Mixing
Why do inequalities flip signs?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Is there a mathematical reason (like a proof) of why this happens? You can do it with examples and it is 'intuitive.' But the proof of why this happens is never shown in pedagogy, we just warn students to remember to flip the inequality when
- multiply or divide by a negative number both sides
$$-2>-3 => 2 < 3$$
- take reciprocals of same sign fractions both sides
$$frac34 > frac12 => frac43 < 2$$
proofs
asked 3 hours ago
Lenny
1987
add a comment |Â
up vote
1
down vote
favorite
Is there a mathematical reason (like a proof) of why this happens? You can do it with examples and it is 'intuitive.' But the proof of why this happens is never shown in pedagogy, we just warn students to remember to flip the inequality when
- multiply or divide by a negative number both sides
$$-2>-3 => 2 < 3$$
- take reciprocals of same sign fractions both sides
$$frac34 > frac12 => frac43 < 2$$
proofs
asked 3 hours ago
Lenny
1987
Does your second example work if we begin with $3 > -4$?
– Nick C
3 hours ago
o no, I gotta specify same sign
– Lenny
3 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there a mathematical reason (like a proof) of why this happens? You can do it with examples and it is 'intuitive.' But the proof of why this happens is never shown in pedagogy, we just warn students to remember to flip the inequality when
- multiply or divide by a negative number both sides
$$-2>-3 => 2 < 3$$
- take reciprocals of same sign fractions both sides
$$frac34 > frac12 => frac43 < 2$$
proofs
asked 3 hours ago
Lenny
1987
Is there a mathematical reason (like a proof) of why this happens? You can do it with examples and it is 'intuitive.' But the proof of why this happens is never shown in pedagogy, we just warn students to remember to flip the inequality when
- multiply or divide by a negative number both sides
$$-2>-3 => 2 < 3$$
- take reciprocals of same sign fractions both sides
$$frac34 > frac12 => frac43 < 2$$
proofs
proofs
asked 3 hours ago
Lenny
1987
asked 3 hours ago
Lenny
1987
edited 3 hours ago
asked 3 hours ago
Lenny
1987
asked 3 hours ago
Lenny
1987
asked 3 hours ago
Lenny
1987
1987
Does your second example work if we begin with $3 > -4$?
– Nick C
3 hours ago
o no, I gotta specify same sign
– Lenny
3 hours ago
add a comment |Â
Does your second example work if we begin with $3 > -4$?
– Nick C
3 hours ago
o no, I gotta specify same sign
– Lenny
3 hours ago
Does your second example work if we begin with $3 > -4$?
– Nick C
3 hours ago
Does your second example work if we begin with $3 > -4$?
– Nick C
3 hours ago
o no, I gotta specify same sign
– Lenny
3 hours ago
o no, I gotta specify same sign
– Lenny
3 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Depending on the context and the previous curriculum, the following might work:
- "less than" means "to the left of" on the number line.
- Multiplying by a negative number flips numbers around 0.
- Thus, "left of" becomes "right of", or "greater than".
answered 2 hours ago
Jasper
43327
add a comment |Â
up vote
1
down vote
I'm slightly concerned that
Is there a mathematical reason (like a proof) of why this happens?
is a purely mathematical question, but since you write "we just warn students" I will assume that this question is purposefully asked here on Math Educators StackExchange.
As to a proof:
Given $a>b$, subtract $a$ from both sides: $0 > b-a$.
Next, subtract $b$ from both sides: $-b > -a$.
Note that this final inequality is equivalent to $-a < -b$.
And so we have proved: If $a > b$, then $-a < -b$.
answered 14 mins ago
Benjamin Dickman
15.4k22791
add a comment |Â
up vote
0
down vote
For multiplying or dividing by -1...
$$beginalign
a&>b\
a-b&>0\
\-(a-b)&<0\
-a&<-b
endalign
$$
(You can then extended to arbitrary negative numbers by multiplying or dividing by the [positive] magnitude.)
For taking reciprocals... assuming $ab>0$
$$beginalign
a&>b\
left(frac1abright)a&>left(frac1abright)b\
frac1b&>frac1a\
frac1a&<frac1b\
endalign
$$
answered 16 mins ago
robphy
101
New contributor
robphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Depending on the context and the previous curriculum, the following might work:
- "less than" means "to the left of" on the number line.
- Multiplying by a negative number flips numbers around 0.
- Thus, "left of" becomes "right of", or "greater than".
answered 2 hours ago
Jasper
43327
add a comment |Â
up vote
2
down vote
Depending on the context and the previous curriculum, the following might work:
- "less than" means "to the left of" on the number line.
- Multiplying by a negative number flips numbers around 0.
- Thus, "left of" becomes "right of", or "greater than".
answered 2 hours ago
Jasper
43327
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Depending on the context and the previous curriculum, the following might work:
- "less than" means "to the left of" on the number line.
- Multiplying by a negative number flips numbers around 0.
- Thus, "left of" becomes "right of", or "greater than".
answered 2 hours ago
Jasper
43327
Depending on the context and the previous curriculum, the following might work:
- "less than" means "to the left of" on the number line.
- Multiplying by a negative number flips numbers around 0.
