How to prove $|x − y| ≤ |x| + |y|$, proof and reasoning [duplicate]
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This question already has an answer here:
Prove: $|x-y|leq |x|+|y|$ [duplicate]
3 answers
Prove that, for all $x, y ∈ mathbb{R}$, we have $|x − y| ≤ |x| + |y|$.
Can I say . $|x − y|^2 ≤ (x − y)^2$, and work from there?
Thank you
inequality real-numbers
edited Dec 18 '18 at 4:24
user 170039
10.5k42466
asked Dec 18 '18 at 4:21
ommomm
2
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marked as duplicate by Eevee Trainer, Lord Shark the Unknown, Chinnapparaj R, user21820, user10354138 Dec 21 '18 at 19:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Prove: $|x-y|leq |x|+|y|$ [duplicate]
3 answers
Prove that, for all $x, y ∈ mathbb{R}$, we have $|x − y| ≤ |x| + |y|$.
Can I say . $|x − y|^2 ≤ (x − y)^2$, and work from there?
Thank you
inequality real-numbers
edited Dec 18 '18 at 4:24
user 170039
10.5k42466
asked Dec 18 '18 at 4:21
ommomm
2
$endgroup$
marked as duplicate by Eevee Trainer, Lord Shark the Unknown, Chinnapparaj R, user21820, user10354138 Dec 21 '18 at 19:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
How exactly do you plan to work from $|x − y|^2 ≤ (x − y)^2$ to the inequality to be proved?
$endgroup$
– user 170039
Dec 18 '18 at 4:24
$begingroup$
In fact $|x-y|^2=(x-y)^2$, so nothing to gain from that. If you're familiar with the triangle inequality, this actually is the triangle inequality for $x$ and $-y$, since $|-y|=|y|$.
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– Matt Samuel
Dec 18 '18 at 4:26
$begingroup$
so does it follow to say |x − y|=|x + (-y)|≤ |x|+|-y|...
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– omm
Dec 18 '18 at 4:28
$begingroup$
@omm: Yes. You are right.
$endgroup$
– user 170039
Dec 18 '18 at 7:04
add a comment |
$begingroup$
This question already has an answer here:
Prove: $|x-y|leq |x|+|y|$ [duplicate]
3 answers
Prove that, for all $x, y ∈ mathbb{R}$, we have $|x − y| ≤ |x| + |y|$.
Can I say . $|x − y|^2 ≤ (x − y)^2$, and work from there?
Thank you
inequality real-numbers
edited Dec 18 '18 at 4:24
user 170039
10.5k42466
asked Dec 18 '18 at 4:21
ommomm
2
$endgroup$
This question already has an answer here:
Prove: $|x-y|leq |x|+|y|$ [duplicate]
3 answers
Prove that, for all $x, y ∈ mathbb{R}$, we have $|x − y| ≤ |x| + |y|$.
Can I say . $|x − y|^2 ≤ (x − y)^2$, and work from there?
Thank you
This question already has an answer here:
Prove: $|x-y|leq |x|+|y|$ [duplicate]
3 answers
inequality real-numbers
inequality real-numbers
edited Dec 18 '18 at 4:24
user 170039
10.5k42466
asked Dec 18 '18 at 4:21
ommomm
2
edited Dec 18 '18 at 4:24
user 170039
10.5k42466
asked Dec 18 '18 at 4:21
ommomm
2
edited Dec 18 '18 at 4:24
user 170039
10.5k42466
edited Dec 18 '18 at 4:24
user 170039
10.5k42466
edited Dec 18 '18 at 4:24
user 170039
10.5k42466
10.5k42466
asked Dec 18 '18 at 4:21
ommomm
2
asked Dec 18 '18 at 4:21
ommomm
2
asked Dec 18 '18 at 4:21
ommomm
2
2
marked as duplicate by Eevee Trainer, Lord Shark the Unknown, Chinnapparaj R, user21820, user10354138 Dec 21 '18 at 19:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Eevee Trainer, Lord Shark the Unknown, Chinnapparaj R, user21820, user10354138 Dec 21 '18 at 19:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
How exactly do you plan to work from $|x − y|^2 ≤ (x − y)^2$ to the inequality to be proved?
