Mixing
Square roots in inequalities
$begingroup$
If I have the inequality
$$x^2 > b^2,$$
is this always equal to $;|x| > |b|quad ?$
inequality
edited Sep 19 '14 at 17:22
amWhy
1
asked Sep 19 '14 at 17:10
MotherMother
5013
$endgroup$
add a comment |
$begingroup$
If I have the inequality
$$x^2 > b^2,$$
is this always equal to $;|x| > |b|quad ?$
inequality
edited Sep 19 '14 at 17:22
amWhy
1
asked Sep 19 '14 at 17:10
MotherMother
5013
$endgroup$
add a comment |
$begingroup$
If I have the inequality
$$x^2 > b^2,$$
is this always equal to $;|x| > |b|quad ?$
inequality
edited Sep 19 '14 at 17:22
amWhy
1
asked Sep 19 '14 at 17:10
MotherMother
5013
$endgroup$
If I have the inequality
$$x^2 > b^2,$$
is this always equal to $;|x| > |b|quad ?$
inequality
inequality
edited Sep 19 '14 at 17:22
amWhy
1
asked Sep 19 '14 at 17:10
MotherMother
5013
edited Sep 19 '14 at 17:22
amWhy
1
asked Sep 19 '14 at 17:10
MotherMother
5013
edited Sep 19 '14 at 17:22
amWhy
1
edited Sep 19 '14 at 17:22
amWhy
1
edited Sep 19 '14 at 17:22
amWhy
1
1
asked Sep 19 '14 at 17:10
MotherMother
5013
asked Sep 19 '14 at 17:10
MotherMother
5013
asked Sep 19 '14 at 17:10
MotherMother
5013
5013
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Just a clarification here in what is happening, and your use of the term "equal":
$$x^2 gt b^2, text{ if and only if } ,|x| > |b|.$$
That is not to say that the inequalities are equal, but yes, they are equivalent inequalities: they share the same solution set. What's true of one, is true of the other.
answered Sep 19 '14 at 17:19
amWhyamWhy
1
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add a comment |
$begingroup$
Yes because both sides are positive and thus the square root can be taken and then you use the fact that $sqrt {x^2}= |x|$
answered Sep 19 '14 at 17:12
Sheheryar ZaidiSheheryar Zaidi
2,339714
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add a comment |
$begingroup$
Technically this is not true if you are talking about complex numbers. For example, let $b$ be $i$ and let $x$ be $-1$. Then $(-1)^2=1 > i^2=-1$, but $mathopen|-1mathclose|=mathopen|imathclose|=1$.
answered Jan 18 at 2:13
Michael WangMichael Wang
188215
$endgroup$
1
$begingroup$
I'm sure that the OP is talking about $b,xinBbb R$ only. In $Bbb C$, then inequality $x^2>b^2$ usually does not make sense, as it is not possible to define a total order $>$ such that $Bbb C$ is an ordered field.
$endgroup$
– TheSimpliFire
Feb 13 at 9:41
$begingroup$
Oh ok thanks for this.
$endgroup$
– Michael Wang
Feb 14 at 4:17
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just a clarification here in what is happening, and your use of the term "equal":
$$x^2 gt b^2, text{ if and only if } ,|x| > |b|.$$
That is not to say that the inequalities are equal, but yes, they are equivalent inequalities: they share the same solution set. What's true of one, is true of the other.
answered Sep 19 '14 at 17:19
amWhyamWhy
1
$endgroup$
add a comment |
$begingroup$
Just a clarification here in what is happening, and your use of the term "equal":
$$x^2 gt b^2, text{ if and only if } ,|x| > |b|.$$
That is not to say that the inequalities are equal, but yes, they are equivalent inequalities: they share the same solution set. What's true of one, is true of the other.
answered Sep 19 '14 at 17:19
amWhyamWhy
1
$endgroup$
add a comment |
$begingroup$
Just a clarification here in what is happening, and your use of the term "equal":
$$x^2 gt b^2, text{ if and only if } ,|x| > |b|.$$
That is not to say that the inequalities are equal, but yes, they are equivalent inequalities: they share the same solution set. What's true of one, is true of the other.
answered Sep 19 '14 at 17:19
amWhyamWhy
1
$endgroup$
Just a clarification here in what is happening, and your use of the term "equal":
$$x^2 gt b^2, text{ if and only if } ,|x| > |b|.$$
That is not to say that the inequalities are equal, but yes, they are equivalent inequalities: they share the same solution set. What's true of one, is true of the other.
