Mixing
![How to factorize $a(y) := xy^3+xy^2+(x+1)y+x in GF(2)[x]_x^2+x+1[y]$](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
How to prove $x^2<4 implies|x|<2$?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
How to prove $x^2<4 implies|x|<2$?
I don't know exactly there is a proof for this or we take this as an axiom. Please help.
What I did so far is, $$x^2-4<0$$
$$(x-2)(x+2)<0$$
$$-2<x<2$$
From this step can we directly say? $|x|<2$
inequality absolute-value
edited Aug 6 at 17:35
Math Lover
12.4k21232
asked Aug 6 at 16:30
emil
324310
add a comment |Â
up vote
1
down vote
favorite
How to prove $x^2<4 implies|x|<2$?
I don't know exactly there is a proof for this or we take this as an axiom. Please help.
What I did so far is, $$x^2-4<0$$
$$(x-2)(x+2)<0$$
$$-2<x<2$$
From this step can we directly say? $|x|<2$
inequality absolute-value
edited Aug 6 at 17:35
Math Lover
12.4k21232
asked Aug 6 at 16:30
emil
324310
2
Yep, your working is alright. You can just directly conclude.
– Karn Watcharasupat
Aug 6 at 16:33
2
$|x|$ means on the cartesian plane the distance of $x $ from origin on the X axis. You got the values of x between -2 to 2. An either way of writing that will be that your solution is the $x $ situated at not more that 2 units distance from origin on the X axis.
– Jasmine
Aug 6 at 17:52
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How to prove $x^2<4 implies|x|<2$?
I don't know exactly there is a proof for this or we take this as an axiom. Please help.
What I did so far is, $$x^2-4<0$$
$$(x-2)(x+2)<0$$
$$-2<x<2$$
From this step can we directly say? $|x|<2$
inequality absolute-value
edited Aug 6 at 17:35
Math Lover
12.4k21232
asked Aug 6 at 16:30
emil
324310
How to prove $x^2<4 implies|x|<2$?
I don't know exactly there is a proof for this or we take this as an axiom. Please help.
What I did so far is, $$x^2-4<0$$
$$(x-2)(x+2)<0$$
$$-2<x<2$$
From this step can we directly say? $|x|<2$
inequality absolute-value
edited Aug 6 at 17:35
Math Lover
12.4k21232
asked Aug 6 at 16:30
emil
324310
edited Aug 6 at 17:35
Math Lover
12.4k21232
edited Aug 6 at 17:35
Math Lover
12.4k21232
edited Aug 6 at 17:35
Math Lover
12.4k21232
12.4k21232
asked Aug 6 at 16:30
emil
324310
asked Aug 6 at 16:30
emil
324310
asked Aug 6 at 16:30
emil
324310
324310
2
Yep, your working is alright. You can just directly conclude.
– Karn Watcharasupat
Aug 6 at 16:33
2
$|x|$ means on the cartesian plane the distance of $x $ from origin on the X axis. You got the values of x between -2 to 2. An either way of writing that will be that your solution is the $x $ situated at not more that 2 units distance from origin on the X axis.
– Jasmine
Aug 6 at 17:52
add a comment |Â
2
Yep, your working is alright. You can just directly conclude.
– Karn Watcharasupat
Aug 6 at 16:33
2
$|x|$ means on the cartesian plane the distance of $x $ from origin on the X axis. You got the values of x between -2 to 2. An either way of writing that will be that your solution is the $x $ situated at not more that 2 units distance from origin on the X axis.
– Jasmine
Aug 6 at 17:52
2
2
Yep, your working is alright. You can just directly conclude.
– Karn Watcharasupat
Aug 6 at 16:33
Yep, your working is alright. You can just directly conclude.
– Karn Watcharasupat
Aug 6 at 16:33
2
2
$|x|$ means on the cartesian plane the distance of $x $ from origin on the X axis. You got the values of x between -2 to 2. An either way of writing that will be that your solution is the $x $ situated at not more that 2 units distance from origin on the X axis.
– Jasmine
Aug 6 at 17:52
$|x|$ means on the cartesian plane the distance of $x $ from origin on the X axis. You got the values of x between -2 to 2. An either way of writing that will be that your solution is the $x $ situated at not more that 2 units distance from origin on the X axis.
