The book mentioned in the video can be found here:
Genius At Play: The Curious Mind of John Horton Conway
Many scientists and research has revealed that the true source of genius comes from both nature and nurture.
Talent is Overrated: What Really Separates World-Class Performers from Everybody Else
GEP Test Dates (August)
In August, Primary 3 pupils in Singapore schools have the opportunity to take the GEP Screening Test, comprising 2 papers: English Language and Mathematics.
Check out the Recommended Books for GEP Test here!
IQ is based on both nature (inheritance of genes), and more importantly nurture (habits, family background, books read as a child, …), hence a logical way to prepare for the GEP is to read some books relevant to the GEP test. Some preparation is always better than zero preparation, as authors of many self-help books have researched and concluded.
It would be a huge advantage for students to be familiar with basic logic quizzes that are found in IQ tests. Seeing this type of question for the first time in the GEP test would not be very conducive as it may lead to nervousness which would affect the logical thinking.
be a measure defined on a 
-algebra X.
) is an increasing sequence in X, then
) is a decreasing sequence in X and if 
, then
. A decreasing sequence means the opposite, i.e. 
.
for some n, then both sides of the equation are 
, and inequality holds. Henceforth, we can just consider the case 
for all n.
and 
for n>1. Then 
is a disjoint sequence of sets in X such that
, 
(since (
) is a disjoint sequence of sets)
, we have that 
for n>1, so the finite series on the right side telescopes to become
, so that 
is an increasing sequence of sets in X.![\begin{aligned} \mu (\bigcup_{n=1}^\infty E_n) &=\lim \mu (E_n)\\ &=\lim [\mu (F_1)-\mu (F_n)]\\ &=\mu (F_1) -\lim \mu (F_n) \end{aligned}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D++++%5Cmu+%28%5Cbigcup_%7Bn%3D1%7D%5E%5Cinfty+E_n%29+%26%3D%5Clim+%5Cmu+%28E_n%29%5C%5C++++%26%3D%5Clim+%5B%5Cmu+%28F_1%29-%5Cmu+%28F_n%29%5D%5C%5C++++%26%3D%5Cmu+%28F_1%29+-%5Clim+%5Cmu+%28F_n%29++++%5Cend%7Baligned%7D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
, it follows that
.
, let 
denote the constant path 
mpping all of I to the point x. If f is a path in X from 
to 
, then ![[f]*[e_{x_1}]=[f]](https://s0.wp.com/latex.php?latex=%5Bf%5D%2A%5Be_%7Bx_1%7D%5D%3D%5Bf%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
and ![[e_{x_0}]*[f]=[f]](https://s0.wp.com/latex.php?latex=%5Be_%7Bx_0%7D%5D%2A%5Bf%5D%3D%5Bf%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
.
be the path defined by 
. ![[f]*[\bar{f}]=[e_{x_0}]](https://s0.wp.com/latex.php?latex=%5Bf%5D%2A%5B%5Cbar%7Bf%7D%5D%3D%5Be_%7Bx_0%7D%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
and ![[\bar{f}]*[f]=[e_{x_1}]](https://s0.wp.com/latex.php?latex=%5B%5Cbar%7Bf%7D%5D%2A%5Bf%5D%3D%5Be_%7Bx_1%7D%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
.
is a continuous map, and if F is a path homotopy in X between the paths f and f’, then 
is a path homotopy in Y between the paths 
and 
.
is a continuous map and if f and g are paths in X with f(1)=g(0), then
, where h=f*g as defined previously.
.
denote the constant path in I at 0, and we let 
denote the identity map, which is a path in I from 0 to 1. Then 
is also a path in I from 0 to 1.
(Straight-line homotopy) Then 
is a path homotopy in X between the paths 
and 
(Lemma 1). Furthermore by Lemma 2, 
which is equivalent to 
.
denotes the constant path at 1, then 
is path homotopic in I to the path i, shows that 
. Then 
is a path in I beginning and ending at 0. The constant path 
is also beginning and ending at 0. Again, because I is convex, there is a path homotopy H in I between 
is path homotopy between 
and 
. Very similarly, we can use the fact that 
is path homotopic in I to 
. It covers the areas of Algebra, Geometry, Counting and Probability, and Number Sense, over 500 examples and problems with fully explained solutions.
and measures should be additive over disjoint sets in X.
for all 
) of sets in X, then
.
