Mixing
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Multiplying inequalities.
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Suppose, I have two inequalities as follows:
$$x^2>1tag1$$
$$y<-1tag2$$
Is it possible to form a new inequality by combining $(1)$ and $(2)$ where the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers?
algebra-precalculus inequality
edited Jul 15 at 10:32
TheSimpliFire
9,69261951
asked Jul 15 at 10:29
yahoo.com
391216
add a comment |Â
up vote
3
down vote
favorite
Suppose, I have two inequalities as follows:
$$x^2>1tag1$$
$$y<-1tag2$$
Is it possible to form a new inequality by combining $(1)$ and $(2)$ where the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers?
algebra-precalculus inequality
edited Jul 15 at 10:32
TheSimpliFire
9,69261951
asked Jul 15 at 10:29
yahoo.com
391216
A common axiom of an ordered filed is "if $0 ≤ a $ and $0 ≤ b$ then $0 ≤ a times b$" which can easily be extended to "if $0 lt a lt c$ and $0 lt b lt d$ then $0 lt a times b lt c times d$" so get your inequalities into that form
– Henry
Jul 15 at 20:04
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose, I have two inequalities as follows:
$$x^2>1tag1$$
$$y<-1tag2$$
Is it possible to form a new inequality by combining $(1)$ and $(2)$ where the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers?
algebra-precalculus inequality
edited Jul 15 at 10:32
TheSimpliFire
9,69261951
asked Jul 15 at 10:29
yahoo.com
391216
Suppose, I have two inequalities as follows:
$$x^2>1tag1$$
$$y<-1tag2$$
Is it possible to form a new inequality by combining $(1)$ and $(2)$ where the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers?
algebra-precalculus inequality
edited Jul 15 at 10:32
TheSimpliFire
9,69261951
asked Jul 15 at 10:29
yahoo.com
391216
edited Jul 15 at 10:32
TheSimpliFire
9,69261951
edited Jul 15 at 10:32
TheSimpliFire
9,69261951
edited Jul 15 at 10:32
TheSimpliFire
9,69261951
9,69261951
asked Jul 15 at 10:29
yahoo.com
391216
asked Jul 15 at 10:29
yahoo.com
391216
asked Jul 15 at 10:29
yahoo.com
391216
391216
A common axiom of an ordered filed is "if $0 ≤ a $ and $0 ≤ b$ then $0 ≤ a times b$" which can easily be extended to "if $0 lt a lt c$ and $0 lt b lt d$ then $0 lt a times b lt c times d$" so get your inequalities into that form
– Henry
Jul 15 at 20:04
add a comment |Â
A common axiom of an ordered filed is "if $0 ≤ a $ and $0 ≤ b$ then $0 ≤ a times b$" which can easily be extended to "if $0 lt a lt c$ and $0 lt b lt d$ then $0 lt a times b lt c times d$" so get your inequalities into that form
– Henry
Jul 15 at 20:04
A common axiom of an ordered filed is "if $0 ≤ a $ and $0 ≤ b$ then $0 ≤ a times b$" which can easily be extended to "if $0 lt a lt c$ and $0 lt b lt d$ then $0 lt a times b lt c times d$" so get your inequalities into that form
– Henry
Jul 15 at 20:04
A common axiom of an ordered filed is "if $0 ≤ a $ and $0 ≤ b$ then $0 ≤ a times b$" which can easily be extended to "if $0 lt a lt c$ and $0 lt b lt d$ then $0 lt a times b lt c times d$" so get your inequalities into that form
– Henry
Jul 15 at 20:04
add a comment |Â
3 Answers
3
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oldest
votes
up vote
8
down vote
accepted
You can multiply $$x^2>1tag1$$ by a positive number without changing the direction of inequality.
In this case since $$ y<-1tag2$$ we have $$-y>1$$ so you may multiply $$x^2>1$$ by $-y$ to get $$-x^2y>-y>1implies-x^2y>1 $$ which is the same as multiplying the two inequalities $$-y>1$$ and $$x^2>1$$
answered Jul 15 at 10:41
Mohammad Riazi-Kermani
27.6k41852
You could then easily get to the question's the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers with $x^2y lt -1$
– Henry
Jul 15 at 20:07
add a comment |Â
up vote
6
down vote
We have $y<-1iff -y>1>0$. So $-y>1$ and $x^2>1$. Notice that both sides of these two inequalities are non-negative, thus we can multiply them. So we have $-y cdot x^2>1 cdot 1=1$.
edited Jul 15 at 12:21
RiaD
641717
answered Jul 15 at 10:37
Le Anh Dung
708318
add a comment |Â
up vote
0
down vote
The two basic axioms are:
If $a < b$ then $a + c < b + c$ for all $c$
If $a < b$ and $c > 0$ then $ac < bc$.
