For every real number $x$, $ lvert 2x - 6 rvert gt x iff lvert x-4 rvert gt 2 $
I was hoping to get a bit of feedback on a proof I've done involving absolute values. This problem is taken from D. Velleman's How to Prove it (#3.5.11). I've only written on one side of the biconditional, but the I believe the other conditional can be structured similarly.
Problem Statement:
Prove that for every real number $x$, $lvert 2x - 6 rvert gt x $ iff $lvert x - 4 rvert gt 2 $.
My question:
Do the statements below constitute a valid proof or do they require supplemental explanation?
($rightarrow$) suppose $ 2x - 6 gt 0$. Per the definition of $ lvert 2x - 6 rvert $ we proceed by cases.
Case 1: $ 2x - 6 geq 0 $.
Since $ 2x - 6 geq 0 $, $ lvert 2x - 6 rvert = 2x - 6.$
Then
$$ lvert 2x - 6 rvert gt x $$
$$= 2x - 6 gt x $$
$$= -6 gt -x $$
$$= 6 lt x $$
$$= 4+2 lt x $$
$$= 2 lt x-4 $$
$$= 2 lt lvert x - 4 rvert $$
$$= lvert x-4 rvert gt 2 $$
Case 2: $ 2x - 6 lt 0 $. Since $ 2x - 6 lt 0 $, $ lvert 2x - 6 rvert = 6 - 2x $.
Then
$$ lvert 2x - 6 rvert gt x $$
$$= 6 - 2x gt x $$
$$= 6 gt 3x $$
$$= 2 gt x $$
$$= 4 > x + 2 $$
$$= 4 - x gt 2 $$
$$= lvert x - 4 rvert gt 2 $$
EDIT
I removed the word 'equivalence' from the problem because it might not be the most appropriate word to use. I'm mostly just asking whether the series of statements I've given provide a valid proof of the conclusion.
inequality self-learning real-numbers absolute-value
edited Dec 2 '18 at 7:39
Tianlalu
3,09621038
asked Dec 1 '18 at 20:23
tom
54
add a comment |
I was hoping to get a bit of feedback on a proof I've done involving absolute values. This problem is taken from D. Velleman's How to Prove it (#3.5.11). I've only written on one side of the biconditional, but the I believe the other conditional can be structured similarly.
Problem Statement:
Prove that for every real number $x$, $lvert 2x - 6 rvert gt x $ iff $lvert x - 4 rvert gt 2 $.
My question:
Do the statements below constitute a valid proof or do they require supplemental explanation?
($rightarrow$) suppose $ 2x - 6 gt 0$. Per the definition of $ lvert 2x - 6 rvert $ we proceed by cases.
Case 1: $ 2x - 6 geq 0 $.
Since $ 2x - 6 geq 0 $, $ lvert 2x - 6 rvert = 2x - 6.$
Then
$$ lvert 2x - 6 rvert gt x $$
$$= 2x - 6 gt x $$
$$= -6 gt -x $$
$$= 6 lt x $$
$$= 4+2 lt x $$
$$= 2 lt x-4 $$
$$= 2 lt lvert x - 4 rvert $$
$$= lvert x-4 rvert gt 2 $$
Case 2: $ 2x - 6 lt 0 $. Since $ 2x - 6 lt 0 $, $ lvert 2x - 6 rvert = 6 - 2x $.
Then
$$ lvert 2x - 6 rvert gt x $$
$$= 6 - 2x gt x $$
$$= 6 gt 3x $$
$$= 2 gt x $$
$$= 4 > x + 2 $$
$$= 4 - x gt 2 $$
$$= lvert x - 4 rvert gt 2 $$
EDIT
I removed the word 'equivalence' from the problem because it might not be the most appropriate word to use. I'm mostly just asking whether the series of statements I've given provide a valid proof of the conclusion.
inequality self-learning real-numbers absolute-value
edited Dec 2 '18 at 7:39
Tianlalu
3,09621038
asked Dec 1 '18 at 20:23
tom
54
1
Tom do your = denote the equivalence?
– user376343
Dec 1 '18 at 20:25
add a comment |
I was hoping to get a bit of feedback on a proof I've done involving absolute values. This problem is taken from D. Velleman's How to Prove it (#3.5.11). I've only written on one side of the biconditional, but the I believe the other conditional can be structured similarly.
