07 · Coordinate calculation07 · 座標計算
First experiment先做這三步
- Press Reset and check that u=(1,2) and v=(2,−1).按「重設」，確認 u=(1,2)、v=(2,−1)。
- Move a from 1 to −1 and keep b at 1. Predict the direction of au first.把 a 從 1 移到 −1，b 保持 1。先猜 au 的方向。
- Compare the diagram with the coordinate equations: each coordinate is multiplied by a, then added to bv.比較圖與座標算式：每個座標都要乘上 a，再與 bv 相加。
Predict → change one value → inspect the calculation → read the key idea.先預測 → 改一個值 → 檢查算式 → 看懂重點概念。
A vector can be scaled, added, and combined.向量可以伸縮、相加，也可以組合。
Start in ℝ² with real scalars. Predict the result, then change one coordinate or scalar. Negative scalars reverse direction; a zero scalar gives the zero vector.先在 ℝ² 中使用實數純量（scalar）。先預測結果，再改變一個座標或純量。負的純量會使方向相反；純量為 0 時得到零向量（zero vector）。
Choose the vectors選擇向量
u =
v =
Integers, decimals and fractions such as 1/2 are accepted. Coordinates: −5 to 5.可輸入整數、小數或分數（例如 1/2）。座標範圍：−5 到 5。
Linear combination線性組合（linear combination）
au + bv, a,b ∈ ℝ.
Both the coefficients and the operations are part of the definition.係數（coefficient）的範圍和運算方式，都是定義的一部分。Head-to-tail addition頭尾相接的向量加法
The dashed sides complete the parallelogram. Coincident arrows can overlap; the labeled coordinates still distinguish them. The axes rescale to keep the vectors visible.虛線補成一個平行四邊形。重合的箭頭可能互相蓋住，但標示的座標仍能區分它們。座標軸會自動縮放，讓向量保持可見。
Why is the result still in ℝ²?為什麼結果仍在 ℝ² 裡？
u=(u₁,u₂), v=(v₁,v₂)
au+bv=(au₁+bv₁, au₂+bv₂) ∈ ℝ².
au+bv=(au₁+bv₁, au₂+bv₂) ∈ ℝ².
Every coordinate is real because real numbers are closed under addition and multiplication. This argument works for every u, v, a and b; the picture shows only particular choices.因為實數對加法和乘法封閉，每個座標都還是實數。這個論證對所有 u、v、a、b 都成立；圖形只呈現某些特定的選擇。
Challenge挑戰
打中目標Hit the target
調整 a 和 b，讓紅色的 au+bv 剛好落在紫色的 ◎ 上。可以拉滑桿，也可以在圖上拖動藍色、綠色箭頭的尖端：au 只能沿著 u 的方向伸縮，bv 只能沿著 v 的方向伸縮。a、b 可以是 0.5 的倍數。Adjust a and b so that the red au+bv lands exactly on the purple ◎. Use the sliders, or drag the tips of the blue and green arrows on the grid: au can only stretch along u, and bv only along v. a and b can be multiples of 0.5.
Explore first, then match what you observe with the equations.先動手探索，再對照算式，看懂你觀察到的現象。
Find a counterexample這次要找的是反例Start by comparing “y=2x” with “y=2x+1”. Check that your starting vectors are in the set S, then see whether adding or scaling takes you outside it. ∈ means “is in”; ∉ means “is not in”.先選「y=2x」與「y=2x+1」比較。檢查起始向量是否在集合 S 內，再看相加或乘純量後是否離開。∈ 表示「屬於」，∉ 表示「不屬於」。
Does the operation keep you inside the set?運算後還留在集合裡嗎？
Every set below uses the usual operations inherited from ℝ² and real scalars. A failed closure test must start with vectors actually in the set.下面每個集合都沿用 ℝ² 的一般運算，純量為實數。要說封閉性（closure）不成立，起始向量必須真的在集合裡。
Try to break closure試著打破封閉性
u =
v =
Coordinates and scalar: −5 to 5; fractions allowed.座標與純量範圍：−5 到 5；可輸入分數。
A passing example is not a proof. Closure requires u+v ∈ S for all u,v ∈ S and au ∈ S for all a ∈ ℝ, u ∈ S. One valid counterexample is enough to disprove it.一個例子成功不等於證明。封閉性要求對所有 u,v ∈ S 都有 u+v ∈ S，而且對所有 a ∈ ℝ、u ∈ S 都有 au ∈ S。只要一個合法的反例（counterexample）就足以否定它。
Reveal the conclusion and its justification顯示結論與理由
The set and your test vectors集合與你的測試向量
Subspace test (in an existing vector space):子空間（subspace）判別法（在已知的向量空間中）：
0 ∈ S; u,v ∈ S ⇒ u+v ∈ S;
a ∈ ℝ, u ∈ S ⇒ au ∈ S.