– Jasmine
Aug 6 at 17:52
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
What I did so far is, $$x^2-4<0$$
$$(x-2)(x+2)<0$$
Since $,x^2=|x|^2,$, you could also factor it as:
$$left(|x|-2right)left(|x|+2right) lt 0$$
Given that $,|x|+2 gt 0,$ it follows that $,|x| - 2lt 0,$.
answered Aug 6 at 16:45
dxiv
54.2k64797
add a comment |Â
up vote
4
down vote
Hint:
- Square root monotonic grows (proof is trivial)
- $sqrtx^2=|x|$ (follows directly from the definition of $sqrt$)
Alternative way, continuing what you started:
- $acdot b lt 0 leftrightarrow (a lt 0 land b gt 0) lor (a gt 0 land b lt 0)$
- So you have to find the interval where $(x-2)(x+2)lt 0$
edited Aug 6 at 17:36
Math Lover
12.4k21232
answered Aug 6 at 16:35
peterh
2,14531631
add a comment |Â
up vote
3
down vote
Saying $-2lt x lt 2$ is exactly the same thing as saying $|x|lt 2$. Here's why: from the definition of absolute value we have that $$|x|=begincasesx&xgt 0\-x&xleq0endcases$$ so the inequality $|x|lt 2$ becomes $$begincasesxlt 2&x>0\-xlt 2& xleq0endcases$$ the second inequality can be simplified changing the sing, which changes the direction of the inequality, so $$xgt -2;;;xleq 0$$ In the end you can take the union of the solutions to have a solution so that the won't be disjoint $$-2lt xleq 0;; cup0lt xlt 2;; implies -2lt xlt2$$
answered Aug 6 at 16:41
Davide Morgante
1,942220
add a comment |Â
up vote
1
down vote
The implication is stronger, indeed we have
$$x^2<4 iff x^2-4<0 iff (x+2)(x-2)<0 iff -2< x < 2 iff |x|<2$$
indeed
for $xge 0$ we have $|x|<2 iff x<2iff 0le x<2$
for $x< 0$ we have $|x|<2 iff -x<2iff x>-2iff -2<x<0$
therefore the last implication follows.
answered Aug 6 at 17:00
gimusi
65.4k73684
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
What I did so far is, $$x^2-4<0$$
$$(x-2)(x+2)<0$$
Since $,x^2=|x|^2,$, you could also factor it as:
$$left(|x|-2right)left(|x|+2right) lt 0$$
Given that $,|x|+2 gt 0,$ it follows that $,|x| - 2lt 0,$.
answered Aug 6 at 16:45
dxiv
54.2k64797
add a comment |Â
up vote
4
down vote
accepted
What I did so far is, $$x^2-4<0$$
$$(x-2)(x+2)<0$$
Since $,x^2=|x|^2,$, you could also factor it as:
$$left(|x|-2right)left(|x|+2right) lt 0$$
Given that $,|x|+2 gt 0,$ it follows that $,|x| - 2lt 0,$.
answered Aug 6 at 16:45
dxiv
54.2k64797
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
What I did so far is, $$x^2-4<0$$
$$(x-2)(x+2)<0$$
Since $,x^2=|x|^2,$, you could also factor it as:
$$left(|x|-2right)left(|x|+2right) lt 0$$
Given that $,|x|+2 gt 0,$ it follows that $,|x| - 2lt 0,$.
answered Aug 6 at 16:45
dxiv
54.2k64797
What I did so far is, $$x^2-4<0$$
$$(x-2)(x+2)<0$$
Since $,x^2=|x|^2,$, you could also factor it as:
$$left(|x|-2right)left(|x|+2right) lt 0$$
Given that $,|x|+2 gt 0,$ it follows that $,|x| - 2lt 0,$.
