For all $x in Bbb R$, if $x > 4$ then $x^2 > 9$
$begingroup$
Can I just say $x > 4 Rightarrow x^2 > 16$ and since $16$ is bigger than $9$, true?
real-analysis inequality
edited Dec 1 '18 at 19:23
Masacroso
13k41746
asked Dec 1 '18 at 19:18
mingming
3205
$endgroup$
add a comment |
$begingroup$
Can I just say $x > 4 Rightarrow x^2 > 16$ and since $16$ is bigger than $9$, true?
real-analysis inequality
edited Dec 1 '18 at 19:23
Masacroso
13k41746
asked Dec 1 '18 at 19:18
mingming
3205
$endgroup$
1
$begingroup$
@Masacroso The $Rightarrow$ sign would be more appropriate here.
$endgroup$
– MisterRiemann
Dec 1 '18 at 19:22
$begingroup$
I edited following the title, anyway the original body of the question is not so clear
$endgroup$
– Masacroso
Dec 1 '18 at 19:25
$begingroup$
Yes, $x>4>0$ implies $x^2>16>9$
$endgroup$
– user376343
Dec 1 '18 at 19:30
add a comment |
$begingroup$
Can I just say $x > 4 Rightarrow x^2 > 16$ and since $16$ is bigger than $9$, true?
real-analysis inequality
edited Dec 1 '18 at 19:23
Masacroso
13k41746
asked Dec 1 '18 at 19:18
mingming
3205
$endgroup$
Can I just say $x > 4 Rightarrow x^2 > 16$ and since $16$ is bigger than $9$, true?
real-analysis inequality
real-analysis inequality
edited Dec 1 '18 at 19:23
Masacroso
13k41746
asked Dec 1 '18 at 19:18
mingming
3205
edited Dec 1 '18 at 19:23
Masacroso
13k41746
asked Dec 1 '18 at 19:18
mingming
3205
edited Dec 1 '18 at 19:23
Masacroso
13k41746
edited Dec 1 '18 at 19:23
Masacroso
13k41746
edited Dec 1 '18 at 19:23
Masacroso
13k41746
13k41746
asked Dec 1 '18 at 19:18
mingming
3205
asked Dec 1 '18 at 19:18
mingming
3205
asked Dec 1 '18 at 19:18
mingming
3205
3205
1
$begingroup$
@Masacroso The $Rightarrow$ sign would be more appropriate here.
$endgroup$
– MisterRiemann
Dec 1 '18 at 19:22
$begingroup$
I edited following the title, anyway the original body of the question is not so clear
$endgroup$
– Masacroso
Dec 1 '18 at 19:25
$begingroup$
Yes, $x>4>0$ implies $x^2>16>9$
$endgroup$
– user376343
Dec 1 '18 at 19:30
add a comment |
1
$begingroup$
@Masacroso The $Rightarrow$ sign would be more appropriate here.
$endgroup$
– MisterRiemann
Dec 1 '18 at 19:22
$begingroup$
I edited following the title, anyway the original body of the question is not so clear
$endgroup$
– Masacroso
Dec 1 '18 at 19:25
$begingroup$
Yes, $x>4>0$ implies $x^2>16>9$
$endgroup$
– user376343
Dec 1 '18 at 19:30
1
1
$begingroup$
@Masacroso The $Rightarrow$ sign would be more appropriate here.
$endgroup$
– MisterRiemann
Dec 1 '18 at 19:22
$begingroup$
@Masacroso The $Rightarrow$ sign would be more appropriate here.
