Norm Verification for Vectors in R^2

[Mark44]

$x+y=(x_1+y_1, x_2+y_2)$

$\| x+y\|_{\#}= \| |x1|+|y1|, 2|x2|+2|y2| \|_{\#}$...?

Did you use the definition?

(i) || ||_#: R^2 defined by ||x||_#=|x_1|+2|x_2|

Doesn't look like it to me.

 

[bugatti79]

Did you use the definition?


Doesn't look like it to me.

Here you're adding x and y correctly, but why don't you just insert the result you've found for x+y into $\|x+y\|_{\#}$, and then use the definition of $\|\ \|_{\#}$? What you wrote doesn't make any sense. The function $\|\ \|_{\#}$ takes a member of $\mathbb R^2$ as input. Here you used a real number as input. That doesn't make sense.

Edit: Why is there a comma in there? Did you mean $\|(|x_1|+|y_1|,2|x_2|+2|y_2|)\|_{\#}$? That doesn't make much sense either. I mean, it it's not a nonsense expression, but I have no idea why you would consider an ordered pair with those components.

Now I am really confused. I know this is a big ask but is it not possible for one of you to write the answer and then I will query why you did it that way?. I have spent 3.5 hours on this trivial problem.

 

[bugatti79]

The function $\|\ \|_{\#}$ takes a member of $\mathbb R^2$ as input. Here you used a real number as input. That doesn't make sense.

OK, should it be

$\| x+y\|_{\#}= \| \|x\|+\|y\| \|_{\#}$

 

[Mark44]

That's against forum rules.

Here's an example, though.
Suppose x = <3, 2> and y = <-1, 5>
||x + y||_# = ||<2, 7>||_# = |2| + 2|7| = 2 + 14 = 16

I can justify every step above. Can you provide a reason for each step?

See if you can use this as a template to calculate a more general result: ||x + y||_#, with x = <x₁, x₂> and y = <y₁, y₂>

 

[Mark44]

OK, should it be

$\| x+y\|_{\#}= \| \|x\|+\|y\| \|_{\#}$

Way off.

 

[bugatti79]

That's against forum rules.

Here's an example, though.
Suppose x = <3, 2> and y = <-1, 5>
||x + y||_# = ||<2, 7>||_# = |2| + 2|7| = 2 + 14 = 16

I can justify every step above. Can you provide a reason for each step?

See if you can use this as a template to calculate a more general result: ||x + y||_#, with x = <x₁, x₂> and y = <y₁, y₂>

x=(x1,x2), y=(y1,y2) both in R^2

line 1 ||x+y||_#=||(x1+y1),(x2+y2)||_# then using definition of the norm || ||_# we have
= |x1+y1|+2|x2+y2|
=|x1|+|y1|+2|x2| +2|y2|
=|x1|+2|x2|+|y1|+2|y2|
= ||x||_#+||y||_#

I am not sure how to justify going from line 2 to line 3...?

At what stage do I play with the inequalities?

 

[micromass]

x=(x1,x2), y=(y1,y2) both in R^2

line 1 ||x+y||_#=||(x1+y1),(x2+y2)||_# then using definition of the norm || ||_# we have
= |x1+y1|+2|x2+y2|
=|x1|+|y1|+2|x2| +2|y2|
=|x1|+2|x2|+|y1|+2|y2|
= ||x||_#+||y||_#

I am not sure how to justify going from line 2 to line 3...?

At what stage do I play with the inequalities?

Line 2 to 3 is wrong. You won't have an [STRIKE]inequality[/STRIKE] equality. But perhaps you can use

|a+b|\leq |a|+|b|

for a,b\in \mathbb{R}.

 

[bugatti79]

That's against forum rules.

Here's an example, though.
Suppose x = <3, 2> and y = <-1, 5>
||x + y||_# = ||<2, 7>||_# = |2| + 2|7| = 2 + 14 = 16

I can justify every step above. Can you provide a reason for each step?

