Mixing
Let $a$ be a real number. If $a$ is positive, then $-a$ is negative. Conversely, if $a$ is negative, then...
$begingroup$
Let $a$ be a real number. If $a$ is positive, then $-a$ is negative. Conversely, if $a$ is negative, then $-a$ is positive.
Having a hard time with intuition and the obvious answer getting in the way of my thought process. We are supposed to "verify" the statement above, given Axiom II -- the order axioms.
Since $ainmathbb{R}^{+}$, then $-(-a)$ is positive for all $ainmathbb{R}^{+}$.
Something about $mathbb{R}^{+}$ is confusing me.
real-analysis axioms
asked Jan 17 at 2:31
RyanRyan
1578
$endgroup$
add a comment |
$begingroup$
Let $a$ be a real number. If $a$ is positive, then $-a$ is negative. Conversely, if $a$ is negative, then $-a$ is positive.
Having a hard time with intuition and the obvious answer getting in the way of my thought process. We are supposed to "verify" the statement above, given Axiom II -- the order axioms.
Since $ainmathbb{R}^{+}$, then $-(-a)$ is positive for all $ainmathbb{R}^{+}$.
Something about $mathbb{R}^{+}$ is confusing me.
real-analysis axioms
asked Jan 17 at 2:31
RyanRyan
1578
$endgroup$
$begingroup$
Perhaps you should be thinking along the lines of: “$-x$” means “$x$ with its sign changed”.
$endgroup$
– Lubin
Jan 17 at 2:55
add a comment |
$begingroup$
Let $a$ be a real number. If $a$ is positive, then $-a$ is negative. Conversely, if $a$ is negative, then $-a$ is positive.
Having a hard time with intuition and the obvious answer getting in the way of my thought process. We are supposed to "verify" the statement above, given Axiom II -- the order axioms.
Since $ainmathbb{R}^{+}$, then $-(-a)$ is positive for all $ainmathbb{R}^{+}$.
Something about $mathbb{R}^{+}$ is confusing me.
real-analysis axioms
asked Jan 17 at 2:31
RyanRyan
1578
$endgroup$
Let $a$ be a real number. If $a$ is positive, then $-a$ is negative. Conversely, if $a$ is negative, then $-a$ is positive.
Having a hard time with intuition and the obvious answer getting in the way of my thought process. We are supposed to "verify" the statement above, given Axiom II -- the order axioms.
Since $ainmathbb{R}^{+}$, then $-(-a)$ is positive for all $ainmathbb{R}^{+}$.
Something about $mathbb{R}^{+}$ is confusing me.
real-analysis axioms
real-analysis axioms
asked Jan 17 at 2:31
RyanRyan
1578
asked Jan 17 at 2:31
RyanRyan
1578
edited Jan 17 at 2:45
Ryan
asked Jan 17 at 2:31
RyanRyan
1578
asked Jan 17 at 2:31
RyanRyan
1578
asked Jan 17 at 2:31
RyanRyan
1578
1578
$begingroup$
Perhaps you should be thinking along the lines of: “$-x$” means “$x$ with its sign changed”.
$endgroup$
– Lubin
Jan 17 at 2:55
add a comment |
$begingroup$
Perhaps you should be thinking along the lines of: “$-x$” means “$x$ with its sign changed”.
$endgroup$
– Lubin
Jan 17 at 2:55
$begingroup$
Perhaps you should be thinking along the lines of: “$-x$” means “$x$ with its sign changed”.
$endgroup$
– Lubin
Jan 17 at 2:55
$begingroup$
Perhaps you should be thinking along the lines of: “$-x$” means “$x$ with its sign changed”.
$endgroup$
– Lubin
Jan 17 at 2:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
- By definition, if $a$ is positive then $a>0$.
- By the order axioms, we have that $a+(-a)>0+(-a)$
- This simplifies to $0>-a$ (by definition of additive inverse and additive identity)
- Therefore $-a<0$ and by definition $-a$ is negative.
answered Jan 17 at 2:55
obscuransobscurans
1,152311
$endgroup$
$begingroup$
Thank you! I wasn't sure if we could use Field Axioms, but that was my own brain fart.
$endgroup$
– Ryan
Jan 22 at 17:41
add a comment |
$begingroup$
According to the order axioms, if $a > b$ then $c + a > c + b$.
Thus if $a > 0$ then $-a + a > -a + 0$, i.e., $0 > -a$, i.e., $-a < 0$.
Conversely, if $a < 0$ then $-a + a < -a + 0$, i.e., $0 < -a$, i.e., $-a > 0$.
answered Jan 17 at 3:01
J. W. TannerJ. W. Tanner
2,3761117
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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$begingroup$
- By definition, if $a$ is positive then $a>0$.