From there there are some basic propositions that can be proven with those two axioms:
If $a > 0$ then $-a < 0$.
If $a < b$ and $c < 0$ then $ac > bc$.
And $1 > 0$. and for $cne 0$ then $c^2 > 0$.
....
So if $x^2 > 1$ and $y < - 1<0$ then $x^2cdot y < 1cdot y = y < -1$.
That's it.
edited Aug 9 at 9:49
TheSimpliFire
9,69261951
answered Jul 15 at 22:06
fleablood
60.5k22575
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
You can multiply $$x^2>1tag1$$ by a positive number without changing the direction of inequality.
In this case since $$ y<-1tag2$$ we have $$-y>1$$ so you may multiply $$x^2>1$$ by $-y$ to get $$-x^2y>-y>1implies-x^2y>1 $$ which is the same as multiplying the two inequalities $$-y>1$$ and $$x^2>1$$
answered Jul 15 at 10:41
Mohammad Riazi-Kermani
27.6k41852
You could then easily get to the question's the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers with $x^2y lt -1$
– Henry
Jul 15 at 20:07
add a comment |Â
up vote
8
down vote
accepted
You can multiply $$x^2>1tag1$$ by a positive number without changing the direction of inequality.
In this case since $$ y<-1tag2$$ we have $$-y>1$$ so you may multiply $$x^2>1$$ by $-y$ to get $$-x^2y>-y>1implies-x^2y>1 $$ which is the same as multiplying the two inequalities $$-y>1$$ and $$x^2>1$$
answered Jul 15 at 10:41
Mohammad Riazi-Kermani
27.6k41852
You could then easily get to the question's the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers with $x^2y lt -1$
– Henry
Jul 15 at 20:07
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
You can multiply $$x^2>1tag1$$ by a positive number without changing the direction of inequality.
In this case since $$ y<-1tag2$$ we have $$-y>1$$ so you may multiply $$x^2>1$$ by $-y$ to get $$-x^2y>-y>1implies-x^2y>1 $$ which is the same as multiplying the two inequalities $$-y>1$$ and $$x^2>1$$
answered Jul 15 at 10:41
Mohammad Riazi-Kermani
27.6k41852
You can multiply $$x^2>1tag1$$ by a positive number without changing the direction of inequality.
In this case since $$ y<-1tag2$$ we have $$-y>1$$ so you may multiply $$x^2>1$$ by $-y$ to get $$-x^2y>-y>1implies-x^2y>1 $$ which is the same as multiplying the two inequalities $$-y>1$$ and $$x^2>1$$
answered Jul 15 at 10:41
Mohammad Riazi-Kermani
27.6k41852
edited Jul 15 at 21:52
answered Jul 15 at 10:41
Mohammad Riazi-Kermani
27.6k41852
answered Jul 15 at 10:41
Mohammad Riazi-Kermani
27.6k41852
answered Jul 15 at 10:41
Mohammad Riazi-Kermani
27.6k41852
27.6k41852
You could then easily get to the question's the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers with $x^2y lt -1$
– Henry
Jul 15 at 20:07
add a comment |Â
You could then easily get to the question's the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers with $x^2y lt -1$
– Henry
Jul 15 at 20:07
You could then easily get to the question's the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers with $x^2y lt -1$
– Henry
Jul 15 at 20:07
You could then easily get to the question's the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers with $x^2y lt -1$
– Henry
Jul 15 at 20:07
add a comment |Â
up vote
6
down vote
We have $y<-1iff -y>1>0$. So $-y>1$ and $x^2>1$. Notice that both sides of these two inequalities are non-negative, thus we can multiply them. So we have $-y cdot x^2>1 cdot 1=1$.
edited Jul 15 at 12:21
RiaD
641717
answered Jul 15 at 10:37
Le Anh Dung
708318
add a comment |Â
up vote
6
down vote
We have $y<-1iff -y>1>0$. So $-y>1$ and $x^2>1$. Notice that both sides of these two inequalities are non-negative, thus we can multiply them. So we have $-y cdot x^2>1 cdot 1=1$.
edited Jul 15 at 12:21
RiaD
641717
answered Jul 15 at 10:37
Le Anh Dung
708318
add a comment |Â
up vote
6
down vote
up vote
6
down vote
We have $y<-1iff -y>1>0$. So $-y>1$ and $x^2>1$. Notice that both sides of these two inequalities are non-negative, thus we can multiply them. So we have $-y cdot x^2>1 cdot 1=1$.