See if you can use this as a template to calculate a more general result: ||x + y||_#, with x = <x₁, x₂> and y = <y₁, y₂>

From ||x + y||_# = ||<2, 7>||_# because we have x+y=(x1+y1, x2+y2)

From ||<2, 7>||_# = |2| + 2|7| because we have the definition of the norm ||x ||_# as |x1 |+ 2 | x2|

 

[Mark44]

x=(x1,x2), y=(y1,y2) both in R^2

line 1 ||x+y||_#=||(x1+y1),(x2+y2)||_# then using definition of the norm || ||_# we have
= |x1+y1|+2|x2+y2|
=|x1|+|y1|+2|x2| +2|y2|
=|x1|+2|x2|+|y1|+2|y2|
= ||x||_#+||y||_#

I am not sure how to justify going from line 2 to line 3...?

As already noted by micromass, it's not justifiable.

Your first couple of steps were a good start, though, giving justifications for those steps. If you think about things from this perspective (giving a justification - definition, theorem, etc. - for each step) you will have made an important step toward thinking mathematically.

 

[Mark44]

From ||x + y||_# = ||<2, 7>||_# because we have x+y=(x1+y1, x2+y2)

From ||<2, 7>||_# = |2| + 2|7| because we have the definition of the norm ||x ||_# as |x1 |+ 2 | x2|

Yes, exactly right. This is precisely what we have been trying to get you to do.

 

[Fredrik]

I have spent 3.5 hours on this trivial problem.

I haven't counted the hours I've spent trying to help you, but I would guess that it's at least that many, just for this thread.

x=(x1,x2), y=(y1,y2) both in R^2

line 1 ||x+y||_#=||(x1+y1),(x2+y2)||_# then using definition of the norm || ||_# we have
= |x1+y1|+2|x2+y2|

Finally! This is what I've been trying to get you to do the whole time. Note that there's nothing fancy going on here. It's just the definition of +, followed by the definition of $\|\ \|_{\#}$.

However, the notation $||(x1+y1),(x2+y2)||_{\#}$ is a bit messed up. It should say $\|(x_1+y_1,x_2+y_2)\|_{\#}$.

At what stage do I play with the inequalities?

When you go from line 2 to line 3. Micromass gave you the details.

 

[bugatti79]

Line 2 to 3 is wrong. You won't have an inequality. But perhaps you can use

|a+b|\leq |a|+|b|

for a,b\in \mathbb{R}.

From ||x + y||_# = ||<2, 7>||_# because we have x+y=(x1+y1, x2+y2)

From ||<2, 7>||_# = |2| + 2|7| because we have the definition of the norm ||x ||_# as |x1 |+ 2 | x2|

As already noted by micromass, it's not justifiable.

Your first couple of steps were a good start, though, giving justifications for those steps. If you think about things from this perspective (giving a justification - definition, theorem, etc. - for each step) you will have made an important step toward thinking mathematically.

Given $x=(x1,x2), y=(y1,y2)$ both in R^2

$\|x+y\|_{\#}=\|(x1+y1,x2+y2)\|_{\#}$ where $x+y=(x1+y1, x2+y2)$ by definition of vector addition in R^2

then using definition of the norm || ||_# given we have
$\|x+y\|_{\#}= |x1+y1|+2|x2+y2|$

but by the triangle inequality we have |a+b| <= |a| +|b| hence

$\|x+y\|_{\#}\le|x1|+|y1|+2|x2| +2|y2|$
$\le|x1|+2|x2|+|y1|+2|y2|$
$\le \|x\|_{\#}+\|y\|_{\#}$ Therefore axiom 4 holds

 

[bugatti79]

Yes, exactly right. This is precisely what we have been trying to get you to do.

Line 2 to 3 is wrong. You won't have an [STRIKE]inequality[/STRIKE] equality. But perhaps you can use

|a+b|\leq |a|+|b|

for a,b\in \mathbb{R}.

I haven't counted the hours I've spent trying to help you, but I would guess that it's at least that many, just for this thread.


Finally! This is what I've been trying to get you to do the whole time. Note that there's nothing fancy going on here. It's just the definition of +, followed by the definition of $\|\ \|_{\#}$.