answered Sep 19 '14 at 17:19
amWhyamWhy
1
answered Sep 19 '14 at 17:19
amWhyamWhy
1
answered Sep 19 '14 at 17:19
amWhyamWhy
1
answered Sep 19 '14 at 17:19
amWhyamWhy
1
1
add a comment |
add a comment |
$begingroup$
Yes because both sides are positive and thus the square root can be taken and then you use the fact that $sqrt {x^2}= |x|$
answered Sep 19 '14 at 17:12
Sheheryar ZaidiSheheryar Zaidi
2,339714
$endgroup$
add a comment |
$begingroup$
Yes because both sides are positive and thus the square root can be taken and then you use the fact that $sqrt {x^2}= |x|$
answered Sep 19 '14 at 17:12
Sheheryar ZaidiSheheryar Zaidi
2,339714
$endgroup$
add a comment |
$begingroup$
Yes because both sides are positive and thus the square root can be taken and then you use the fact that $sqrt {x^2}= |x|$
answered Sep 19 '14 at 17:12
Sheheryar ZaidiSheheryar Zaidi
2,339714
$endgroup$
Yes because both sides are positive and thus the square root can be taken and then you use the fact that $sqrt {x^2}= |x|$
answered Sep 19 '14 at 17:12
Sheheryar ZaidiSheheryar Zaidi
2,339714
answered Sep 19 '14 at 17:12
Sheheryar ZaidiSheheryar Zaidi
2,339714
answered Sep 19 '14 at 17:12
Sheheryar ZaidiSheheryar Zaidi
2,339714
answered Sep 19 '14 at 17:12
Sheheryar ZaidiSheheryar Zaidi
2,339714
2,339714
add a comment |
add a comment |
$begingroup$
Technically this is not true if you are talking about complex numbers. For example, let $b$ be $i$ and let $x$ be $-1$. Then $(-1)^2=1 > i^2=-1$, but $mathopen|-1mathclose|=mathopen|imathclose|=1$.
answered Jan 18 at 2:13
Michael WangMichael Wang
188215
$endgroup$
1
$begingroup$
I'm sure that the OP is talking about $b,xinBbb R$ only. In $Bbb C$, then inequality $x^2>b^2$ usually does not make sense, as it is not possible to define a total order $>$ such that $Bbb C$ is an ordered field.
$endgroup$
– TheSimpliFire
Feb 13 at 9:41
$begingroup$
Oh ok thanks for this.
$endgroup$
– Michael Wang
Feb 14 at 4:17
add a comment |
$begingroup$
Technically this is not true if you are talking about complex numbers. For example, let $b$ be $i$ and let $x$ be $-1$. Then $(-1)^2=1 > i^2=-1$, but $mathopen|-1mathclose|=mathopen|imathclose|=1$.
answered Jan 18 at 2:13
Michael WangMichael Wang
188215
$endgroup$
1
$begingroup$
I'm sure that the OP is talking about $b,xinBbb R$ only. In $Bbb C$, then inequality $x^2>b^2$ usually does not make sense, as it is not possible to define a total order $>$ such that $Bbb C$ is an ordered field.
$endgroup$
– TheSimpliFire
Feb 13 at 9:41
$begingroup$
Oh ok thanks for this.
$endgroup$
– Michael Wang
Feb 14 at 4:17
add a comment |
$begingroup$
Technically this is not true if you are talking about complex numbers. For example, let $b$ be $i$ and let $x$ be $-1$. Then $(-1)^2=1 > i^2=-1$, but $mathopen|-1mathclose|=mathopen|imathclose|=1$.
answered Jan 18 at 2:13
Michael WangMichael Wang
188215
$endgroup$
Technically this is not true if you are talking about complex numbers. For example, let $b$ be $i$ and let $x$ be $-1$. Then $(-1)^2=1 > i^2=-1$, but $mathopen|-1mathclose|=mathopen|imathclose|=1$.
answered Jan 18 at 2:13
Michael WangMichael Wang
188215
answered Jan 18 at 2:13
Michael WangMichael Wang
188215
answered Jan 18 at 2:13
Michael WangMichael Wang
188215
answered Jan 18 at 2:13
Michael WangMichael Wang
188215
188215
1
$begingroup$
I'm sure that the OP is talking about $b,xinBbb R$ only. In $Bbb C$, then inequality $x^2>b^2$ usually does not make sense, as it is not possible to define a total order $>$ such that $Bbb C$ is an ordered field.
$endgroup$
– TheSimpliFire
Feb 13 at 9:41
$begingroup$
Oh ok thanks for this.
$endgroup$
– Michael Wang
Feb 14 at 4:17
add a comment |
1
$begingroup$
I'm sure that the OP is talking about $b,xinBbb R$ only. In $Bbb C$, then inequality $x^2>b^2$ usually does not make sense, as it is not possible to define a total order $>$ such that $Bbb C$ is an ordered field.
$endgroup$
– TheSimpliFire
Feb 13 at 9:41
$begingroup$
Oh ok thanks for this.
$endgroup$
– Michael Wang
Feb 14 at 4:17
1
1
$begingroup$
I'm sure that the OP is talking about $b,xinBbb R$ only. In $Bbb C$, then inequality $x^2>b^2$ usually does not make sense, as it is not possible to define a total order $>$ such that $Bbb C$ is an ordered field.
$endgroup$
– TheSimpliFire
Feb 13 at 9:41
$begingroup$
I'm sure that the OP is talking about $b,xinBbb R$ only. In $Bbb C$, then inequality $x^2>b^2$ usually does not make sense, as it is not possible to define a total order $>$ such that $Bbb C$ is an ordered field.
$endgroup$
– TheSimpliFire
Feb 13 at 9:41
$begingroup$
Oh ok thanks for this.
$endgroup$
– Michael Wang
Feb 14 at 4:17
$begingroup$
Oh ok thanks for this.
$endgroup$
– Michael Wang
Feb 14 at 4:17
add a comment |
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