$endgroup$
– user 170039
Dec 18 '18 at 4:24
$begingroup$
In fact $|x-y|^2=(x-y)^2$, so nothing to gain from that. If you're familiar with the triangle inequality, this actually is the triangle inequality for $x$ and $-y$, since $|-y|=|y|$.
$endgroup$
– Matt Samuel
Dec 18 '18 at 4:26
$begingroup$
so does it follow to say |x − y|=|x + (-y)|≤ |x|+|-y|...
$endgroup$
– omm
Dec 18 '18 at 4:28
$begingroup$
@omm: Yes. You are right.
$endgroup$
– user 170039
Dec 18 '18 at 7:04
add a comment |
$begingroup$
How exactly do you plan to work from $|x − y|^2 ≤ (x − y)^2$ to the inequality to be proved?
$endgroup$
– user 170039
Dec 18 '18 at 4:24
$begingroup$
In fact $|x-y|^2=(x-y)^2$, so nothing to gain from that. If you're familiar with the triangle inequality, this actually is the triangle inequality for $x$ and $-y$, since $|-y|=|y|$.
$endgroup$
– Matt Samuel
Dec 18 '18 at 4:26
$begingroup$
so does it follow to say |x − y|=|x + (-y)|≤ |x|+|-y|...
$endgroup$
– omm
Dec 18 '18 at 4:28
$begingroup$
@omm: Yes. You are right.
$endgroup$
– user 170039
Dec 18 '18 at 7:04
$begingroup$
How exactly do you plan to work from $|x − y|^2 ≤ (x − y)^2$ to the inequality to be proved?
$endgroup$
– user 170039
Dec 18 '18 at 4:24
$begingroup$
How exactly do you plan to work from $|x − y|^2 ≤ (x − y)^2$ to the inequality to be proved?
$endgroup$
– user 170039
Dec 18 '18 at 4:24
$begingroup$
In fact $|x-y|^2=(x-y)^2$, so nothing to gain from that. If you're familiar with the triangle inequality, this actually is the triangle inequality for $x$ and $-y$, since $|-y|=|y|$.
$endgroup$
– Matt Samuel
Dec 18 '18 at 4:26
$begingroup$
In fact $|x-y|^2=(x-y)^2$, so nothing to gain from that. If you're familiar with the triangle inequality, this actually is the triangle inequality for $x$ and $-y$, since $|-y|=|y|$.
$endgroup$
– Matt Samuel
Dec 18 '18 at 4:26
$begingroup$
so does it follow to say |x − y|=|x + (-y)|≤ |x|+|-y|...
$endgroup$
– omm
Dec 18 '18 at 4:28
$begingroup$
so does it follow to say |x − y|=|x + (-y)|≤ |x|+|-y|...
$endgroup$
– omm
Dec 18 '18 at 4:28
$begingroup$
@omm: Yes. You are right.
$endgroup$
– user 170039
Dec 18 '18 at 7:04
$begingroup$
@omm: Yes. You are right.
$endgroup$
– user 170039
Dec 18 '18 at 7:04
add a comment |
1 Answer
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$begingroup$
Let $x, y in mathbb{R}$.
Consider by contradiction that $|x-y| > |x| + |y|$. Then, multiplying both sides by $|x-y|$, we have that
$$|x-y|^2 > (|x| + |y|)(|x-y|)$$
$$(x-y)^2 > (|x| + |y|)(|x-y|).$$
So we have that
$$ x^2 - 2xy + y^2 > (|x| + |y|)(|x-y|) > (|x| + |y|)(|x| + |y|), $$
by assumption. It follows that
$$ x^2 - 2xy + y^2 > |x|^2 + 2|x||y| + |y|^2 $$
$$ x^2 - 2xy + y^2 > x^2 + 2|x||y| + y^2 $$
$$ -xy > |x||y| $$
which is a contradiction. So, we have that $|x-y| leq |x| + |y|$, as required.
answered Dec 18 '18 at 5:46
alexsieusahaialexsieusahai
638
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $x, y in mathbb{R}$.