However, the notation $||(x1+y1),(x2+y2)||_{\#}$ is a bit messed up. It should say $\|(x_1+y_1,x_2+y_2)\|_{\#}$.


When you go from line 2 to line 3. Micromass gave you the details.

Ok guys, thanks for your help. I really appreciate it considering you could be spending your time on less trivial tasks. I will tackle part 2 in thread 1 tomorrow. night!

 

[Fredrik]

Given $x=(x1,x2), y=(y1,y2)$ both in R^2

$\|x+y\|_{\#}=\|(x1+y1,x2+y2)\|_{\#}$ where $x+y=(x1+y1, x2+y2)$ by definition of vector addition in R^2

then using definition of the norm || ||_# given we have
$\|x+y\|_{\#}= |x1+y1|+2|x2+y2|$

but by the triangle inequality we have |a+b| <= |a| +|b| hence

$\|x+y\|_{\#}\le|x1|+|y1|+2|x2| +2|y2|$
$\le|x1|+2|x2|+|y1|+2|y2|$
$\le \|x\|_{\#}+\|y\|_{\#}$ Therefore axiom 4 holds

Correct. Here's how I would do it:

For all $x,y\in\mathbb R^2$, we have
\begin{align}
\|x+y\|_{\#} &=\|(x_1+y_1,x_2+y_2)\|_{\#}=|x_1+y_1|+2|x_2+y_2| \leq|x_1|+|y_1|+2|x_2|+2|y_2|=\|x\|_{\#}+\|y\|_{\#}.
\end{align} As you can see, it's not so hard if you just use the definitions and do one step at a time.

 

[bugatti79]

(ii) || ||_3:R^2 defined by ||x||=3|x_1|

Is || ||_3 a special function like || ||_1, || ||_2 and || ||_infty etc?
Or do I proceed as usual to test the axioms?

 

[Mark44]

(ii) || ||_3:R^2 defined by ||x||=3|x_1|

The last part should be ||x||_3 = 3|x_1|. That line is defining how this norm works

Is || ||_3 a special function like || ||_1, || ||_2 and || ||_infty etc?

No. The || ||₁ and || ||₂ norms are special functions. This norm is what is being defined in the top line here.

Or do I proceed as usual to test the axioms?

 

[bugatti79]

Decide whether or not the following is a norm defined on R^2, hence verify if it is or prove by counter example if it is not.
Q2 ii
(ii) || ||_3: R^2 to R defined by || x ||_3= 3 |x_1| for x=(x_1,x_2) in R^2

a) Let x be an element in R^2.


Since |x_1| >=0 then by definition we have ||x||_3=3|x_1| >=0 for x_1 in R.

b) ||x||_3=3|x_1|=0 iff x=0

c) Let a be an element in R, then

||ax||_3=3|ax_1|
= |a|(3|x_1|)
=|a| ||x||_3

d) let y=(y_1) be an element in R^2, then

||x+y||_3=||(x_1+y_1)||_3=3|x_1+y_1| we have |a+b|<=|a|+|b| therefore
<=3(|x_1|+|y_1|)
<=3|x_1|+3|y_1|
<=||x||_3+||y||_3

Therefore all 4 axioms hold implies this norm is defined in R^2...?

 

[Fredrik]

a) Correct
b) If what you're saying is true, then (0,1)=(0,0). This would imply that 1=0.
c) Correct. I would break up the first step into two, for clarity: $\|ax\|_3=\|(ax_1,ax_2)\|_3=3|ax_1|=\cdots$.
d) What do you mean by y=(y_1)? An ordered pair isn't equal to its first component. Also, x+y is not equal to x_1+y_1. The latter is a real number, not a vector, so you also can't use it as input for the function $\|\ \|_3$. Also, the last two inequalities are actually equalities, so replace <= with = there.

 

[bugatti79]

Decide whether or not the following is a norm defined on R^2, hence verify if it is or prove by counter example if it is not.
Q2 ii
(ii) || ||_3: R^2 to R defined by || x ||_3= 3 |x_1| for x=(x_1,x_2) in R^2

a) Let x be an element in R^2.