0 ∈ S; u,v ∈ S ⇒ u+v ∈ S;
a ∈ ℝ, u ∈ S ⇒ au ∈ S.
These conditions suffice here because the other axioms are inherited from ℝ². For newly defined operations, closure alone is not enough.這裡檢查這三個條件就夠了，因為其他公理（axiom）都從 ℝ² 繼承而來。如果運算是新定義的，只有封閉性並不夠。
Read one rule at a time不用一次背完公理First understand two ideas: after adding or scaling you stay in the set, and there must be a zero vector. Then match each rule to its equation. The polynomial example below shows that a vector need not be an arrow.先理解「加完、乘完還在集合裡」（封閉性 closure）和「要有零向量」。再逐條對照公理（axiom）的式子。下方的多項式例子顯示：向量不一定是箭頭。
A vector space is a set with rules.向量空間是一個帶有規則的集合。
Specify a nonempty set V, a scalar field F, addition V × V → V, and scalar multiplication F × V → V. Here F is usually ℝ; ℂ is another possible field.要指定一個非空集合 V、一個純量體（scalar field）F、加法 V × V → V，以及純量乘法 F × V → V。這裡 F 通常是 ℝ；ℂ 也是一種可能的體。
For every u,v,w ∈ V and a,b ∈ F對所有 u,v,w ∈ V 及 a,b ∈ F
Closure under addition加法封閉性u+v ∈ V
Closure under scaling純量乘法封閉性av ∈ V
Commutativity交換律u+v = v+u
Associativity of addition加法結合律(u+v)+w = u+(v+w)
A zero vector exists存在零向量∃0 ∈ V, ∀v ∈ V: v+0 = v
Additive inverses exist存在加法反元素∀v ∈ V, ∃(−v) ∈ V: v+(−v) = 0
Compatible scalar multiplication純量乘法相容性(ab)v = a(bv)
Scalar identity純量單位元素1v = v
Distribute over vectors對向量加法的分配律a(u+v) = au+av
Distribute over scalars對純量加法的分配律(a+b)v = av+bv
Consequences: zero scaling and negatives推論：乘以 0 與負向量
0v = (0+0)v = 0v+0v ⇒ 0v = 0.
(−1)v+v = (−1+1)v = 0v = 0 ⇒ (−1)v = −v.
(−1)v+v = (−1+1)v = 0v = 0 ⇒ (−1)v = −v.
In the first implication, add the additive inverse of 0v to both sides. The scalar 0 and the vector 0 play different roles.第一個推論中，兩邊同時加上 0v 的加法反元素（additive inverse）。純量 0 和零向量 0 扮演不同的角色。
The same rules for different objects同樣的規則，不同的物件
| Object物件 | constant常數項 | t | t² |
|---|---|---|---|
| p / f | |||
| q / g |
More examples from the notes講義中的更多例子
ℝⁿ: (u+v)ⱼ = uⱼ+vⱼ, (au)ⱼ = auⱼ.
P(ℝ): all real polynomials, with ordinary addition and real scalar multiplication.所有實係數多項式，使用一般的加法與實數純量乘法。
C([r,s],ℝ): continuous functions, with連續函數，運算為 (f+g)(t)=f(t)+g(t), (af)(t)=af(t).
