Is this a valid way to prove that $x > 4$ $Rightarrow$ $x^2 > 9$
up vote
1
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$x > 4$
$rightarrow$ $x - 1 > 3$
$rightarrow$ $(x - 1)^2 > 9$
and obviously if $(x - 1)^2 > 9$ then $x^2 > 9$
algebra-precalculus
edited Nov 18 at 4:51
gt6989b
32.1k22351
asked Nov 18 at 4:43
ming
374
add a comment |
up vote
1
down vote
favorite
$x > 4$
$rightarrow$ $x - 1 > 3$
$rightarrow$ $(x - 1)^2 > 9$
and obviously if $(x - 1)^2 > 9$ then $x^2 > 9$
algebra-precalculus
edited Nov 18 at 4:51
gt6989b
32.1k22351
asked Nov 18 at 4:43
ming
374
See: math.stackexchange.com/questions/568780/…
– NoChance
Nov 18 at 4:50
It's not true that $(x-1)^{2}>9$ implies that $x^{2}>9$. For example, if $x=-2.5$, then $(x-1)^{2}>9$ but $x^{2}<9$.
– Brian Borchers
Nov 18 at 4:52
@BrianBorchers $x>4$ from the start.
– Rócherz
Nov 18 at 4:53
1
I guess somewhere you need to use that $x>0$ in an essential way.
– Andres Mejia
Nov 18 at 4:53
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$x > 4$
$rightarrow$ $x - 1 > 3$
$rightarrow$ $(x - 1)^2 > 9$
and obviously if $(x - 1)^2 > 9$ then $x^2 > 9$
algebra-precalculus
edited Nov 18 at 4:51
gt6989b
32.1k22351
asked Nov 18 at 4:43
ming
374
$x > 4$
$rightarrow$ $x - 1 > 3$
$rightarrow$ $(x - 1)^2 > 9$
and obviously if $(x - 1)^2 > 9$ then $x^2 > 9$
algebra-precalculus
algebra-precalculus
edited Nov 18 at 4:51
gt6989b
32.1k22351
asked Nov 18 at 4:43
ming
374
edited Nov 18 at 4:51
gt6989b
32.1k22351
asked Nov 18 at 4:43
ming
374
edited Nov 18 at 4:51
gt6989b
32.1k22351
edited Nov 18 at 4:51
gt6989b
32.1k22351
edited Nov 18 at 4:51
gt6989b
32.1k22351
32.1k22351
asked Nov 18 at 4:43
ming
374
asked Nov 18 at 4:43
ming
374
asked Nov 18 at 4:43
ming
374
374
See: math.stackexchange.com/questions/568780/…
– NoChance
Nov 18 at 4:50
It's not true that $(x-1)^{2}>9$ implies that $x^{2}>9$. For example, if $x=-2.5$, then $(x-1)^{2}>9$ but $x^{2}<9$.
– Brian Borchers
Nov 18 at 4:52
@BrianBorchers $x>4$ from the start.
– Rócherz
Nov 18 at 4:53
1
I guess somewhere you need to use that $x>0$ in an essential way.
– Andres Mejia
Nov 18 at 4:53
add a comment |
See: math.stackexchange.com/questions/568780/…
– NoChance
Nov 18 at 4:50
It's not true that $(x-1)^{2}>9$ implies that $x^{2}>9$. For example, if $x=-2.5$, then $(x-1)^{2}>9$ but $x^{2}<9$.
– Brian Borchers
Nov 18 at 4:52
@BrianBorchers $x>4$ from the start.
– Rócherz
Nov 18 at 4:53
1
I guess somewhere you need to use that $x>0$ in an essential way.
– Andres Mejia
Nov 18 at 4:53
See: math.stackexchange.com/questions/568780/…
– NoChance
Nov 18 at 4:50
See: math.stackexchange.com/questions/568780/…
– NoChance
Nov 18 at 4:50
It's not true that $(x-1)^{2}>9$ implies that $x^{2}>9$. For example, if $x=-2.5$, then $(x-1)^{2}>9$ but $x^{2}<9$.
– Brian Borchers
Nov 18 at 4:52
It's not true that $(x-1)^{2}>9$ implies that $x^{2}>9$. For example, if $x=-2.5$, then $(x-1)^{2}>9$ but $x^{2}<9$.
– Brian Borchers
Nov 18 at 4:52
@BrianBorchers $x>4$ from the start.
– Rócherz
Nov 18 at 4:53
@BrianBorchers $x>4$ from the start.
– Rócherz
Nov 18 at 4:53
1
1
I guess somewhere you need to use that $x>0$ in an essential way.
– Andres Mejia
Nov 18 at 4:53
I guess somewhere you need to use that $x>0$ in an essential way.
– Andres Mejia
Nov 18 at 4:53
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
Why don't you use
$$x>4implies x^2>16>9$$
Note: How did we arrive at the second step from the first one?
answered Nov 18 at 4:47
tatan
5,52462555
Oh that's probably better haha. But does mine make sense at least? Like if I used it on an exam would I get full marks?
– ming
Nov 18 at 5:30
@ming Your argument is also correct however to be on the safe side in an exam, you should write that "...since,we know, $x^2$ is an increasing function, hence the result follows...". You should explicitly mention that $x^2$ is an increasing function.
– tatan
Nov 18 at 15:50
add a comment |
up vote
0
down vote
There are two underlying facts in what you label as "obvious" in your last step.
$x-1>3 implies x-1>0$.
$x>x-1$ (this one is universal) and $x-1>0 implies x^2>(x-1)^2$.