$endgroup$
– MisterRiemann
Dec 1 '18 at 19:22
$begingroup$
I edited following the title, anyway the original body of the question is not so clear
$endgroup$
– Masacroso
Dec 1 '18 at 19:25
$begingroup$
I edited following the title, anyway the original body of the question is not so clear
$endgroup$
– Masacroso
Dec 1 '18 at 19:25
$begingroup$
Yes, $x>4>0$ implies $x^2>16>9$
$endgroup$
– user376343
Dec 1 '18 at 19:30
$begingroup$
Yes, $x>4>0$ implies $x^2>16>9$
$endgroup$
– user376343
Dec 1 '18 at 19:30
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Let $x in Bbb R$, if $x > 4$, then as $x mapsto x^2$ is increasing over $Bbb R_+$,
$x^2 > 16 > 9$.
answered Dec 1 '18 at 19:31
Euler PythagorasEuler Pythagoras
52110
$endgroup$
add a comment |
$begingroup$
Yes of course that's correct indeed since $f: x mapsto x^2$ is increasing for $x gt 0,$
$$x>4 implies x^2>16>9$$
What is not true is that for example
$$x^2>9 implies x>3$$
answered Dec 1 '18 at 19:30
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
An alternative approach is $x>4>0implies x^2>4x>4^2=16>9$.
answered Dec 1 '18 at 19:34
J.G.J.G.
23.9k22539
$endgroup$
add a comment |
$begingroup$
Here is another take.
Write $x=3+y$,with $y>1$. Then $x^2=(3+y)^2=9+6y+y^2>9$, since $y>1>0$.
answered Dec 1 '18 at 20:07
lhflhf
163k10168390
$endgroup$
add a comment |
$begingroup$
More or less.
You need some concept that $x^2$ is increasing for positive numbers so that if $0 < a < b$ then $a^2 < b^2$ (and so $x >4$ then $x^2 > 16 > 9$).
If you are doing calculus you can point out that for $f(x) = x^2$ then $f'(x) = 2x$ and if $x > 0$ then $f'(x) > 0$ so $f$ is increasing on all positives.
Or if you are doing ordered field axioms you can do via axiom: If $a < b$ and $c > 0$ then $ac < bc$. And therefore if $0 < a < b$ then $acdot a < bcdot a$ and $a^2cdot b < bcdot b^2$.
So one way or another, yes, you can say it. (But you should be prepared to defend it if anyone questions.)
....
And there are "tricks".
$x^2 - 9 = (x-3)(x+3)$ and $x - 3 > 4-3 = 1> 0$ and $x+3 > 4+3 = 7 > 0$ so $x^2 - 9 > 0$ so $x^ > 9$ and other cute little things like that.
answered Dec 1 '18 at 20:17
fleabloodfleablood
68.7k22685
$endgroup$
$begingroup$
Do you think I'd get full marks on a first year uni algebra proofs course? Like is that proof I wrote not enough you think?
$endgroup$
– ming
Dec 1 '18 at 21:46
$begingroup$
I honestly don't know. I can't see this being asked as a question in it's own right. As a step in a bigger problem I was accept "as $x > 4$ then $x^2 > 4^2 > 9$" as enough.
$endgroup$
– fleablood
Dec 1 '18 at 22:05
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $x in Bbb R$, if $x > 4$, then as $x mapsto x^2$ is increasing over $Bbb R_+$,
$x^2 > 16 > 9$.
answered Dec 1 '18 at 19:31
Euler PythagorasEuler Pythagoras
52110
$endgroup$
add a comment |
$begingroup$
Let $x in Bbb R$, if $x > 4$, then as $x mapsto x^2$ is increasing over $Bbb R_+$,
$x^2 > 16 > 9$.
answered Dec 1 '18 at 19:31
Euler PythagorasEuler Pythagoras
52110
$endgroup$
add a comment |
$begingroup$
Let $x in Bbb R$, if $x > 4$, then as $x mapsto x^2$ is increasing over $Bbb R_+$,
$x^2 > 16 > 9$.
answered Dec 1 '18 at 19:31
Euler PythagorasEuler Pythagoras
52110
$endgroup$
Let $x in Bbb R$, if $x > 4$, then as $x mapsto x^2$ is increasing over $Bbb R_+$,
$x^2 > 16 > 9$.