edited Jul 15 at 12:21
RiaD
641717
answered Jul 15 at 10:37
Le Anh Dung
708318
We have $y<-1iff -y>1>0$. So $-y>1$ and $x^2>1$. Notice that both sides of these two inequalities are non-negative, thus we can multiply them. So we have $-y cdot x^2>1 cdot 1=1$.
edited Jul 15 at 12:21
RiaD
641717
answered Jul 15 at 10:37
Le Anh Dung
708318
edited Jul 15 at 12:21
RiaD
641717
edited Jul 15 at 12:21
RiaD
641717
edited Jul 15 at 12:21
RiaD
641717
641717
answered Jul 15 at 10:37
Le Anh Dung
708318
answered Jul 15 at 10:37
Le Anh Dung
708318
answered Jul 15 at 10:37
Le Anh Dung
708318
708318
add a comment |Â
add a comment |Â
up vote
0
down vote
The two basic axioms are:
If $a < b$ then $a + c < b + c$ for all $c$
If $a < b$ and $c > 0$ then $ac < bc$.
From there there are some basic propositions that can be proven with those two axioms:
If $a > 0$ then $-a < 0$.
If $a < b$ and $c < 0$ then $ac > bc$.
And $1 > 0$. and for $cne 0$ then $c^2 > 0$.
....
So if $x^2 > 1$ and $y < - 1<0$ then $x^2cdot y < 1cdot y = y < -1$.
That's it.
edited Aug 9 at 9:49
TheSimpliFire
9,69261951
answered Jul 15 at 22:06
fleablood
60.5k22575
add a comment |Â
up vote
0
down vote
The two basic axioms are:
If $a < b$ then $a + c < b + c$ for all $c$
If $a < b$ and $c > 0$ then $ac < bc$.
From there there are some basic propositions that can be proven with those two axioms:
If $a > 0$ then $-a < 0$.
If $a < b$ and $c < 0$ then $ac > bc$.
And $1 > 0$. and for $cne 0$ then $c^2 > 0$.
....
So if $x^2 > 1$ and $y < - 1<0$ then $x^2cdot y < 1cdot y = y < -1$.
That's it.
edited Aug 9 at 9:49
TheSimpliFire
9,69261951
answered Jul 15 at 22:06
fleablood
60.5k22575
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The two basic axioms are:
If $a < b$ then $a + c < b + c$ for all $c$
If $a < b$ and $c > 0$ then $ac < bc$.
From there there are some basic propositions that can be proven with those two axioms:
If $a > 0$ then $-a < 0$.
If $a < b$ and $c < 0$ then $ac > bc$.
And $1 > 0$. and for $cne 0$ then $c^2 > 0$.
....
So if $x^2 > 1$ and $y < - 1<0$ then $x^2cdot y < 1cdot y = y < -1$.
That's it.
edited Aug 9 at 9:49
TheSimpliFire
9,69261951
answered Jul 15 at 22:06
fleablood
60.5k22575
The two basic axioms are:
If $a < b$ then $a + c < b + c$ for all $c$
If $a < b$ and $c > 0$ then $ac < bc$.
From there there are some basic propositions that can be proven with those two axioms:
If $a > 0$ then $-a < 0$.
If $a < b$ and $c < 0$ then $ac > bc$.
And $1 > 0$. and for $cne 0$ then $c^2 > 0$.
....
So if $x^2 > 1$ and $y < - 1<0$ then $x^2cdot y < 1cdot y = y < -1$.
That's it.
edited Aug 9 at 9:49
TheSimpliFire
9,69261951
answered Jul 15 at 22:06
fleablood
60.5k22575
edited Aug 9 at 9:49
TheSimpliFire
9,69261951
edited Aug 9 at 9:49
TheSimpliFire
9,69261951
edited Aug 9 at 9:49
TheSimpliFire
9,69261951
9,69261951
answered Jul 15 at 22:06
fleablood
60.5k22575
answered Jul 15 at 22:06
fleablood
60.5k22575
answered Jul 15 at 22:06
fleablood
60.5k22575
60.5k22575
add a comment |Â
add a comment |Â
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A common axiom of an ordered filed is "if $0 ≤ a $ and $0 ≤ b$ then $0 ≤ a times b$" which can easily be extended to "if $0 lt a lt c$ and $0 lt b lt d$ then $0 lt a times b lt c times d$" so get your inequalities into that form
– Henry
Jul 15 at 20:04