Is the converse of this implication true?
$begingroup$
If we have,
begin{align}
|x-2|leq {1} iff & -1leq{x-2}leq1\ iff & 4leq {x+3}leq6 \ iff & 5leq {x+4}leq7.end{align}
Then In particular,
$ |x-2|leq {1} implies 4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7.$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
$4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7 implies |x-2|leq {1}$.
This is a question in my textbook. I feel like the implication can be reversed.
inequality logic proof-explanation
$endgroup$
add a comment |
$begingroup$
If we have,
begin{align}
|x-2|leq {1} iff & -1leq{x-2}leq1\ iff & 4leq {x+3}leq6 \ iff & 5leq {x+4}leq7.end{align}
Then In particular,
$ |x-2|leq {1} implies 4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7.$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
$4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7 implies |x-2|leq {1}$.
This is a question in my textbook. I feel like the implication can be reversed.
inequality logic proof-explanation
$endgroup$
1
$begingroup$
Did you mean $|x-2|leq 1$?
$endgroup$
– Lucas Corrêa
Dec 29 '18 at 2:51
2
$begingroup$
Yes, that's what I meant.
$endgroup$
– user503154
Dec 29 '18 at 2:53
add a comment |
$begingroup$
If we have,
begin{align}
|x-2|leq {1} iff & -1leq{x-2}leq1\ iff & 4leq {x+3}leq6 \ iff & 5leq {x+4}leq7.end{align}
Then In particular,
$ |x-2|leq {1} implies 4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7.$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
$4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7 implies |x-2|leq {1}$.
This is a question in my textbook. I feel like the implication can be reversed.
inequality logic proof-explanation
$endgroup$
If we have,
begin{align}
|x-2|leq {1} iff & -1leq{x-2}leq1\ iff & 4leq {x+3}leq6 \ iff & 5leq {x+4}leq7.end{align}
Then In particular,
$ |x-2|leq {1} implies 4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7.$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
$4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7 implies |x-2|leq {1}$.
This is a question in my textbook. I feel like the implication can be reversed.
inequality logic proof-explanation
inequality logic proof-explanation
edited Dec 29 '18 at 3:13
asked Dec 29 '18 at 2:34
user503154
1
$begingroup$
Did you mean $|x-2|leq 1$?
$endgroup$
– Lucas Corrêa
Dec 29 '18 at 2:51
2
$begingroup$
Yes, that's what I meant.
$endgroup$
– user503154
Dec 29 '18 at 2:53
add a comment |
1
$begingroup$
Did you mean $|x-2|leq 1$?
$endgroup$
– Lucas Corrêa
Dec 29 '18 at 2:51
2
$begingroup$
Yes, that's what I meant.
$endgroup$
– user503154
Dec 29 '18 at 2:53
1
1
$begingroup$
Did you mean $|x-2|leq 1$?
$endgroup$
– Lucas Corrêa
Dec 29 '18 at 2:51
$begingroup$
Did you mean $|x-2|leq 1$?
$endgroup$
– Lucas Corrêa
Dec 29 '18 at 2:51
2
2
$begingroup$
Yes, that's what I meant.
$endgroup$
– user503154
Dec 29 '18 at 2:53
$begingroup$
Yes, that's what I meant.
$endgroup$
– user503154
Dec 29 '18 at 2:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For the first question: suppose that $|x-2|leq 1$. So,
$$|x+k| = |(x-2) + (k+2)| leq |x-2| + |k+2| leq 1 + |k+2|.$$
For the second:
$$4 leq |x+3| leq 6 Longrightarrow x in [1,3]cup[-9,-7]$$
$$5 leq |x+4| leq 7 Longrightarrow x in [1,3]cup[-11,-9]$$
then if $x$ satisfies both, $x in [1,3]cup{-9}$. Take $x=-9$,
$$|-9-2| = |-11| = 11 > 1$$
answered Dec 29 '18 at 3:13
Lucas CorrêaLucas Corrêa
1,5751321
$endgroup$
add a comment |
$begingroup$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
Note that:
$$begin{align}4le x+3le 6 &Rightarrow 4le |x+3|le 6, text{but} \
4le |x+3|le 6 ¬Rightarrow 4le x+3le 6, end{align}$$
because $4le x+3le 6$ has a solution $xin [1,3]$:
whereas $4le|x+3|le 6$ has a solution $xin [-9,-7]cup [1,3]$:
answered Dec 29 '18 at 8:35
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
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$begingroup$