answered Dec 1 '18 at 19:31
Euler PythagorasEuler Pythagoras
52110
answered Dec 1 '18 at 19:31
Euler PythagorasEuler Pythagoras
52110
answered Dec 1 '18 at 19:31
Euler PythagorasEuler Pythagoras
52110
answered Dec 1 '18 at 19:31
Euler PythagorasEuler Pythagoras
52110
52110
add a comment |
add a comment |
$begingroup$
Yes of course that's correct indeed since $f: x mapsto x^2$ is increasing for $x gt 0,$
$$x>4 implies x^2>16>9$$
What is not true is that for example
$$x^2>9 implies x>3$$
answered Dec 1 '18 at 19:30
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
Yes of course that's correct indeed since $f: x mapsto x^2$ is increasing for $x gt 0,$
$$x>4 implies x^2>16>9$$
What is not true is that for example
$$x^2>9 implies x>3$$
answered Dec 1 '18 at 19:30
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
Yes of course that's correct indeed since $f: x mapsto x^2$ is increasing for $x gt 0,$
$$x>4 implies x^2>16>9$$
What is not true is that for example
$$x^2>9 implies x>3$$
answered Dec 1 '18 at 19:30
gimusigimusi
1
$endgroup$
Yes of course that's correct indeed since $f: x mapsto x^2$ is increasing for $x gt 0,$
$$x>4 implies x^2>16>9$$
What is not true is that for example
$$x^2>9 implies x>3$$
answered Dec 1 '18 at 19:30
gimusigimusi
1
edited Dec 1 '18 at 20:03
answered Dec 1 '18 at 19:30
gimusigimusi
1
answered Dec 1 '18 at 19:30
gimusigimusi
1
answered Dec 1 '18 at 19:30
gimusigimusi
1
1
add a comment |
add a comment |
$begingroup$
An alternative approach is $x>4>0implies x^2>4x>4^2=16>9$.
answered Dec 1 '18 at 19:34
J.G.J.G.
23.9k22539
$endgroup$
add a comment |
$begingroup$
An alternative approach is $x>4>0implies x^2>4x>4^2=16>9$.
answered Dec 1 '18 at 19:34
J.G.J.G.
23.9k22539
$endgroup$
add a comment |
$begingroup$
An alternative approach is $x>4>0implies x^2>4x>4^2=16>9$.
answered Dec 1 '18 at 19:34
J.G.J.G.
23.9k22539
$endgroup$
An alternative approach is $x>4>0implies x^2>4x>4^2=16>9$.
answered Dec 1 '18 at 19:34
J.G.J.G.
23.9k22539
answered Dec 1 '18 at 19:34
J.G.J.G.
23.9k22539
answered Dec 1 '18 at 19:34
J.G.J.G.
23.9k22539
answered Dec 1 '18 at 19:34
J.G.J.G.
23.9k22539
23.9k22539
add a comment |
add a comment |
$begingroup$
Here is another take.
Write $x=3+y$,with $y>1$. Then $x^2=(3+y)^2=9+6y+y^2>9$, since $y>1>0$.
answered Dec 1 '18 at 20:07
lhflhf
163k10168390
$endgroup$
add a comment |
$begingroup$
Here is another take.
Write $x=3+y$,with $y>1$. Then $x^2=(3+y)^2=9+6y+y^2>9$, since $y>1>0$.
answered Dec 1 '18 at 20:07
lhflhf
163k10168390
$endgroup$
add a comment |
$begingroup$
Here is another take.
Write $x=3+y$,with $y>1$. Then $x^2=(3+y)^2=9+6y+y^2>9$, since $y>1>0$.
answered Dec 1 '18 at 20:07
lhflhf
163k10168390
$endgroup$
Here is another take.
Write $x=3+y$,with $y>1$. Then $x^2=(3+y)^2=9+6y+y^2>9$, since $y>1>0$.
answered Dec 1 '18 at 20:07
lhflhf
163k10168390
answered Dec 1 '18 at 20:07
lhflhf
163k10168390
answered Dec 1 '18 at 20:07
lhflhf
163k10168390
answered Dec 1 '18 at 20:07
lhflhf
163k10168390
163k10168390
add a comment |
add a comment |
$begingroup$
More or less.