Problem Statement:
Prove that for every real number $x$, $lvert 2x - 6 rvert gt x $ iff $lvert x - 4 rvert gt 2 $.
My question:
Do the statements below constitute a valid proof or do they require supplemental explanation?
($rightarrow$) suppose $ 2x - 6 gt 0$. Per the definition of $ lvert 2x - 6 rvert $ we proceed by cases.
Case 1: $ 2x - 6 geq 0 $.
Since $ 2x - 6 geq 0 $, $ lvert 2x - 6 rvert = 2x - 6.$
Then
$$ lvert 2x - 6 rvert gt x $$
$$= 2x - 6 gt x $$
$$= -6 gt -x $$
$$= 6 lt x $$
$$= 4+2 lt x $$
$$= 2 lt x-4 $$
$$= 2 lt lvert x - 4 rvert $$
$$= lvert x-4 rvert gt 2 $$
Case 2: $ 2x - 6 lt 0 $. Since $ 2x - 6 lt 0 $, $ lvert 2x - 6 rvert = 6 - 2x $.
Then
$$ lvert 2x - 6 rvert gt x $$
$$= 6 - 2x gt x $$
$$= 6 gt 3x $$
$$= 2 gt x $$
$$= 4 > x + 2 $$
$$= 4 - x gt 2 $$
$$= lvert x - 4 rvert gt 2 $$
EDIT
I removed the word 'equivalence' from the problem because it might not be the most appropriate word to use. I'm mostly just asking whether the series of statements I've given provide a valid proof of the conclusion.
inequality self-learning real-numbers absolute-value
edited Dec 2 '18 at 7:39
Tianlalu
3,09621038
asked Dec 1 '18 at 20:23
tom
54
I was hoping to get a bit of feedback on a proof I've done involving absolute values. This problem is taken from D. Velleman's How to Prove it (#3.5.11). I've only written on one side of the biconditional, but the I believe the other conditional can be structured similarly.
Problem Statement:
Prove that for every real number $x$, $lvert 2x - 6 rvert gt x $ iff $lvert x - 4 rvert gt 2 $.
My question:
Do the statements below constitute a valid proof or do they require supplemental explanation?
($rightarrow$) suppose $ 2x - 6 gt 0$. Per the definition of $ lvert 2x - 6 rvert $ we proceed by cases.
Case 1: $ 2x - 6 geq 0 $.
Since $ 2x - 6 geq 0 $, $ lvert 2x - 6 rvert = 2x - 6.$
Then
$$ lvert 2x - 6 rvert gt x $$
$$= 2x - 6 gt x $$
$$= -6 gt -x $$
$$= 6 lt x $$
$$= 4+2 lt x $$
$$= 2 lt x-4 $$
$$= 2 lt lvert x - 4 rvert $$
$$= lvert x-4 rvert gt 2 $$
Case 2: $ 2x - 6 lt 0 $. Since $ 2x - 6 lt 0 $, $ lvert 2x - 6 rvert = 6 - 2x $.
Then
$$ lvert 2x - 6 rvert gt x $$
$$= 6 - 2x gt x $$
$$= 6 gt 3x $$
$$= 2 gt x $$
$$= 4 > x + 2 $$
$$= 4 - x gt 2 $$
$$= lvert x - 4 rvert gt 2 $$
EDIT
I removed the word 'equivalence' from the problem because it might not be the most appropriate word to use. I'm mostly just asking whether the series of statements I've given provide a valid proof of the conclusion.
inequality self-learning real-numbers absolute-value
inequality self-learning real-numbers absolute-value
edited Dec 2 '18 at 7:39
Tianlalu
3,09621038
asked Dec 1 '18 at 20:23
tom
54
edited Dec 2 '18 at 7:39
Tianlalu
3,09621038
asked Dec 1 '18 at 20:23
tom
54
edited Dec 2 '18 at 7:39
Tianlalu
3,09621038
edited Dec 2 '18 at 7:39
Tianlalu
3,09621038
edited Dec 2 '18 at 7:39
Tianlalu
3,09621038
3,09621038
asked Dec 1 '18 at 20:23
tom
54
asked Dec 1 '18 at 20:23
tom
54
asked Dec 1 '18 at 20:23
tom
54
54
1
Tom do your = denote the equivalence?