Sums and scalar multiples of continuous functions remain continuous. The zero vector is the function t ↦ 0. In P(ℝ), multiplying by the variable t is not scalar multiplication over ℝ.連續函數的和與純量倍仍是連續函數。零向量是函數 t ↦ 0。在 P(ℝ) 中，乘上變數 t 並不是 ℝ 上的純量乘法。
Why the operations matter為什麼運算很重要
Closure alone can mislead.只看封閉性可能被誤導。
On ℝ², invent u ⊕ v = u+v+(1,0), but keep ordinary scalar multiplication. Both operations stay in ℝ².在 ℝ² 上自訂加法 u ⊕ v = u+v+(1,0)，純量乘法維持一般的方式。兩種運算的結果都還在 ℝ² 裡。
Try distributivity with a=0 and u=v=(0,0):用 a=0、u=v=(0,0) 檢查分配律：
0(u ⊕ v) = (0,0),
(0u) ⊕ (0v) = (0,0)+(0,0)+(1,0) = (1,0).
(0u) ⊕ (0v) = (0,0)+(0,0)+(1,0) = (1,0).
Since the two results differ, these operations do not define a vector space.兩個結果不同，所以這組運算不能構成向量空間。
The subspace shortcut applies to inherited operations. A newly defined addition must satisfy the full axioms.子空間的簡便判別法只適用於繼承來的運算。新定義的加法必須逐條滿足所有公理。
Is the additive identity necessarily (0,0)?加法單位元素一定是 (0,0) 嗎？
For this invented addition, the identity would be (−1,0), because v ⊕ (−1,0)=v. The failed distributive law still rules out a vector space with the chosen scaling.對這個自訂加法來說，單位元素會是 (−1,0)，因為 v ⊕ (−1,0)=v。但在這種純量乘法下分配律失敗，所以仍然不是向量空間。
Challenge挑戰
公理偵探：哪一條規則壞了？Axiom detective: which rule breaks?
上面的例子是別人找到的反例，現在換你來找。下面每一種運算都把結果留在 ℝ² 裡（封閉性成立），但至少有一條公理會失敗。改變 u、v、w、a、b，看哪一列出現「≠」，然後勾選所有會失敗的公理。小心：某些公理只在特定數值下才露出破綻。Above, someone else found the counterexample; now it is your turn. Each operation below keeps results in ℝ² (closure holds), but at least one axiom fails. Change u, v, w, a and b, watch for a row showing “≠”, then tick every axiom that fails. Careful: some axioms only break for particular values.
u = (, )v = (, )w = (, )a = b =
Predict before checking先自己想，再檢查Do one question at a time. Choose an answer or enter numbers, then press Check. If you are wrong, read the explanation, edit your answer and try again. Nothing is marked against you.一次做一題。選答案或填入數值，再按檢查。答錯時看中文說明，修改後可以再試；沒有扣分。
Try it. Check it. Learn the key idea.動手做、核對答案、看懂重點。
Allow 25–30 minutes for exploration and practice. Start with your own calculation; feedback identifies the relevant definition and gives an equation or counterexample. Retrying is encouraged.預留 25–30 分鐘探索與練習。先自己計算；回饋會指出相關的定義，並給出算式或反例。鼓勵你多試幾次。
08 · Build your own counterexample08 · 自己造一個反例
Show that S = {(x,y): xy=0} is not closed under addition.說明 S = {(x,y): xy=0} 對加法不封閉：找兩個集合內的向量，使它們的和跑到集合外。
Enter two members u,v whose sum lies outside S. Many answers are possible.輸入 S 中的兩個向量 u、v，使 u+v 不在 S 裡。答案不只一種。
Keep evidence: one computed linear combination, one valid closure counterexample, and a short explanation of why checking a few examples cannot prove an axiom for every vector.留下學習紀錄：一個算好的線性組合、一個合法的封閉性反例，以及一段簡短說明：為什麼只檢查幾個例子，無法證明公理對每個向量都成立。