Let $x$ and $y$ be real numbers. If $xy=0,$ then $x=0$ or $y=0.$
$begingroup$
When solving this proof would it be okay to say
Proof: Let $x,y in mathbb{R}.$ We will do a proof by case
- Case 1. If $x=0,$ then $xy=0y=0$
- Case 2. If $y=0,$ then $xy=0x=0$
In each case, $xy=0.$
Or would you prove by contradiction saying:
If $xneq0$ and $yneq0$ so $x^{-1}$ and $y^{-1}$ must exist.
Then $frac{1}{xy}xy=0frac{1}{xy}.$
$1=0$ which is a contradiction.
proof-verification
edited Dec 6 '18 at 23:40
user376343
3,3933826
asked Dec 6 '18 at 15:40
caroline drummondcaroline drummond
105
$endgroup$
add a comment |
$begingroup$
When solving this proof would it be okay to say
Proof: Let $x,y in mathbb{R}.$ We will do a proof by case
- Case 1. If $x=0,$ then $xy=0y=0$
- Case 2. If $y=0,$ then $xy=0x=0$
In each case, $xy=0.$
Or would you prove by contradiction saying:
If $xneq0$ and $yneq0$ so $x^{-1}$ and $y^{-1}$ must exist.
Then $frac{1}{xy}xy=0frac{1}{xy}.$
$1=0$ which is a contradiction.
proof-verification
edited Dec 6 '18 at 23:40
user376343
3,3933826
asked Dec 6 '18 at 15:40
caroline drummondcaroline drummond
105
$endgroup$
add a comment |
$begingroup$
When solving this proof would it be okay to say
Proof: Let $x,y in mathbb{R}.$ We will do a proof by case
- Case 1. If $x=0,$ then $xy=0y=0$
- Case 2. If $y=0,$ then $xy=0x=0$
In each case, $xy=0.$
Or would you prove by contradiction saying:
If $xneq0$ and $yneq0$ so $x^{-1}$ and $y^{-1}$ must exist.
Then $frac{1}{xy}xy=0frac{1}{xy}.$
$1=0$ which is a contradiction.
proof-verification
edited Dec 6 '18 at 23:40
user376343
3,3933826
asked Dec 6 '18 at 15:40
caroline drummondcaroline drummond
105
$endgroup$
When solving this proof would it be okay to say
Proof: Let $x,y in mathbb{R}.$ We will do a proof by case
- Case 1. If $x=0,$ then $xy=0y=0$
- Case 2. If $y=0,$ then $xy=0x=0$
In each case, $xy=0.$
Or would you prove by contradiction saying:
If $xneq0$ and $yneq0$ so $x^{-1}$ and $y^{-1}$ must exist.
Then $frac{1}{xy}xy=0frac{1}{xy}.$
$1=0$ which is a contradiction.
proof-verification
proof-verification
edited Dec 6 '18 at 23:40
user376343
3,3933826
asked Dec 6 '18 at 15:40
caroline drummondcaroline drummond
105
edited Dec 6 '18 at 23:40
user376343
3,3933826
asked Dec 6 '18 at 15:40
caroline drummondcaroline drummond
105
edited Dec 6 '18 at 23:40
user376343
3,3933826
edited Dec 6 '18 at 23:40
user376343
3,3933826
edited Dec 6 '18 at 23:40
user376343
3,3933826
3,3933826
asked Dec 6 '18 at 15:40
caroline drummondcaroline drummond
105
asked Dec 6 '18 at 15:40
caroline drummondcaroline drummond
105
asked Dec 6 '18 at 15:40
caroline drummondcaroline drummond
105
105
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Proving by cases is a valid way.
Case I: when only $x=0$
Case II: when only $y=0$,
Case III: when both $x=0$ and $y=0$.
answered Dec 6 '18 at 15:42
Hussain-AlqatariHussain-Alqatari
3187
$endgroup$
$begingroup$
But in that way we prove that $$x=0 quad lor quad y=0 implies xy=0$$
$endgroup$
– gimusi
Dec 6 '18 at 15:49
add a comment |
$begingroup$
For a complete proof we can consider four cases
- $x=0 land yneq 0$
- $y=0 land xneq 0$
- $x=0 land y=0$
- $xneq 0 land yneq 0$
to conclude that
$$xy=0 iff x=0 quad lor quad y=0$$
and then of course
$$xy=0 implies x=0 quad lor quad y=0$$
answered Dec 6 '18 at 15:45
gimusigimusi
92.8k84494
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Proving by cases is a valid way.