- Thus, "left of" becomes "right of", or "greater than".
answered 2 hours ago
Jasper
43327
answered 2 hours ago
Jasper
43327
answered 2 hours ago
Jasper
43327
answered 2 hours ago
Jasper
43327
43327
add a comment |Â
add a comment |Â
up vote
1
down vote
I'm slightly concerned that
Is there a mathematical reason (like a proof) of why this happens?
is a purely mathematical question, but since you write "we just warn students" I will assume that this question is purposefully asked here on Math Educators StackExchange.
As to a proof:
Given $a>b$, subtract $a$ from both sides: $0 > b-a$.
Next, subtract $b$ from both sides: $-b > -a$.
Note that this final inequality is equivalent to $-a < -b$.
And so we have proved: If $a > b$, then $-a < -b$.
answered 14 mins ago
Benjamin Dickman
15.4k22791
add a comment |Â
up vote
1
down vote
I'm slightly concerned that
Is there a mathematical reason (like a proof) of why this happens?
is a purely mathematical question, but since you write "we just warn students" I will assume that this question is purposefully asked here on Math Educators StackExchange.
As to a proof:
Given $a>b$, subtract $a$ from both sides: $0 > b-a$.
Next, subtract $b$ from both sides: $-b > -a$.
Note that this final inequality is equivalent to $-a < -b$.
And so we have proved: If $a > b$, then $-a < -b$.
answered 14 mins ago
Benjamin Dickman
15.4k22791
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I'm slightly concerned that
Is there a mathematical reason (like a proof) of why this happens?
is a purely mathematical question, but since you write "we just warn students" I will assume that this question is purposefully asked here on Math Educators StackExchange.
As to a proof:
Given $a>b$, subtract $a$ from both sides: $0 > b-a$.
Next, subtract $b$ from both sides: $-b > -a$.
Note that this final inequality is equivalent to $-a < -b$.
And so we have proved: If $a > b$, then $-a < -b$.
answered 14 mins ago
Benjamin Dickman
15.4k22791
I'm slightly concerned that
Is there a mathematical reason (like a proof) of why this happens?
is a purely mathematical question, but since you write "we just warn students" I will assume that this question is purposefully asked here on Math Educators StackExchange.
As to a proof:
Given $a>b$, subtract $a$ from both sides: $0 > b-a$.
Next, subtract $b$ from both sides: $-b > -a$.
Note that this final inequality is equivalent to $-a < -b$.
And so we have proved: If $a > b$, then $-a < -b$.
answered 14 mins ago
Benjamin Dickman
15.4k22791
answered 14 mins ago
Benjamin Dickman
15.4k22791
answered 14 mins ago
Benjamin Dickman
15.4k22791
answered 14 mins ago
Benjamin Dickman
15.4k22791
15.4k22791
add a comment |Â
add a comment |Â
up vote
0
down vote
For multiplying or dividing by -1...
$$beginalign
a&>b\
a-b&>0\
\-(a-b)&<0\
-a&<-b
endalign
$$
(You can then extended to arbitrary negative numbers by multiplying or dividing by the [positive] magnitude.)
For taking reciprocals... assuming $ab>0$
$$beginalign
a&>b\
left(frac1abright)a&>left(frac1abright)b\
frac1b&>frac1a\
frac1a&<frac1b\
endalign
$$
answered 16 mins ago
robphy
101
New contributor
robphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
For multiplying or dividing by -1...
$$beginalign
a&>b\
a-b&>0\
\-(a-b)&<0\
-a&<-b
endalign
$$
(You can then extended to arbitrary negative numbers by multiplying or dividing by the [positive] magnitude.)
For taking reciprocals... assuming $ab>0$
$$beginalign
a&>b\
left(frac1abright)a&>left(frac1abright)b\
frac1b&>frac1a\
frac1a&<frac1b\
endalign
$$
answered 16 mins ago
robphy
101
New contributor
robphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For multiplying or dividing by -1...
$$beginalign
a&>b\
a-b&>0\
\-(a-b)&<0\
-a&<-b
endalign
$$
(You can then extended to arbitrary negative numbers by multiplying or dividing by the [positive] magnitude.)
For taking reciprocals... assuming $ab>0$
$$beginalign
a&>b\
left(frac1abright)a&>left(frac1abright)b\
frac1b&>frac1a\
frac1a&<frac1b\
endalign
$$
answered 16 mins ago
robphy
101
New contributor
robphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For multiplying or dividing by -1...
$$beginalign
a&>b\
a-b&>0\
\-(a-b)&<0\
-a&<-b
endalign
$$
(You can then extended to arbitrary negative numbers by multiplying or dividing by the [positive] magnitude.)
For taking reciprocals... assuming $ab>0$
$$beginalign
a&>b\
left(frac1abright)a&>left(frac1abright)b\
frac1b&>frac1a\
frac1a&<frac1b\
endalign
$$
answered 16 mins ago
robphy
101
New contributor
robphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 16 mins ago
robphy
101
New contributor
robphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 16 mins ago
robphy
101
answered 16 mins ago
robphy
101
101
New contributor
robphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
robphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
robphy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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%2fmatheducators.stackexchange.com%2fquestions%2f14751%2fwhy-do-inequalities-flip-signs%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
Does your second example work if we begin with $3 > -4$?
– Nick C
3 hours ago
o no, I gotta specify same sign
– Lenny
3 hours ago