- By the order axioms, we have that $a+(-a)>0+(-a)$
- This simplifies to $0>-a$ (by definition of additive inverse and additive identity)
- Therefore $-a<0$ and by definition $-a$ is negative.
answered Jan 17 at 2:55
obscuransobscurans
1,152311
$endgroup$
$begingroup$
Thank you! I wasn't sure if we could use Field Axioms, but that was my own brain fart.
$endgroup$
– Ryan
Jan 22 at 17:41
add a comment |
$begingroup$
- By definition, if $a$ is positive then $a>0$.
- By the order axioms, we have that $a+(-a)>0+(-a)$
- This simplifies to $0>-a$ (by definition of additive inverse and additive identity)
- Therefore $-a<0$ and by definition $-a$ is negative.
answered Jan 17 at 2:55
obscuransobscurans
1,152311
$endgroup$
$begingroup$
Thank you! I wasn't sure if we could use Field Axioms, but that was my own brain fart.
$endgroup$
– Ryan
Jan 22 at 17:41
add a comment |
$begingroup$
- By definition, if $a$ is positive then $a>0$.
- By the order axioms, we have that $a+(-a)>0+(-a)$
- This simplifies to $0>-a$ (by definition of additive inverse and additive identity)
- Therefore $-a<0$ and by definition $-a$ is negative.
answered Jan 17 at 2:55
obscuransobscurans
1,152311
$endgroup$
- By definition, if $a$ is positive then $a>0$.
- By the order axioms, we have that $a+(-a)>0+(-a)$
- This simplifies to $0>-a$ (by definition of additive inverse and additive identity)
- Therefore $-a<0$ and by definition $-a$ is negative.
answered Jan 17 at 2:55
obscuransobscurans
1,152311
answered Jan 17 at 2:55
obscuransobscurans
1,152311
answered Jan 17 at 2:55
obscuransobscurans
1,152311
answered Jan 17 at 2:55
obscuransobscurans
1,152311
1,152311
$begingroup$
Thank you! I wasn't sure if we could use Field Axioms, but that was my own brain fart.
$endgroup$
– Ryan
Jan 22 at 17:41
add a comment |
$begingroup$
Thank you! I wasn't sure if we could use Field Axioms, but that was my own brain fart.
$endgroup$
– Ryan
Jan 22 at 17:41
$begingroup$
Thank you! I wasn't sure if we could use Field Axioms, but that was my own brain fart.
$endgroup$
– Ryan
Jan 22 at 17:41
$begingroup$
Thank you! I wasn't sure if we could use Field Axioms, but that was my own brain fart.
$endgroup$
– Ryan
Jan 22 at 17:41
add a comment |
$begingroup$
According to the order axioms, if $a > b$ then $c + a > c + b$.
Thus if $a > 0$ then $-a + a > -a + 0$, i.e., $0 > -a$, i.e., $-a < 0$.
Conversely, if $a < 0$ then $-a + a < -a + 0$, i.e., $0 < -a$, i.e., $-a > 0$.
answered Jan 17 at 3:01
J. W. TannerJ. W. Tanner
2,3761117
$endgroup$
add a comment |
$begingroup$
According to the order axioms, if $a > b$ then $c + a > c + b$.
Thus if $a > 0$ then $-a + a > -a + 0$, i.e., $0 > -a$, i.e., $-a < 0$.
Conversely, if $a < 0$ then $-a + a < -a + 0$, i.e., $0 < -a$, i.e., $-a > 0$.
answered Jan 17 at 3:01
J. W. TannerJ. W. Tanner
2,3761117
$endgroup$
add a comment |
$begingroup$
According to the order axioms, if $a > b$ then $c + a > c + b$.
Thus if $a > 0$ then $-a + a > -a + 0$, i.e., $0 > -a$, i.e., $-a < 0$.
Conversely, if $a < 0$ then $-a + a < -a + 0$, i.e., $0 < -a$, i.e., $-a > 0$.
answered Jan 17 at 3:01
J. W. TannerJ. W. Tanner
2,3761117
$endgroup$
According to the order axioms, if $a > b$ then $c + a > c + b$.
Thus if $a > 0$ then $-a + a > -a + 0$, i.e., $0 > -a$, i.e., $-a < 0$.
Conversely, if $a < 0$ then $-a + a < -a + 0$, i.e., $0 < -a$, i.e., $-a > 0$.
answered Jan 17 at 3:01
J. W. TannerJ. W. Tanner
2,3761117
answered Jan 17 at 3:01
J. W. TannerJ. W. Tanner
2,3761117
answered Jan 17 at 3:01
J. W. TannerJ. W. Tanner
2,3761117
answered Jan 17 at 3:01
J. W. TannerJ. W. Tanner
2,3761117
2,3761117
add a comment |
add a comment |
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$begingroup$
Perhaps you should be thinking along the lines of: “$-x$” means “$x$ with its sign changed”.
$endgroup$
– Lubin
Jan 17 at 2:55