點解覺得數學公式難背?

咁咪直情玩死埋啲science仔
下下都要mathematical proof 淨係俾你地Phys maths佬讀得啦
利申：由calculus開始無視哂啲proof

無讀過proof就唔好話自己讀數學

無話自己讀過calculus
如果唔係maths compu應該好多人揀唔讀

dse core 要玩proof唔會好難,不過耐冇用就會唔記得點prove。我中學連proof都明埋。

Calculus 我都無視哂d proof, 知咩情況做咩但再深入就唔知。坦白講，yr1 d in 你識做真係夠做,佢又唔會問到好深入。讀緊surface integral, multiple integral , div thm etc, d proof 煩左睇都唔想睇，知點用同點做就算。

non math major既話根本冇所謂

一知半解就得

講多句

dse core以至小學數學既proof根本一d都唔容易

小學嘅proof係最難

e.g. we have a+b+c = c+b+a
but a-b-c =/= c-b-a
That is, addition and multiplication of real numbers are commutative, but not for subtraction and division. I wonder why.

Commutativity is the property a*b=b*a for any a,b in A and binary operation * on A, so what you should start with is “a+b=b+a but NOT a-b=b-a”.

So first thing first, why would we even consider commutativity? My guess is that commutativity is a special property. For example, finitely generated commutative groups (aka abelian groups) are classified by the fundamental theorem of abelian groups, which is an undergraduate level theorem; on the other hand classification of finite simple groups (where the abelian ones are cyclic of prime order) requires years of work. In short, commutative things tend to behave well.

Second, if commutativity is nice, then why would we consider subtraction and division, which are non-commutative? Because we need to solve equations like “a+x=b” and “ax=b”(for a non-zero). Here another question arises: whether having subtraction and division (by non-zero) makes sense? The answer is, unsurprisingly, yes. To see this we define natural numbers, addition and multiplication using Peano axioms, which turns out to be a commutative monoid under addition, allowing us to construct its Grothendieck group, which is the set of all integers. Now we need to define multiplication on integers by filling the missing part of our multiplication table, which is... easy considering every integer is either a natural number or the negative of a natural number, so we do it with the obvious way. Next we take the quotient field of integers to get the set of all rational numbers. Finally we use the method of Dedekind cut to construct the set of all real numbers. In short, “God made the natural numbers; all else is the work of man.” And subtraction and division (by non-zero) constructed really make sense, that is, can be used to solve “a+x=b” and “ax=b”.

That’s why even though subtraction and division don’t behave well, we still consider them.

次次數學post都見到烏大龜
幾時去證明自己啊？

無話自己讀過calculus
如果唔係maths compu應該好多人揀唔讀

dse core 要玩proof唔會好難,不過耐冇用就會唔記得點prove。我中學連proof都明埋。

Calculus 我都無視哂d proof, 知咩情況做咩但再深入就唔知。坦白講，yr1 d in 你識做真係夠做,佢又唔會問到好深入。讀緊surface integral, multiple integral , div thm etc, d proof 煩左睇都唔想睇，知點用同點做就算。

non math major既話根本冇所謂

一知半解就得

講多句

dse core以至小學數學既proof根本一d都唔容易

小學嘅proof係最難

e.g. we have a+b+c = c+b+a
but a-b-c =/= c-b-a
That is, addition and multiplication of real numbers are commutative, but not for subtraction and division. I wonder why.

Commutativity is the property a*b=b*a for any a,b in A and binary operation * on A, so what you should start with is “a+b=b+a but NOT a-b=b-a”.

So first thing first, why would we even consider commutativity? My guess is that commutativity is a special property. For example, finitely generated commutative groups (aka abelian groups) are classified by the fundamental theorem of abelian groups, which is an undergraduate level theorem; on the other hand classification of finite simple groups (where the abelian ones are cyclic of prime order) requires years of work. In short, commutative things tend to behave well.

Second, if commutativity is nice, then why would we consider subtraction and division, which are non-commutative? Because we need to solve equations like “a+x=b” and “ax=b”(for a non-zero). Here another question arises: whether having subtraction and division (by non-zero) makes sense? The answer is, unsurprisingly, yes. To see this we define natural numbers, addition and multiplication using Peano axioms, which turns out to be a commutative monoid under addition, allowing us to construct its Grothendieck group, which is the set of all integers. Now we need to define multiplication on integers by filling the missing part of our multiplication table, which is... easy considering every integer is either a natural number or the negative of a natural number, so we do it with the obvious way. Next we take the quotient field of integers to get the set of all rational numbers. Finally we use the method of Dedekind cut to construct the set of all real numbers. In short, “God made the natural numbers; all else is the work of man.” And subtraction and division (by non-zero) constructed really make sense, that is, can be used to solve “a+x=b” and “ax=b”.

That’s why even though subtraction and division don’t behave well, we still consider them.

Some ug guy suggests that:

a-b = a+ (-b)
a/b = a x (1/b)

The existence of x such that b+x=0 is just a special case of solving “b+x=a”, and we use -b to denote such x, similarly we can define 1/b for b non-zero.

Now given such special case if we define a-b=a+(-b) then b+x=a implies x=-b+b+x=-b+a=a+(-b) which is by definition a-b. Similarly we can solve bx=a for b non-zero.

In short, we only need additive inverse (-b) and multiplicative inverse (1/b, for b non-zero) to guarantee we can “undo” addition and multiplication.

有冇高手求救 睇完佢個proof 同google 完都唔明點解 Z+ x Z+ 係countable

<img src="https://img.eservice-hk.net/upload/2017/09/28/141949_e6005ee141287fb833ea91b0e4a8f2aa.jpg" />


有冇高手求救 睇完佢個proof 同google 完都唔明點解 Z+ x Z+ 係countable

subset of N and N is countable

<img src="https://img.eservice-hk.net/upload/2017/09/28/141949_e6005ee141287fb833ea91b0e4a8f2aa.jpg" />


有冇高手求救 睇完佢個proof 同google 完都唔明點解 Z+ x Z+ 係countable

樓主開個post都係想打救下DSE雞姐
做乜要係咁響到拋書包呢

樓主開個post都係想打救下DSE雞姐
做乜要係咁響到拋書包呢

讀數嘅朋友冇幾可可以show off而又有人理

唔使記噶
即刻諗就惦

今晚先學完 2派ft
都唔知講乜q

網上見到呢條唔識計 求教

相似三角形？2個都90度角
上面三角形可以搵哂邊長

[quote]
網上見到呢條唔識計 求教

相似三角形？2個都90度角
上面三角形可以搵哂邊長

好撚耐了 中學成日做 比返哂老師