Consider by contradiction that $|x-y| > |x| + |y|$. Then, multiplying both sides by $|x-y|$, we have that
$$|x-y|^2 > (|x| + |y|)(|x-y|)$$
$$(x-y)^2 > (|x| + |y|)(|x-y|).$$
So we have that
$$ x^2 - 2xy + y^2 > (|x| + |y|)(|x-y|) > (|x| + |y|)(|x| + |y|), $$
by assumption. It follows that
$$ x^2 - 2xy + y^2 > |x|^2 + 2|x||y| + |y|^2 $$
$$ x^2 - 2xy + y^2 > x^2 + 2|x||y| + y^2 $$
$$ -xy > |x||y| $$
which is a contradiction. So, we have that $|x-y| leq |x| + |y|$, as required.
answered Dec 18 '18 at 5:46
alexsieusahaialexsieusahai
638
$endgroup$
add a comment |
$begingroup$
Let $x, y in mathbb{R}$.
Consider by contradiction that $|x-y| > |x| + |y|$. Then, multiplying both sides by $|x-y|$, we have that
$$|x-y|^2 > (|x| + |y|)(|x-y|)$$
$$(x-y)^2 > (|x| + |y|)(|x-y|).$$
So we have that
$$ x^2 - 2xy + y^2 > (|x| + |y|)(|x-y|) > (|x| + |y|)(|x| + |y|), $$
by assumption. It follows that
$$ x^2 - 2xy + y^2 > |x|^2 + 2|x||y| + |y|^2 $$
$$ x^2 - 2xy + y^2 > x^2 + 2|x||y| + y^2 $$
$$ -xy > |x||y| $$
which is a contradiction. So, we have that $|x-y| leq |x| + |y|$, as required.
answered Dec 18 '18 at 5:46
alexsieusahaialexsieusahai
638
$endgroup$
add a comment |
$begingroup$
Let $x, y in mathbb{R}$.
Consider by contradiction that $|x-y| > |x| + |y|$. Then, multiplying both sides by $|x-y|$, we have that
$$|x-y|^2 > (|x| + |y|)(|x-y|)$$
$$(x-y)^2 > (|x| + |y|)(|x-y|).$$
So we have that
$$ x^2 - 2xy + y^2 > (|x| + |y|)(|x-y|) > (|x| + |y|)(|x| + |y|), $$
by assumption. It follows that
$$ x^2 - 2xy + y^2 > |x|^2 + 2|x||y| + |y|^2 $$
$$ x^2 - 2xy + y^2 > x^2 + 2|x||y| + y^2 $$
$$ -xy > |x||y| $$
which is a contradiction. So, we have that $|x-y| leq |x| + |y|$, as required.
answered Dec 18 '18 at 5:46
alexsieusahaialexsieusahai
638
$endgroup$
Let $x, y in mathbb{R}$.
Consider by contradiction that $|x-y| > |x| + |y|$. Then, multiplying both sides by $|x-y|$, we have that
$$|x-y|^2 > (|x| + |y|)(|x-y|)$$
$$(x-y)^2 > (|x| + |y|)(|x-y|).$$
So we have that
$$ x^2 - 2xy + y^2 > (|x| + |y|)(|x-y|) > (|x| + |y|)(|x| + |y|), $$
by assumption. It follows that
$$ x^2 - 2xy + y^2 > |x|^2 + 2|x||y| + |y|^2 $$
$$ x^2 - 2xy + y^2 > x^2 + 2|x||y| + y^2 $$
$$ -xy > |x||y| $$
which is a contradiction. So, we have that $|x-y| leq |x| + |y|$, as required.
answered Dec 18 '18 at 5:46
alexsieusahaialexsieusahai
638
answered Dec 18 '18 at 5:46
alexsieusahaialexsieusahai
638
answered Dec 18 '18 at 5:46
alexsieusahaialexsieusahai
638
answered Dec 18 '18 at 5:46
alexsieusahaialexsieusahai
638
638
add a comment |
add a comment |
$begingroup$
How exactly do you plan to work from $|x − y|^2 ≤ (x − y)^2$ to the inequality to be proved?
$endgroup$
– user 170039
Dec 18 '18 at 4:24
$begingroup$
In fact $|x-y|^2=(x-y)^2$, so nothing to gain from that. If you're familiar with the triangle inequality, this actually is the triangle inequality for $x$ and $-y$, since $|-y|=|y|$.
$endgroup$
– Matt Samuel
Dec 18 '18 at 4:26
$begingroup$
so does it follow to say |x − y|=|x + (-y)|≤ |x|+|-y|...
$endgroup$
– omm
Dec 18 '18 at 4:28
$begingroup$
@omm: Yes. You are right.
$endgroup$
– user 170039
Dec 18 '18 at 7:04