Since |x_1| >=0 then by definition we have ||x||_3=3|x_1| >=0 for x_1 in R.

b) ||x||_3=3|x_1|=0 iff x=0

c) Let a be an element in R, then

||ax||_3=3|ax_1|
= |a|(3|x_1|)
=|a| ||x||_3

d) let y=(y_1) be an element in R^2, then

||x+y||_3=||(x_1+y_1)||_3=3|x_1+y_1| we have |a+b|<=|a|+|b| therefore
<=3(|x_1|+|y_1|)
<=3|x_1|+3|y_1|
<=||x||_3+||y||_3

Therefore all 4 axioms hold implies this norm is defined in R^2...?

a) Correct
b) If what you're saying is true, then (0,1)=(0,0). This would imply that 1=0.
c) Correct. I would break up the first step into two, for clarity: $\|ax\|_3=\|(ax_1,ax_2)\|_3=3|ax_1|=\cdots$.
d) What do you mean by y=(y_1)? An ordered pair isn't equal to its first component. Also, x+y is not equal to x_1+y_1. The latter is a real number, not a vector, so you also can't use it as input for the function $\|\ \|_3$. Also, the last two inequalities are actually equalities, so replace <= with = there.

b) I am actually in 2 conflicting thoughts here and I don't know which is wrong. My first thought is this:
ok, so this axiom does not hold therefore I need to demonstrate by counter example. So if I use this example instead of yours.
since x=(x1,x2). Let x=(1,0) then ||x||_3=||(x1)||_3=3|1|=3. Implies ||x||_3 not = 0 for all x_1 in R

My other thought is this:
Based on what I originally wrote which is this ||x||_3=3|x_1|=0 iff x=0
Wouldn't ||x||_3 =0 always because when I say x=0 I mean x=(x_1,x_2)=0, ie x_1 and x_2 are both 0 in R

d) Ah, because we are dealing in R^2, y should be y=(y_1,y_2).

||x+y||_3=||(x_1+y_1, x_2+y_2)||_3=3|x_1+y_1| we have |a+b|<=|a|+|b| therefore
=3(|x_1|+|y_1|)
=3|x_1|+3|y_1|
=||x||_3+||y||_3

I don't understand why you leave out the <?

 

[Fredrik]

since x=(x1,x2). Let x=(1,0) then ||x||_3=||(x1)||_3=3|1|=3. Implies ||x||_3 not = 0 for all x_1 in R

I don't know why you write $\|x\|_3=\|(x_1)\|_3$. x is an ordered pair of real numbers. x₁ is a real number. Also, once you have specified that x=(1,0), then there's no need to use the notation x₁. If x=(1,0), then clearly $\|x\|=\|(1,0)\|$. That should be your starting point if you want to try to use (1,0) as a counterexample.

Also, if you have already specified that x=(1,0), which implies that x₁=1, you can't say "for all x₁ in ℝ".

Based on what I originally wrote which is this ||x||_3=3|x_1|=0 iff x=0
Wouldn't ||x||_3 =0 always because when I say x=0 I mean x=(x_1,x_2)=0, ie x_1 and x_2 are both 0 in R

Clearly x=0 implies $\|x\|_3=\|(0,0)\|_3=3|0|=0$. What you need to answer after making this observation is if it's also true that $\|x\|_3=0$ implies that $x=0$.

d) Ah, because we are dealing in R^2, y should be y=(y_1,y_2).

Yes.

||x+y||_3=||(x_1+y_1, x_2+y_2)||_3=3|x_1+y_1| we have |a+b|<=|a|+|b| therefore
=3(|x_1|+|y_1|)
=3|x_1|+3|y_1|
=||x||_3+||y||_3

I don't understand why you leave out the <?

You don't. I said "the last two inequalities", not "the last three inequalities". What I'm saying is that it looks weird to write a≤b when you know for sure that a=b. For example, 5≤3+2 is correct, but it still looks weird. So I'm suggesting that you use equality signs when you know that the two numbers are equal.

