17 '18 at 20:33
add a comment |
$begingroup$
Since
$$ xyle|x||y|$$
then you have
$$ -2|x||y|le -2xy. $$
answered Dec 17 '18 at 20:32
xpaulxpaul
23.3k24655
$endgroup$
Since
$$ xyle|x||y|$$
then you have
$$ -2|x||y|le -2xy. $$
answered Dec 17 '18 at 20:32
xpaulxpaul
23.3k24655
answered Dec 17 '18 at 20:32
xpaulxpaul
23.3k24655
answered Dec 17 '18 at 20:32
xpaulxpaul
23.3k24655
answered Dec 17 '18 at 20:32
xpaulxpaul
23.3k24655
23.3k24655
$begingroup$
Thankyou so much!
$endgroup$
– PolynomialC
Dec 17 '18 at 20:33
add a comment |
$begingroup$
Thankyou so much!
$endgroup$
– PolynomialC
Dec 17 '18 at 20:33
$begingroup$
Thankyou so much!
$endgroup$
– PolynomialC
Dec 17 '18 at 20:33
$begingroup$
Thankyou so much!
$endgroup$
– PolynomialC
Dec 17 '18 at 20:33
add a comment |
$begingroup$
You divided by $-2$ but did not change the sign. Another approach.
Observe $$|x|=|x-y+y|leq|x-y|+|y|implies |x|-|y|leq|x-y|$$
Simillarly we have $|y|-|x|leq|x-y|$.
Hence $$-|x-y|leq|x|-|y|leq|x-y|implies big||x|-|y|big|leq|x-y|$$
answered Dec 17 '18 at 20:34
Yadati KiranYadati Kiran
2,0331621
$endgroup$
add a comment |
$begingroup$
You divided by $-2$ but did not change the sign. Another approach.
Observe $$|x|=|x-y+y|leq|x-y|+|y|implies |x|-|y|leq|x-y|$$
Simillarly we have $|y|-|x|leq|x-y|$.
Hence $$-|x-y|leq|x|-|y|leq|x-y|implies big||x|-|y|big|leq|x-y|$$
answered Dec 17 '18 at 20:34
Yadati KiranYadati Kiran
2,0331621
$endgroup$
add a comment |
$begingroup$
You divided by $-2$ but did not change the sign. Another approach.
Observe $$|x|=|x-y+y|leq|x-y|+|y|implies |x|-|y|leq|x-y|$$
Simillarly we have $|y|-|x|leq|x-y|$.
Hence $$-|x-y|leq|x|-|y|leq|x-y|implies big||x|-|y|big|leq|x-y|$$
answered Dec 17 '18 at 20:34
Yadati KiranYadati Kiran
2,0331621
$endgroup$
You divided by $-2$ but did not change the sign. Another approach.
Observe $$|x|=|x-y+y|leq|x-y|+|y|implies |x|-|y|leq|x-y|$$
Simillarly we have $|y|-|x|leq|x-y|$.
Hence $$-|x-y|leq|x|-|y|leq|x-y|implies big||x|-|y|big|leq|x-y|$$
answered Dec 17 '18 at 20:34
Yadati KiranYadati Kiran
2,0331621
answered Dec 17 '18 at 20:34
Yadati KiranYadati Kiran
2,0331621
answered Dec 17 '18 at 20:34
Yadati KiranYadati Kiran
2,0331621
answered Dec 17 '18 at 20:34
Yadati KiranYadati Kiran
2,0331621
2,0331621
add a comment |
add a comment |
$begingroup$
First, in line 3 you are already using the fact that $||x|-|y||leq|x-y|$.
However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows:
$-2|x||y|leq-2xy$ or equivalently $-|x||y|leq -xy$.
There, you multiply both sides of the inequality by $-1$, inverting the inequality $bigl((aleq b)rightarrow (-bleq -a)bigr)$ ending up with $xyleq |x||y|$.
answered Dec 17 '18 at 20:50
Jesús Miguel Martínez CamarenaJesús Miguel Martínez Camarena
363
$endgroup$
add a comment |
$begingroup$
First, in line 3 you are already using the fact that $||x|-|y||leq|x-y|$.
However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows:
$-2|x||y|leq-2xy$ or equivalently $-|x||y|leq -xy$.
There, you multiply both sides of the inequality by $-1$, inverting the inequality $bigl((aleq b)rightarrow (-bleq -a)bigr)$ ending up with $xyleq |x||y|$.
answered Dec 17 '18 at 20:50
Jesús Miguel Martínez CamarenaJesús Miguel Martínez Camarena
363
$endgroup$
add a comment |
$begingroup$
First, in line 3 you are already using the fact that $||x|-|y||leq|x-y|$.
However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows:
$-2|x||y|leq-2xy$ or equivalently $-|x||y|leq -xy$.
There, you multiply both sides of the inequality by $-1$, inverting the inequality $bigl((aleq b)rightarrow (-bleq -a)bigr)$ ending up with $xyleq |x||y|$.
answered Dec 17 '18 at 20:50
Jesús Miguel Martínez CamarenaJesús Miguel Martínez Camarena
363
$endgroup$
First, in line 3 you are already using the fact that $||x|-|y||leq|x-y|$.
However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows:
$-2|x||y|leq-2xy$ or equivalently $-|x||y|leq -xy$.
There, you multiply both sides of the inequality by $-1$, inverting the inequality $bigl((aleq b)rightarrow (-bleq -a)bigr)$ ending up with $xyleq |x||y|$.
answered Dec 17 '18 at 20:50
Jesús Miguel Martínez CamarenaJesús Miguel Martínez Camarena
363
answered Dec 17 '18 at 20:50
Jesús Miguel Martínez CamarenaJesús Miguel Martínez Camarena
363
answered Dec 17 '18 at 20:50
Jesús Miguel Martínez CamarenaJesús Miguel Martínez Camarena
363
answered Dec 17 '18 at 20:50
Jesús Miguel Martínez CamarenaJesús Miguel Martínez Camarena
363
363
add a comment |
add a comment |
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2
$begingroup$
You are dividing at the last step by $-2$. This changes the $le$ into a $ge$, giving you a correct statement.
$endgroup$
– Crostul
Dec 17 '18 at 20:32
1
$begingroup$
You divided by $-2$ but didn't change the inequality direction.
$endgroup$
– orange
Dec 17 '18 at 20:33