answered Aug 6 at 16:45
dxiv
54.2k64797
answered Aug 6 at 16:45
dxiv
54.2k64797
answered Aug 6 at 16:45
dxiv
54.2k64797
answered Aug 6 at 16:45
dxiv
54.2k64797
54.2k64797
add a comment |Â
add a comment |Â
up vote
4
down vote
Hint:
- Square root monotonic grows (proof is trivial)
- $sqrtx^2=|x|$ (follows directly from the definition of $sqrt$)
Alternative way, continuing what you started:
- $acdot b lt 0 leftrightarrow (a lt 0 land b gt 0) lor (a gt 0 land b lt 0)$
- So you have to find the interval where $(x-2)(x+2)lt 0$
edited Aug 6 at 17:36
Math Lover
12.4k21232
answered Aug 6 at 16:35
peterh
2,14531631
add a comment |Â
up vote
4
down vote
Hint:
- Square root monotonic grows (proof is trivial)
- $sqrtx^2=|x|$ (follows directly from the definition of $sqrt$)
Alternative way, continuing what you started:
- $acdot b lt 0 leftrightarrow (a lt 0 land b gt 0) lor (a gt 0 land b lt 0)$
- So you have to find the interval where $(x-2)(x+2)lt 0$
edited Aug 6 at 17:36
Math Lover
12.4k21232
answered Aug 6 at 16:35
peterh
2,14531631
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Hint:
- Square root monotonic grows (proof is trivial)
- $sqrtx^2=|x|$ (follows directly from the definition of $sqrt$)
Alternative way, continuing what you started:
- $acdot b lt 0 leftrightarrow (a lt 0 land b gt 0) lor (a gt 0 land b lt 0)$
- So you have to find the interval where $(x-2)(x+2)lt 0$
edited Aug 6 at 17:36
Math Lover
12.4k21232
answered Aug 6 at 16:35
peterh
2,14531631
Hint:
- Square root monotonic grows (proof is trivial)
- $sqrtx^2=|x|$ (follows directly from the definition of $sqrt$)
Alternative way, continuing what you started:
- $acdot b lt 0 leftrightarrow (a lt 0 land b gt 0) lor (a gt 0 land b lt 0)$
- So you have to find the interval where $(x-2)(x+2)lt 0$
edited Aug 6 at 17:36
Math Lover
12.4k21232
answered Aug 6 at 16:35
peterh
2,14531631
edited Aug 6 at 17:36
Math Lover
12.4k21232
edited Aug 6 at 17:36
Math Lover
12.4k21232
edited Aug 6 at 17:36
Math Lover
12.4k21232
12.4k21232
answered Aug 6 at 16:35
peterh
2,14531631
answered Aug 6 at 16:35
peterh
2,14531631
answered Aug 6 at 16:35
peterh
2,14531631
2,14531631
add a comment |Â
add a comment |Â
up vote
3
down vote
Saying $-2lt x lt 2$ is exactly the same thing as saying $|x|lt 2$. Here's why: from the definition of absolute value we have that $$|x|=begincasesx&xgt 0\-x&xleq0endcases$$ so the inequality $|x|lt 2$ becomes $$begincasesxlt 2&x>0\-xlt 2& xleq0endcases$$ the second inequality can be simplified changing the sing, which changes the direction of the inequality, so $$xgt -2;;;xleq 0$$ In the end you can take the union of the solutions to have a solution so that the won't be disjoint $$-2lt xleq 0;; cup0lt xlt 2;; implies -2lt xlt2$$
answered Aug 6 at 16:41
Davide Morgante
1,942220
add a comment |Â
up vote
3
down vote
Saying $-2lt x lt 2$ is exactly the same thing as saying $|x|lt 2$. Here's why: from the definition of absolute value we have that $$|x|=begincasesx&xgt 0\-x&xleq0endcases$$ so the inequality $|x|lt 2$ becomes $$begincasesxlt 2&x>0\-xlt 2& xleq0endcases$$ the second inequality can be simplified changing the sing, which changes the direction of the inequality, so $$xgt -2;;;xleq 0$$ In the end you can take the union of the solutions to have a solution so that the won't be disjoint $$-2lt xleq 0;; cup0lt xlt 2;; implies -2lt xlt2$$
answered Aug 6 at 16:41
Davide Morgante
1,942220