, we say it is finite. More generally, if there exists a sequence 
and such that 
for all n, then we say that 
be definied on X by 
, for all 
. We can see that 
be defined by 
, 
if 
. 
defined on B which coincides with length on open intervals. I.e. if E is the nonempty interval (a,b), then 
. This measure is usually called Lebesgue measure (or sometimes Borel measure). It is not a finite measure since 
. But it is 
, we define the product f*g of f and g to be he path h given by the equations![h(s)=\begin{cases}f(2s) &\text{for }s\in [0,\frac{1}{2}], \\ g(2s-1)& \text{for }s\in[\frac{1}{2}, 1]\end{cases}](https://s0.wp.com/latex.php?latex=h%28s%29%3D%5Cbegin%7Bcases%7Df%282s%29+%26%5Ctext%7Bfor+%7Ds%5Cin+%5B0%2C%5Cfrac%7B1%7D%7B2%7D%5D%2C+%5C%5C+g%282s-1%29%26+%5Ctext%7Bfor+%7Ds%5Cin%5B%5Cfrac%7B1%7D%7B2%7D%2C+1%5D%5Cend%7Bcases%7D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
![[f]*[g]=[f*g]](https://s0.wp.com/latex.php?latex=%5Bf%5D%2A%5Bg%5D%3D%5Bf%2Ag%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
![H(s,t)=\begin{cases}F(2s,t) &\text{for }s\in[0,\frac{1}{2}],\\ G(2s-1,t)&\text{for }s\in[\frac{1}{2},1]\end{cases}](https://s0.wp.com/latex.php?latex=H%28s%2Ct%29%3D%5Cbegin%7Bcases%7DF%282s%2Ct%29+%26%5Ctext%7Bfor+%7Ds%5Cin%5B0%2C%5Cfrac%7B1%7D%7B2%7D%5D%2C%5C%5C+G%282s-1%2Ct%29%26%5Ctext%7Bfor+%7Ds%5Cin%5B%5Cfrac%7B1%7D%7B2%7D%2C1%5D%5Cend%7Bcases%7D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
.
belongs to X for every set E belonging to Y.
the set 
belongs to X.
for every Borel set E.
.
if f and f’ are path homotopic.
and 
are equivalence relations.
. The map F(x,t) is the required homotopy.
. Let F be a homotopy between f and f’. We can then verify that G(x,t) = F(x, 1-t) is a homotopy between f’ and f.
, we show that 
. Let F be a homotopy between f and f’, and let F’ be a homotopy between f’ and f”. This time, we need to define a slightly more complicated homotopy G: X x I -> Y by the equation![G(x,t) = \begin{cases} F(x,2t) &\text{for }t\in [0,\frac{1}{2}],\\ F'(x, 2t-1) &\text{for } t\in [\frac{1}{2}, 1].\end{cases}](https://s0.wp.com/latex.php?latex=G%28x%2Ct%29+%3D+%5Cbegin%7Bcases%7D+F%28x%2C2t%29+%26%5Ctext%7Bfor+%7Dt%5Cin+%5B0%2C%5Cfrac%7B1%7D%7B2%7D%5D%2C%5C%5C+F%27%28x%2C+2t-1%29+%26%5Ctext%7Bfor+%7D+t%5Cin+%5B%5Cfrac%7B1%7D%7B2%7D%2C+1%5D.%5Cend%7Bcases%7D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
) in M(X,X) such that:
for 
.
for each 
, let 
be the set
.
, let 
.
are disjoint.
on 
is true.
, i.e. 
, thus 
for each 
are also in M(X,X).
, let 
be the “truncation of f” defined by 
be defined similarly. We will work out the proof that 
to 
.
is measurable.
, and using an earlier corollary that says that if a sequence 
is in M(X,X) converges to f on X, then f is also in M(X,X), we have that 
belongs to M(X,X).
, and hence fg also belongs to M(X,X).
implies 
.
consists of a half-line which is either of the form 
or the form 
. (We will show later that both cases can occur.) Thus, the set will belong to the Borel algebra B which is the 
, then when 
, the set will be 
.
, 
belong to X and the real-valued function 
defined by 
is measurable.
and 
, then we have that 
which is in X since it is the complement of the union of A and 
.
, then 
, which is a union of two sets in X and hence also in X.
and 
when 
, and 
when 
, due to a similar reason as above. Therefore f is measurable!
belongs to X.
belongs to X,
belongs to X,
belongs to X,
belongs to X,
and 
are complements of each other, hence statement (a) is equivalent to statement (b). This is due to one of the properties of 
belongs to X for each n.
, it follows that 
. Thus, (a) implies (c).
, and hence if (c) is true, each 
is in X, and the union of them is also in X (definition for 
, or even “<” and nothing would change!
by Stephen Covey (Highly recommended to read), there are four types of activities:
.
Mathtuition88 