Proof. (a) By (M3), a · 1 = a. Hence a + a · 0 = a · 1 + a · 0 = a · (1 + 0) = a · 1 = a, and Theorem 4.2(a) forces a · 0 = 0.
這一步的所求：證 a · 0 是零元的化身。辦法不是直接算它，而是讓它滿足零元的判定等式：湊出 a + a · 0 = a。湊法從 a = a · 1 出發，把 a · 1 + a · 0 用分配律 D 倒著收成 a · (1 + 0)，括號裡 A3 歸 1，於是整串等於 a。等式湊齊，4.2(a) 點名：能讓「加了等於沒加」的只有 0，所以 a · 0 = 0。這也兌現了 §4-2 的預告：0 乘任何數都到不了 1，M4 拒發倒數有理有據。
(b) By (M3) and (D), a + (−1) · a = 1 · a + (−1) · a = (1 + (−1)) · a = 0 · a = 0, where the last step uses part (a) together with (M1). Theorem 4.3(a) then identifies (−1) · a as −a.
所求：證 (−1) · a 正是 a 的反元素。同一套辦法：讓它與 a 相加歸到 0——先把 a 換裝成 1 · a，分配律倒收成 (1 + (−1)) · a = 0 · a，再由 (a) 配 M1 歸 0。等式 a + (−1) · a = 0 到手，4.3(a) 判定身分：(−1) · a = −a。從此「取負號」不再是獨立動作，而是「乘上 −1」的縮寫——後面 (c)、(e) 全吃這條翻譯的紅利。
(c) Using part (b) twice and (D), −(a + b) = (−1) · (a + b) = (−1) · a + (−1) · b = (−a) + (−b).
(d) By (A4), (−a) + a = 0. Reading this equation through Theorem 4.3(a) — with −a in the leading role — shows that a is the additive inverse of −a; that is, a = −(−a).
這一步一行字卻值得放慢：A4 的等式 (−a) + a = 0 本來是替 a 立反元素，換個代入對象重讀——把 −a 放到 4.3(a) 裡 a 的位置——同一條等式就變成「a 與 −a 相加得 0」的身分證明，4.3(a) 判定 a = −(−a)。一條等式，兩種讀法，一次都沒重算。
(e) Put a = −1 in part (b): −(−1) = (−1) · (−1). But part (d) with a = 1 reads −(−1) = 1. Therefore (−1) · (−1) = 1.
(a) If a ≠ 0, then 1/a ≠ 0 and 1/(1/a) = a. (b) If a · b = 0, then a = 0 or b = 0. (c) (−a) · (−b) = a · b for all real a, b. (d) If a ≠ 0, then 1/(−a) = −(1/a).
Proof. (a) Suppose a ≠ 0, so 1/a exists. Were 1/a = 0, then 1 = a · (1/a) = a · 0 = 0 by 4.5(a) — contradicting (M3). Hence 1/a ≠ 0. Since (1/a) · a = 1, Theorem 4.3(b) — applied to the non-zero number 1/a — identifies a as 1/(1/a).
前半是小反證：倒數若是 0，等式 a · (1/a) = 1 的左端會被 4.5(a)（乘 0 歸 0）壓成 0，逼出 1 = 0——M3 明文禁止。後半又是「一條等式換個代入對象重讀」：(1/a) · a = 1 說明 a 與 1/a 相乘得 1，把 1/a 放到 4.3(b) 裡 a 的位置、判定 a 就是它的倒數——1/(1/a) = a，例行核對。
(b) Suppose a · b = 0 and a ≠ 0; we show b must vanish. Multiplying by 1/a, b = 1 · b = ((1/a) · a) · b = (1/a) · (a · b) = (1/a) · 0 = 0, where the last equality is 4.5(a). If instead b ≠ 0, the same argument with 1/b shows a = 0.
「或」的證法是任揀一個活口：若 a ≠ 0，同乘 1/a——M3 換裝、M4 相消、M2 挪括號、代入假設 a · b = 0、最後 4.5(a) 收尾——b 被一路押到 0。這條定理的日常名字叫「因式分解解方程」：乘積歸零，因子必有一個歸零，所以 (x − 2)(x − 3) = 0 可以放心拆成兩個小方程。數字對照：3 · b = 0 同乘 1/3，得 b = 0。
(c) By 4.5(b), −a = (−1) · a and −b = (−1) · b. Using (M1) and (M2) to regroup, and then 4.5(e), (−a) · (−b) = ((−1) · a) · ((−1) · b) = ((−1) · (−1)) · (a · b) = 1 · (a · b) = a · b.
(d) Suppose a ≠ 0; then −a ≠ 0 as well, for −a = 0 would give a = −(−a) = −0 = 0 by 4.5(d). Since part (c) yields (−a) · (−(1/a)) = a · (1/a) = 1, Theorem 4.3(b) identifies −(1/a) as 1/(−a).
Proof. Suppose, to the contrary, that (p/q)² = 2 for some integers p, q. Without loss of generality we may take p and q to have no common integral factor — otherwise cancel it first.
Clearing denominators turns the hypothesis into p² = 2q², so p² is even. Then p itself must be even: were p = 2k + 1 odd, its square p² = 4k² + 4k + 1 = 2(2k² + 2k) + 1 would be odd — impossible. Write p = 2k.
兩邊同乘 q²，假設變成 p² = 2q²：右端明擺著是偶數，所以 p² 是偶數。接著補上關鍵的一小步——為什麼 p² 偶就逼得 p 偶？看反面：假如 p 是奇數，寫成 2k + 1，平方展開 4k² + 4k + 1，前兩項都能被 2 整除、尾巴多出一個 1——奇數的平方必是奇數。數字對照：7² = 49 奇、6² = 36 偶——平方不會改變奇偶。所以 p 只能是偶數，記 p = 2k。
Substituting p = 2k gives 4k² = 2q², hence q² = 2k². By the same parity argument, q is even as well.