You need some concept that $x^2$ is increasing for positive numbers so that if $0 < a < b$ then $a^2 < b^2$ (and so $x >4$ then $x^2 > 16 > 9$).
If you are doing calculus you can point out that for $f(x) = x^2$ then $f'(x) = 2x$ and if $x > 0$ then $f'(x) > 0$ so $f$ is increasing on all positives.
Or if you are doing ordered field axioms you can do via axiom: If $a < b$ and $c > 0$ then $ac < bc$. And therefore if $0 < a < b$ then $acdot a < bcdot a$ and $a^2cdot b < bcdot b^2$.
So one way or another, yes, you can say it. (But you should be prepared to defend it if anyone questions.)
....
And there are "tricks".
$x^2 - 9 = (x-3)(x+3)$ and $x - 3 > 4-3 = 1> 0$ and $x+3 > 4+3 = 7 > 0$ so $x^2 - 9 > 0$ so $x^ > 9$ and other cute little things like that.
answered Dec 1 '18 at 20:17
fleabloodfleablood
68.7k22685
$endgroup$
$begingroup$
Do you think I'd get full marks on a first year uni algebra proofs course? Like is that proof I wrote not enough you think?
$endgroup$
– ming
Dec 1 '18 at 21:46
$begingroup$
I honestly don't know. I can't see this being asked as a question in it's own right. As a step in a bigger problem I was accept "as $x > 4$ then $x^2 > 4^2 > 9$" as enough.
$endgroup$
– fleablood
Dec 1 '18 at 22:05
add a comment |
$begingroup$
More or less.
You need some concept that $x^2$ is increasing for positive numbers so that if $0 < a < b$ then $a^2 < b^2$ (and so $x >4$ then $x^2 > 16 > 9$).
If you are doing calculus you can point out that for $f(x) = x^2$ then $f'(x) = 2x$ and if $x > 0$ then $f'(x) > 0$ so $f$ is increasing on all positives.
Or if you are doing ordered field axioms you can do via axiom: If $a < b$ and $c > 0$ then $ac < bc$. And therefore if $0 < a < b$ then $acdot a < bcdot a$ and $a^2cdot b < bcdot b^2$.
So one way or another, yes, you can say it. (But you should be prepared to defend it if anyone questions.)
....
And there are "tricks".
$x^2 - 9 = (x-3)(x+3)$ and $x - 3 > 4-3 = 1> 0$ and $x+3 > 4+3 = 7 > 0$ so $x^2 - 9 > 0$ so $x^ > 9$ and other cute little things like that.
answered Dec 1 '18 at 20:17
fleabloodfleablood
68.7k22685
$endgroup$
$begingroup$
Do you think I'd get full marks on a first year uni algebra proofs course? Like is that proof I wrote not enough you think?
$endgroup$
– ming
Dec 1 '18 at 21:46
$begingroup$
I honestly don't know. I can't see this being asked as a question in it's own right. As a step in a bigger problem I was accept "as $x > 4$ then $x^2 > 4^2 > 9$" as enough.
$endgroup$
– fleablood
Dec 1 '18 at 22:05
add a comment |
$begingroup$
More or less.
You need some concept that $x^2$ is increasing for positive numbers so that if $0 < a < b$ then $a^2 < b^2$ (and so $x >4$ then $x^2 > 16 > 9$).
If you are doing calculus you can point out that for $f(x) = x^2$ then $f'(x) = 2x$ and if $x > 0$ then $f'(x) > 0$ so $f$ is increasing on all positives.
Or if you are doing ordered field axioms you can do via axiom: If $a < b$ and $c > 0$ then $ac < bc$. And therefore if $0 < a < b$ then $acdot a < bcdot a$ and $a^2cdot b < bcdot b^2$.
So one way or another, yes, you can say it. (But you should be prepared to defend it if anyone questions.)
....
And there are "tricks".