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Saying $-2lt x lt 2$ is exactly the same thing as saying $|x|lt 2$. Here's why: from the definition of absolute value we have that $$|x|=begincasesx&xgt 0\-x&xleq0endcases$$ so the inequality $|x|lt 2$ becomes $$begincasesxlt 2&x>0\-xlt 2& xleq0endcases$$ the second inequality can be simplified changing the sing, which changes the direction of the inequality, so $$xgt -2;;;xleq 0$$ In the end you can take the union of the solutions to have a solution so that the won't be disjoint $$-2lt xleq 0;; cup0lt xlt 2;; implies -2lt xlt2$$
answered Aug 6 at 16:41
Davide Morgante
1,942220
Saying $-2lt x lt 2$ is exactly the same thing as saying $|x|lt 2$. Here's why: from the definition of absolute value we have that $$|x|=begincasesx&xgt 0\-x&xleq0endcases$$ so the inequality $|x|lt 2$ becomes $$begincasesxlt 2&x>0\-xlt 2& xleq0endcases$$ the second inequality can be simplified changing the sing, which changes the direction of the inequality, so $$xgt -2;;;xleq 0$$ In the end you can take the union of the solutions to have a solution so that the won't be disjoint $$-2lt xleq 0;; cup0lt xlt 2;; implies -2lt xlt2$$
answered Aug 6 at 16:41
Davide Morgante
1,942220
answered Aug 6 at 16:41
Davide Morgante
1,942220
answered Aug 6 at 16:41
Davide Morgante
1,942220
answered Aug 6 at 16:41
Davide Morgante
1,942220
1,942220
add a comment |Â
add a comment |Â
up vote
1
down vote
The implication is stronger, indeed we have
$$x^2<4 iff x^2-4<0 iff (x+2)(x-2)<0 iff -2< x < 2 iff |x|<2$$
indeed
for $xge 0$ we have $|x|<2 iff x<2iff 0le x<2$
for $x< 0$ we have $|x|<2 iff -x<2iff x>-2iff -2<x<0$
therefore the last implication follows.
answered Aug 6 at 17:00
gimusi
65.4k73684
add a comment |Â
up vote
1
down vote
The implication is stronger, indeed we have
$$x^2<4 iff x^2-4<0 iff (x+2)(x-2)<0 iff -2< x < 2 iff |x|<2$$
indeed
for $xge 0$ we have $|x|<2 iff x<2iff 0le x<2$
for $x< 0$ we have $|x|<2 iff -x<2iff x>-2iff -2<x<0$
therefore the last implication follows.
answered Aug 6 at 17:00
gimusi
65.4k73684
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The implication is stronger, indeed we have
$$x^2<4 iff x^2-4<0 iff (x+2)(x-2)<0 iff -2< x < 2 iff |x|<2$$
indeed
for $xge 0$ we have $|x|<2 iff x<2iff 0le x<2$
for $x< 0$ we have $|x|<2 iff -x<2iff x>-2iff -2<x<0$
therefore the last implication follows.
answered Aug 6 at 17:00
gimusi
65.4k73684
The implication is stronger, indeed we have
$$x^2<4 iff x^2-4<0 iff (x+2)(x-2)<0 iff -2< x < 2 iff |x|<2$$
indeed
for $xge 0$ we have $|x|<2 iff x<2iff 0le x<2$
for $x< 0$ we have $|x|<2 iff -x<2iff x>-2iff -2<x<0$
therefore the last implication follows.
answered Aug 6 at 17:00
gimusi
65.4k73684
edited Aug 6 at 17:53
answered Aug 6 at 17:00
gimusi
65.4k73684
answered Aug 6 at 17:00
gimusi
65.4k73684
answered Aug 6 at 17:00
gimusi
65.4k73684
65.4k73684
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2874062%2fhow-to-prove-x24-impliesx2%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Yep, your working is alright. You can just directly conclude.
– Karn Watcharasupat
Aug 6 at 16:33
2
$|x|$ means on the cartesian plane the distance of $x $ from origin on the X axis. You got the values of x between -2 to 2. An either way of writing that will be that your solution is the $x $ situated at not more that 2 units distance from origin on the X axis.
– Jasmine
Aug 6 at 17:52