 

[bugatti79]

I don't know why you write $\|x\|_3=\|(x_1)\|_3$. x is an ordered pair of real numbers. x₁ is a real number. Also, once you have specified that x=(1,0), then there's no need to use the notation x₁. If x=(1,0), then clearly $\|x\|=\|(1,0)\|$. That should be your starting point if you want to try to use (1,0) as a counterexample.

Also, if you have already specified that x=(1,0), which implies that x₁=1, you can't say "for all x₁ in ℝ".

b) since x=(x1,x2). Let x=(1,0) then ||x||_3=||(1,0)||_3=3|1|=3. Implies ||x||_3 not = 0
Therefore, this axiom does not hold for all x in R^2

...if it's also true that $\|x\|_3=0$ implies that $x=0$.

but that is just the reverse of what you have shown 'when x=0 etc implies...' right?

You don't. I said "the last two inequalities", not "the last three inequalities". What I'm saying is that it looks weird to write a≤b when you know for sure that a=b. For example, 5≤3+2 is correct, but it still looks weird. So I'm suggesting that you use equality signs when you know that the two numbers are equal.

ok

 

[Mark44]

b) since x=(x1,x2). Let x=(1,0) then ||x||_3=||(1,0)||_3=3|1|=3. Implies ||x||_3 not = 0
Therefore, this axiom does not hold for all x in R^2

First off, this is not a counterexample. The axiom is
|| x ||₃ = 0 iff x = 0
What you are saying here is that if x = <1, 0>, then || x ||₃ = 3. Your vector here is not the zero vector in R², and its norm (|| ||₃ norm) isn't zero either.

A counterexample would be either
a) x = <0, 0> having a norm that is nonzero, OR
b) x is nonzero, but its norm is zero.

Can you think of some other vector in R² that isn't the zero vector, but its norm is zero?

Second, don't say "since x = (x1, x2). Let x = (1, 0) ..."
Being that this is the 2nd of four axioms, the context here should already be clear that you are dealing with vectors in R².

Just say, "Let x = <1, 0> ... " (or whatever), and continue from there.

but that is just the reverse of what you have shown 'when x=0 etc implies...' right?



ok

 

[Fredrik]

b) since x=(x1,x2). Let x=(1,0) then ||x||_3=||(1,0)||_3=3|1|=3. Implies ||x||_3 not = 0
Therefore, this axiom does not hold for all x in R^2

This time, the calculation is correct and the notation is fine, but you have only proved that there's an x such that $\|x\|_3\neq 0$. As Mark 44 has already said, what you want to prove is that there's an x such that $\|x\|_3=0$ and x≠0. The reason is that "A implies B" is equivalent to "not (A and not B)". The negation of this statement is "A and not B". So if you want to prove that $\|x\|_3\Rightarrow x=0$ is false, you do it by finding an x such that $\|x\|_3=0$ and x≠0.

but that is just the reverse of what you have shown 'when x=0 etc implies...' right?

Yes, but I'll add that the technical term is converse, not "reverse". The converse of an implication $A\Rightarrow B$ is the implication $B\Rightarrow A$.

 

[bugatti79]

First off, this is not a counterexample. The axiom is
|| x ||₃ = 0 iff x = 0
What you are saying here is that if x = <1, 0>, then || x ||₃ = 3. Your vector here is not the zero vector in R², and its norm (|| ||₃ norm) isn't zero either.

A counterexample would be either
a) x = <0, 0> having a norm that is nonzero, OR
b) x is nonzero, but its norm is zero.

Can you think of some other vector in R² that isn't the zero vector, but its norm is zero?

Second, don't say "since x = (x1, x2). Let x = (1, 0) ..."
Being that this is the 2nd of four axioms, the context here should already be clear that you are dealing with vectors in R².

Just say, "Let x = <1, 0> ... " (or whatever), and continue from there.

I guess fredrik provided a counterexample, ie x equal (0,1). So this is sufficient to show axiom 2 does not hold? I can't think of any other example. So i guess this question is reasonably answered.

 

[Fredrik]

Yes, (0,1) is a counterexample to the statement "For all x, $\|x\|_3=0\Leftrightarrow x=0$, because $\|x\|_3=0$ and x≠0. So $\|\ \|_3$ is not a norm. You got that right.