$x^2 - 9 = (x-3)(x+3)$ and $x - 3 > 4-3 = 1> 0$ and $x+3 > 4+3 = 7 > 0$ so $x^2 - 9 > 0$ so $x^ > 9$ and other cute little things like that.
answered Dec 1 '18 at 20:17
fleabloodfleablood
68.7k22685
$endgroup$
More or less.
You need some concept that $x^2$ is increasing for positive numbers so that if $0 < a < b$ then $a^2 < b^2$ (and so $x >4$ then $x^2 > 16 > 9$).
If you are doing calculus you can point out that for $f(x) = x^2$ then $f'(x) = 2x$ and if $x > 0$ then $f'(x) > 0$ so $f$ is increasing on all positives.
Or if you are doing ordered field axioms you can do via axiom: If $a < b$ and $c > 0$ then $ac < bc$. And therefore if $0 < a < b$ then $acdot a < bcdot a$ and $a^2cdot b < bcdot b^2$.
So one way or another, yes, you can say it. (But you should be prepared to defend it if anyone questions.)
....
And there are "tricks".
$x^2 - 9 = (x-3)(x+3)$ and $x - 3 > 4-3 = 1> 0$ and $x+3 > 4+3 = 7 > 0$ so $x^2 - 9 > 0$ so $x^ > 9$ and other cute little things like that.
answered Dec 1 '18 at 20:17
fleabloodfleablood
68.7k22685
answered Dec 1 '18 at 20:17
fleabloodfleablood
68.7k22685
answered Dec 1 '18 at 20:17
fleabloodfleablood
68.7k22685
answered Dec 1 '18 at 20:17
fleabloodfleablood
68.7k22685
68.7k22685
$begingroup$
Do you think I'd get full marks on a first year uni algebra proofs course? Like is that proof I wrote not enough you think?
$endgroup$
– ming
Dec 1 '18 at 21:46
$begingroup$
I honestly don't know. I can't see this being asked as a question in it's own right. As a step in a bigger problem I was accept "as $x > 4$ then $x^2 > 4^2 > 9$" as enough.
$endgroup$
– fleablood
Dec 1 '18 at 22:05
add a comment |
$begingroup$
Do you think I'd get full marks on a first year uni algebra proofs course? Like is that proof I wrote not enough you think?
$endgroup$
– ming
Dec 1 '18 at 21:46
$begingroup$
I honestly don't know. I can't see this being asked as a question in it's own right. As a step in a bigger problem I was accept "as $x > 4$ then $x^2 > 4^2 > 9$" as enough.
$endgroup$
– fleablood
Dec 1 '18 at 22:05
$begingroup$
Do you think I'd get full marks on a first year uni algebra proofs course? Like is that proof I wrote not enough you think?
$endgroup$
– ming
Dec 1 '18 at 21:46
$begingroup$
Do you think I'd get full marks on a first year uni algebra proofs course? Like is that proof I wrote not enough you think?
$endgroup$
– ming
Dec 1 '18 at 21:46
$begingroup$
I honestly don't know. I can't see this being asked as a question in it's own right. As a step in a bigger problem I was accept "as $x > 4$ then $x^2 > 4^2 > 9$" as enough.
$endgroup$
– fleablood
Dec 1 '18 at 22:05
$begingroup$
I honestly don't know. I can't see this being asked as a question in it's own right. As a step in a bigger problem I was accept "as $x > 4$ then $x^2 > 4^2 > 9$" as enough.
$endgroup$
– fleablood
Dec 1 '18 at 22:05
add a comment |
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1
$begingroup$
@Masacroso The $Rightarrow$ sign would be more appropriate here.
$endgroup$
– MisterRiemann
Dec 1 '18 at 19:22
$begingroup$
I edited following the title, anyway the original body of the question is not so clear
$endgroup$
– Masacroso
Dec 1 '18 at 19:25
$begingroup$
Yes, $x>4>0$ implies $x^2>16>9$
$endgroup$
– user376343
Dec 1 '18 at 19:30