But of course the fastest way is as Andrés Mejía and tatan suggested.
answered Nov 18 at 4:59
Rócherz
2,6162720
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Why don't you use
$$x>4implies x^2>16>9$$
Note: How did we arrive at the second step from the first one?
answered Nov 18 at 4:47
tatan
5,52462555
Oh that's probably better haha. But does mine make sense at least? Like if I used it on an exam would I get full marks?
– ming
Nov 18 at 5:30
@ming Your argument is also correct however to be on the safe side in an exam, you should write that "...since,we know, $x^2$ is an increasing function, hence the result follows...". You should explicitly mention that $x^2$ is an increasing function.
– tatan
Nov 18 at 15:50
add a comment |
up vote
4
down vote
Why don't you use
$$x>4implies x^2>16>9$$
Note: How did we arrive at the second step from the first one?
answered Nov 18 at 4:47
tatan
5,52462555
Oh that's probably better haha. But does mine make sense at least? Like if I used it on an exam would I get full marks?
– ming
Nov 18 at 5:30
@ming Your argument is also correct however to be on the safe side in an exam, you should write that "...since,we know, $x^2$ is an increasing function, hence the result follows...". You should explicitly mention that $x^2$ is an increasing function.
– tatan
Nov 18 at 15:50
add a comment |
up vote
4
down vote
up vote
4
down vote
Why don't you use
$$x>4implies x^2>16>9$$
Note: How did we arrive at the second step from the first one?
answered Nov 18 at 4:47
tatan
5,52462555
Why don't you use
$$x>4implies x^2>16>9$$
Note: How did we arrive at the second step from the first one?
answered Nov 18 at 4:47
tatan
5,52462555
answered Nov 18 at 4:47
tatan
5,52462555
answered Nov 18 at 4:47
tatan
5,52462555
answered Nov 18 at 4:47
tatan
5,52462555
5,52462555
Oh that's probably better haha. But does mine make sense at least? Like if I used it on an exam would I get full marks?
– ming
Nov 18 at 5:30
@ming Your argument is also correct however to be on the safe side in an exam, you should write that "...since,we know, $x^2$ is an increasing function, hence the result follows...". You should explicitly mention that $x^2$ is an increasing function.
– tatan
Nov 18 at 15:50
add a comment |
Oh that's probably better haha. But does mine make sense at least? Like if I used it on an exam would I get full marks?
– ming
Nov 18 at 5:30
@ming Your argument is also correct however to be on the safe side in an exam, you should write that "...since,we know, $x^2$ is an increasing function, hence the result follows...". You should explicitly mention that $x^2$ is an increasing function.
– tatan
Nov 18 at 15:50
Oh that's probably better haha. But does mine make sense at least? Like if I used it on an exam would I get full marks?
– ming
Nov 18 at 5:30
Oh that's probably better haha. But does mine make sense at least? Like if I used it on an exam would I get full marks?
– ming
Nov 18 at 5:30
@ming Your argument is also correct however to be on the safe side in an exam, you should write that "...since,we know, $x^2$ is an increasing function, hence the result follows...". You should explicitly mention that $x^2$ is an increasing function.
– tatan
Nov 18 at 15:50
@ming Your argument is also correct however to be on the safe side in an exam, you should write that "...since,we know, $x^2$ is an increasing function, hence the result follows...". You should explicitly mention that $x^2$ is an increasing function.
– tatan
Nov 18 at 15:50
add a comment |
up vote
0
down vote
There are two underlying facts in what you label as "obvious" in your last step.
$x-1>3 implies x-1>0$.
$x>x-1$ (this one is universal) and $x-1>0 implies x^2>(x-1)^2$.
But of course the fastest way is as Andrés Mejía and tatan suggested.
answered Nov 18 at 4:59
Rócherz
2,6162720
add a comment |
up vote
0
down vote
There are two underlying facts in what you label as "obvious" in your last step.
$x-1>3 implies x-1>0$.
$x>x-1$ (this one is universal) and $x-1>0 implies x^2>(x-1)^2$.
But of course the fastest way is as Andrés Mejía and tatan suggested.
answered Nov 18 at 4:59
Rócherz
2,6162720
add a comment |
up vote
0
down vote
up vote
0
down vote
There are two underlying facts in what you label as "obvious" in your last step.
$x-1>3 implies x-1>0$.
$x>x-1$ (this one is universal) and $x-1>0 implies x^2>(x-1)^2$.
But of course the fastest way is as Andrés Mejía and tatan suggested.
answered Nov 18 at 4:59
Rócherz
2,6162720
There are two underlying facts in what you label as "obvious" in your last step.
$x-1>3 implies x-1>0$.
$x>x-1$ (this one is universal) and $x-1>0 implies x^2>(x-1)^2$.
But of course the fastest way is as Andrés Mejía and tatan suggested.
answered Nov 18 at 4:59
Rócherz
2,6162720
edited Nov 18 at 5:10
answered Nov 18 at 4:59
Rócherz
2,6162720
answered Nov 18 at 4:59
Rócherz
2,6162720
answered Nov 18 at 4:59
Rócherz
2,6162720
2,6162720
add a comment |
add a comment |
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See: math.stackexchange.com/questions/568780/…
– NoChance
Nov 18 at 4:50
It's not true that $(x-1)^{2}>9$ implies that $x^{2}>9$. For example, if $x=-2.5$, then $(x-1)^{2}>9$ but $x^{2}<9$.
– Brian Borchers
Nov 18 at 4:52
@BrianBorchers $x>4$ from the start.
– Rócherz
Nov 18 at 4:53
1
I guess somewhere you need to use that $x>0$ in an essential way.
– Andres Mejia
Nov 18 at 4:53