leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

InteGrand · Feb 1, 2016

leehuan said:
Why is everyone infesting my cry-for-help thread.

They mistook it as a general thread for everyone to post their Q's in I think.

leehuan · Feb 1, 2016

InteGrand said:
They mistook it as a general thread for everyone to post their Q's in I think.

I realised actually. Though I would've hoped that with the earlier posts by Speed and Drsoccerball people would've realised.

InteGrand · Feb 1, 2016

leehuan said:
I realised actually. Though I would've hoped that with the earlier posts by Speed and Drsoccerball people would've realised.

They probably didn't read through the whole thread carefully.

Drsoccerball · Feb 1, 2016

leehuan said:
I realised actually. Though I would've hoped that with the earlier posts by Speed and Drsoccerball people would've realised.

Nop, this is basically the confessions forum for maths.

DatAtarLyfe · Feb 2, 2016

Yahh, a new thread to derail

leehuan · Feb 2, 2016

leehuan · Feb 2, 2016

More general questions please

$\\ $Is it convention to refer to the unit vectors using specifically the letters $i, j, k$? i.e. $ \bf{i}=\left( 1,0,0 \right) , \bf{j} =\left( 0,1,0 \right) , \bf{k}=\left( 0,0,1 \right)$
This is mainly cause so far I've only read one textbook.

$\\ $So now that I know for vectors $\bf{a},\bf{b}$, \\ $\bf{a} \cdot \bf{b} = \left\| \bf{a} \right\| \left\| \bf{b} \right\| \cos{\theta}$, \\ would my assumption be that the dot product is useful to determine the angle between vectors?$

seanieg89 · Feb 2, 2016

leehuan said:
$\\ $The concept that for a vector $\bf{a}$, we have $\left\| \bf{a} + \alpha \bf{a} \right\| = \left\| \bf{a} \right\| + \left\| \alpha \bf{a} \right\| , \alpha >0$ only makes sense to me through a geometric visualisation in my head. How would you prove this result?$

Comes from the absolute homogeneity of norms. $(\|cx\|=|c|\|x\|)$

So:

$\|a+\alpha a\|=\|(1+\alpha)a\|=|1+\alpha|\|a\|=(1+|\alpha|)\| a \|=\|a\|+\|\alpha a\|.$

leehuan · Feb 2, 2016

seanieg89 said:
Comes from the absolute homogeneity of norms. $(\|cx\|=|c|\|x\|)$

So:

$\|a+\alpha a\|=\|(1+\alpha)a\|=|1+\alpha|\|a\|=(1+|\alpha|)\| a \|=\|a\|+\|\alpha a\|.$

Right, the first part makes sense. But I got lost here:

$|1+\alpha|\|a\|=(1+|\alpha|)\| a \|$

seanieg89 · Feb 2, 2016

leehuan said:
More general questions please

$\\ $Is it convention to refer to the unit vectors using specifically the letters $i, j, k$? i.e. $ \bf{i}=\left( 1,0,0 \right) , \bf{j} =\left( 0,1,0 \right) , \bf{k}=\left( 0,0,1 \right)$
This is mainly cause so far I've only read one textbook.

$\\ $So now that I know for vectors $\bf{a},\bf{b}$, \\ $\bf{a} \cdot \bf{b} = \left\| \bf{a} \right\| \left\| \bf{b} \right\| \cos{\theta}$, \\ would my assumption be that the dot product is useful to determine the angle between vectors?$

1. Yes, this is a pretty common convention. Another one is $e_i$ to denote the coordinate vector whose only nonzero entry is it's i-th one, which is equal to 1.

2. Yep. In fact the dot product is actually a more fundamental notion than that of an angle. We use the dot product (or more generally an inner product) to DEFINE the concept of an angle between two vectors in an abstract vector space that has an inner product.

seanieg89 · Feb 2, 2016

leehuan said:
Right, the first part makes sense. But I got lost here:

$|1+\alpha|\|a\|=(1+|\alpha|)\| a \|$

Because $|1+\alpha|=1+\alpha=1+|\alpha|$

noting that taking the modulus does not change a positive number, and also using the fact that alpha is positive by assumption.

leehuan · Feb 2, 2016

seanieg89 said:
Because $|1+\alpha|=1+\alpha=1+|\alpha|$

noting that taking the modulus does not change a positive number, and also using the fact that alpha is positive by assumption.

Oh I'm so silly. Thanks!

seanieg89 · Feb 2, 2016

No worries

.

leehuan · Feb 2, 2016

InteGrand · Feb 2, 2016

leehuan said:
$\\ $Wait, so restricting the domain is actually required to determine inverse hyperbolic cosine as well? Wow...$

$\noindent Yes. This is because the graph of $y = \cosh x$ fails the Horizontal Line Test.$

Paradoxica · Feb 2, 2016

InteGrand said:
$\noindent Yes. This is because the graph of $y = \cosh x$ fails the Horizontal Line Test.$

Or to put it more succinctly, no even function can possibly have a complete inverse.

leehuan · Feb 2, 2016

Too many even functions out there I swear.

drsabz101 · Feb 3, 2016

How do I prove that log base 2 to the power of 3 is irrational ? Can u go step by step in the method?

Thankyou

InteGrand · Feb 3, 2016

drsabz101 said:
How do I prove that log base 2 to the power of 3 is irrational ? Can u go step by step in the method?

Thankyou

$\noindent What do you mean by that? Do you mean the number $\log _ 2 3$ (we shouldn't be saying `to the power of' 3)? Generally for these Q's, the approach is a proof by contradiction.$

$\noindent Assuming you meant this, assume by way of contradiction that $\log_2 3 =\frac{p}{q}$ for some positive integers $p$ and $q$ (i.e. it is rational (obviously $\log_2 3$ is positive so we can take $p$ and $q$ to both be positive)). Then $3 = 2^{\frac{p}{q}} \Rightarrow 3^q = 2^p$. But the left-hand side here is an odd number (any positive integer power of any odd number is odd) whilst the right-hand side is even (any positive integer power of an even number is even), so we have the desired contradiction.$

drsabz101 · Feb 3, 2016

Yes thankyou

leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

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