For the first question: suppose that $|x-2|leq 1$. So,
$$|x+k| = |(x-2) + (k+2)| leq |x-2| + |k+2| leq 1 + |k+2|.$$
For the second:
$$4 leq |x+3| leq 6 Longrightarrow x in [1,3]cup[-9,-7]$$
$$5 leq |x+4| leq 7 Longrightarrow x in [1,3]cup[-11,-9]$$
then if $x$ satisfies both, $x in [1,3]cup{-9}$. Take $x=-9$,
$$|-9-2| = |-11| = 11 > 1$$
answered Dec 29 '18 at 3:13
Lucas CorrêaLucas Corrêa
1,5751321
$endgroup$
add a comment |
$begingroup$
For the first question: suppose that $|x-2|leq 1$. So,
$$|x+k| = |(x-2) + (k+2)| leq |x-2| + |k+2| leq 1 + |k+2|.$$
For the second:
$$4 leq |x+3| leq 6 Longrightarrow x in [1,3]cup[-9,-7]$$
$$5 leq |x+4| leq 7 Longrightarrow x in [1,3]cup[-11,-9]$$
then if $x$ satisfies both, $x in [1,3]cup{-9}$. Take $x=-9$,
$$|-9-2| = |-11| = 11 > 1$$
answered Dec 29 '18 at 3:13
Lucas CorrêaLucas Corrêa
1,5751321
$endgroup$
add a comment |
$begingroup$
For the first question: suppose that $|x-2|leq 1$. So,
$$|x+k| = |(x-2) + (k+2)| leq |x-2| + |k+2| leq 1 + |k+2|.$$
For the second:
$$4 leq |x+3| leq 6 Longrightarrow x in [1,3]cup[-9,-7]$$
$$5 leq |x+4| leq 7 Longrightarrow x in [1,3]cup[-11,-9]$$
then if $x$ satisfies both, $x in [1,3]cup{-9}$. Take $x=-9$,
$$|-9-2| = |-11| = 11 > 1$$
answered Dec 29 '18 at 3:13
Lucas CorrêaLucas Corrêa
1,5751321
$endgroup$
For the first question: suppose that $|x-2|leq 1$. So,
$$|x+k| = |(x-2) + (k+2)| leq |x-2| + |k+2| leq 1 + |k+2|.$$
For the second:
$$4 leq |x+3| leq 6 Longrightarrow x in [1,3]cup[-9,-7]$$
$$5 leq |x+4| leq 7 Longrightarrow x in [1,3]cup[-11,-9]$$
then if $x$ satisfies both, $x in [1,3]cup{-9}$. Take $x=-9$,
$$|-9-2| = |-11| = 11 > 1$$
answered Dec 29 '18 at 3:13
Lucas CorrêaLucas Corrêa
1,5751321
answered Dec 29 '18 at 3:13
Lucas CorrêaLucas Corrêa
1,5751321
answered Dec 29 '18 at 3:13
Lucas CorrêaLucas Corrêa
1,5751321
answered Dec 29 '18 at 3:13
Lucas CorrêaLucas Corrêa
1,5751321
1,5751321
add a comment |
add a comment |
$begingroup$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
Note that:
$$begin{align}4le x+3le 6 &Rightarrow 4le |x+3|le 6, text{but} \
4le |x+3|le 6 ¬Rightarrow 4le x+3le 6, end{align}$$
because $4le x+3le 6$ has a solution $xin [1,3]$:
whereas $4le|x+3|le 6$ has a solution $xin [-9,-7]cup [1,3]$:
answered Dec 29 '18 at 8:35
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
$begingroup$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
Note that:
$$begin{align}4le x+3le 6 &Rightarrow 4le |x+3|le 6, text{but} \
4le |x+3|le 6 ¬Rightarrow 4le x+3le 6, end{align}$$
because $4le x+3le 6$ has a solution $xin [1,3]$:
whereas $4le|x+3|le 6$ has a solution $xin [-9,-7]cup [1,3]$:
answered Dec 29 '18 at 8:35
farruhotafarruhota
20.4k2739
$endgroup$
add a comment |
$begingroup$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
Note that:
$$begin{align}4le x+3le 6 &Rightarrow 4le |x+3|le 6, text{but} \
4le |x+3|le 6 ¬Rightarrow 4le x+3le 6, end{align}$$
because $4le x+3le 6$ has a solution $xin [1,3]$:
whereas $4le|x+3|le 6$ has a solution $xin [-9,-7]cup [1,3]$:
answered Dec 29 '18 at 8:35
farruhotafarruhota
20.4k2739
$endgroup$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
Note that:
$$begin{align}4le x+3le 6 &Rightarrow 4le |x+3|le 6, text{but} \
4le |x+3|le 6 ¬Rightarrow 4le x+3le 6, end{align}$$
because $4le x+3le 6$ has a solution $xin [1,3]$:
whereas $4le|x+3|le 6$ has a solution $xin [-9,-7]cup [1,3]$:
answered Dec 29 '18 at 8:35
farruhotafarruhota
20.4k2739
answered Dec 29 '18 at 8:35
farruhotafarruhota
20.4k2739
answered Dec 29 '18 at 8:35
farruhotafarruhota
20.4k2739
answered Dec 29 '18 at 8:35
farruhotafarruhota
20.4k2739
20.4k2739
add a comment |
add a comment |
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1
$begingroup$
Did you mean $|x-2|leq 1$?
$endgroup$
– Lucas Corrêa
Dec 29 '18 at 2:51
2
$begingroup$
Yes, that's what I meant.
$endgroup$
– user503154
Dec 29 '18 at 2:53