– user376343
Dec 1 '18 at 20:25
add a comment |
1
Tom do your = denote the equivalence?
– user376343
Dec 1 '18 at 20:25
1
1
Tom do your = denote the equivalence?
– user376343
Dec 1 '18 at 20:25
Tom do your = denote the equivalence?
– user376343
Dec 1 '18 at 20:25
add a comment |
1 Answer
1
active
oldest
votes
Looks pretty good to me. But there is an important point you should change:
- As pointed out in the comments, all the "=" preceding the line by line inequalities in your proof should be replaced by an equivalence sign "<=>". Except at the lines where the proof only goes in one direction (when you add the absolute values again to the inequality for instance); at that moment you only put an implies sign "=>"
Concerning the proof line by line (if you want to dot every "i"):
- Case 1: the last line is obvious, no need for it.
- Case 2: you could also add a short explanation about why the transition from the second to last line to the last line is valid, it's not immediately obvious. But math textbooks often skip more steps than that from one line to the next, so I'm not shocked in any way.
answered Dec 1 '18 at 21:13
Jeffery
834
add a comment |
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Looks pretty good to me. But there is an important point you should change:
- As pointed out in the comments, all the "=" preceding the line by line inequalities in your proof should be replaced by an equivalence sign "<=>". Except at the lines where the proof only goes in one direction (when you add the absolute values again to the inequality for instance); at that moment you only put an implies sign "=>"
Concerning the proof line by line (if you want to dot every "i"):
- Case 1: the last line is obvious, no need for it.
- Case 2: you could also add a short explanation about why the transition from the second to last line to the last line is valid, it's not immediately obvious. But math textbooks often skip more steps than that from one line to the next, so I'm not shocked in any way.
answered Dec 1 '18 at 21:13
Jeffery
834
add a comment |
Looks pretty good to me. But there is an important point you should change:
- As pointed out in the comments, all the "=" preceding the line by line inequalities in your proof should be replaced by an equivalence sign "<=>". Except at the lines where the proof only goes in one direction (when you add the absolute values again to the inequality for instance); at that moment you only put an implies sign "=>"
Concerning the proof line by line (if you want to dot every "i"):
- Case 1: the last line is obvious, no need for it.
- Case 2: you could also add a short explanation about why the transition from the second to last line to the last line is valid, it's not immediately obvious. But math textbooks often skip more steps than that from one line to the next, so I'm not shocked in any way.
answered Dec 1 '18 at 21:13
Jeffery
834
add a comment |
Looks pretty good to me. But there is an important point you should change:
- As pointed out in the comments, all the "=" preceding the line by line inequalities in your proof should be replaced by an equivalence sign "<=>". Except at the lines where the proof only goes in one direction (when you add the absolute values again to the inequality for instance); at that moment you only put an implies sign "=>"
Concerning the proof line by line (if you want to dot every "i"):
- Case 1: the last line is obvious, no need for it.
- Case 2: you could also add a short explanation about why the transition from the second to last line to the last line is valid, it's not immediately obvious. But math textbooks often skip more steps than that from one line to the next, so I'm not shocked in any way.
answered Dec 1 '18 at 21:13
Jeffery
834
Looks pretty good to me. But there is an important point you should change:
- As pointed out in the comments, all the "=" preceding the line by line inequalities in your proof should be replaced by an equivalence sign "<=>". Except at the lines where the proof only goes in one direction (when you add the absolute values again to the inequality for instance); at that moment you only put an implies sign "=>"
Concerning the proof line by line (if you want to dot every "i"):
- Case 1: the last line is obvious, no need for it.
- Case 2: you could also add a short explanation about why the transition from the second to last line to the last line is valid, it's not immediately obvious. But math textbooks often skip more steps than that from one line to the next, so I'm not shocked in any way.
answered Dec 1 '18 at 21:13
Jeffery
834
edited Dec 1 '18 at 21:20
answered Dec 1 '18 at 21:13
Jeffery
834
answered Dec 1 '18 at 21:13
Jeffery
834
answered Dec 1 '18 at 21:13
Jeffery
834
834
add a comment |
add a comment |
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1
Tom do your = denote the equivalence?
– user376343
Dec 1 '18 at 20:25