Case I: when only $x=0$
Case II: when only $y=0$,
Case III: when both $x=0$ and $y=0$.
answered Dec 6 '18 at 15:42
Hussain-AlqatariHussain-Alqatari
3187
$endgroup$
$begingroup$
But in that way we prove that $$x=0 quad lor quad y=0 implies xy=0$$
$endgroup$
– gimusi
Dec 6 '18 at 15:49
add a comment |
$begingroup$
Proving by cases is a valid way.
Case I: when only $x=0$
Case II: when only $y=0$,
Case III: when both $x=0$ and $y=0$.
answered Dec 6 '18 at 15:42
Hussain-AlqatariHussain-Alqatari
3187
$endgroup$
$begingroup$
But in that way we prove that $$x=0 quad lor quad y=0 implies xy=0$$
$endgroup$
– gimusi
Dec 6 '18 at 15:49
add a comment |
$begingroup$
Proving by cases is a valid way.
Case I: when only $x=0$
Case II: when only $y=0$,
Case III: when both $x=0$ and $y=0$.
answered Dec 6 '18 at 15:42
Hussain-AlqatariHussain-Alqatari
3187
$endgroup$
Proving by cases is a valid way.
Case I: when only $x=0$
Case II: when only $y=0$,
Case III: when both $x=0$ and $y=0$.
answered Dec 6 '18 at 15:42
Hussain-AlqatariHussain-Alqatari
3187
answered Dec 6 '18 at 15:42
Hussain-AlqatariHussain-Alqatari
3187
answered Dec 6 '18 at 15:42
Hussain-AlqatariHussain-Alqatari
3187
answered Dec 6 '18 at 15:42
Hussain-AlqatariHussain-Alqatari
3187
3187
$begingroup$
But in that way we prove that $$x=0 quad lor quad y=0 implies xy=0$$
$endgroup$
– gimusi
Dec 6 '18 at 15:49
add a comment |
$begingroup$
But in that way we prove that $$x=0 quad lor quad y=0 implies xy=0$$
$endgroup$
– gimusi
Dec 6 '18 at 15:49
$begingroup$
But in that way we prove that $$x=0 quad lor quad y=0 implies xy=0$$
$endgroup$
– gimusi
Dec 6 '18 at 15:49
$begingroup$
But in that way we prove that $$x=0 quad lor quad y=0 implies xy=0$$
$endgroup$
– gimusi
Dec 6 '18 at 15:49
add a comment |
$begingroup$
For a complete proof we can consider four cases
- $x=0 land yneq 0$
- $y=0 land xneq 0$
- $x=0 land y=0$
- $xneq 0 land yneq 0$
to conclude that
$$xy=0 iff x=0 quad lor quad y=0$$
and then of course
$$xy=0 implies x=0 quad lor quad y=0$$
answered Dec 6 '18 at 15:45
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
For a complete proof we can consider four cases
- $x=0 land yneq 0$
- $y=0 land xneq 0$
- $x=0 land y=0$
- $xneq 0 land yneq 0$
to conclude that
$$xy=0 iff x=0 quad lor quad y=0$$
and then of course
$$xy=0 implies x=0 quad lor quad y=0$$
answered Dec 6 '18 at 15:45
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
For a complete proof we can consider four cases
- $x=0 land yneq 0$
- $y=0 land xneq 0$
- $x=0 land y=0$
- $xneq 0 land yneq 0$
to conclude that
$$xy=0 iff x=0 quad lor quad y=0$$
and then of course
$$xy=0 implies x=0 quad lor quad y=0$$
answered Dec 6 '18 at 15:45
gimusigimusi
92.8k84494
$endgroup$
For a complete proof we can consider four cases
- $x=0 land yneq 0$
- $y=0 land xneq 0$
- $x=0 land y=0$
- $xneq 0 land yneq 0$
to conclude that
$$xy=0 iff x=0 quad lor quad y=0$$
and then of course
$$xy=0 implies x=0 quad lor quad y=0$$
answered Dec 6 '18 at 15:45
gimusigimusi
92.8k84494
answered Dec 6 '18 at 15:45
gimusigimusi
92.8k84494
answered Dec 6 '18 at 15:45
gimusigimusi
92.8k84494
answered Dec 6 '18 at 15:45
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
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