T.7 · Set images, equivalence relations and logical negation
GRE Subject Test · GRE Mathematics Topic 20 3:05 English narration · English + 中文 subtitles burned in
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This unit proves image identities and distinguishes inclusion from equality, checks reflexivity, symmetry and transitivity with explicit cases, and negates implications and quantified statements without changing their scope.
本单元证明像的恒等式并区分包含与相等，用具体情形检验自反性、对称性和传递性，并在不改变辖域的情况下否定蕴涵式和量化命题。
An image of a union is the union of images.
并集的像等于像的并集。
For an intersection only one inclusion is automatic, because the same output may come from different inputs: with x squared, minus 1 and 1 give an empty left side but 1 on the right.
对交集只有一个包含关系自动成立，因为同一输出可能来自不同输入：对 x²，取 {−1} 和 {1}，左边为空，右边是 {1}。
Preimages preserve unions, intersections and complements.
原像保持并、交和补。
If f is injective, image intersections become equal.
若 f 是单射，像的交集就相等。
A relation is an equivalence when it is reflexive, symmetric and transitive; its classes partition the set.
自反、对称且传递的关系是等价关系；它的等价类划分整个集合。
Same remainder modulo four gives four classes.
模 4 同余给出四个类。
The relation x minus y at most 1 is reflexive and symmetric but not transitive: 0 relates to 1 and 1 to 2, but 0 does not relate to 2.
关系 |x − y| ≤ 1 自反且对称，但不传递：0 与 1 相关，1 与 2 相关，但 0 与 2 不相关。
An implication is false exactly when its hypothesis is true and its conclusion false.
蕴涵式当且仅当前提真而结论假时为假。
So the negation of P implies Q and R is P and not Q or not R.
所以 P ⇒ (Q ∧ R) 的否定是 P ∧ (¬Q ∨ ¬R)。
Every becomes there exists, and there exists becomes every, with the quantifiers kept in their order.
“所有”变成“存在”，“存在”变成“所有”，量词顺序不变。
A counterexample disproves a universal claim; examples never prove one.
反例可以否定全称命题；举例永远证明不了全称命题。
Let f map both a and b to label L.
设 f 把 a 和 b 都映到标签 L。
With A equals a and B equals b, the sets do not meet, so f of A intersect B is empty, but f of A intersect f of B is L.
取 A = {a}、B = {b}，两集合不相交，所以 f(A ∩ B) 为空，但 f(A) ∩ f(B) = {L}。
And a false claim that every stored file has a hash means at least one file has none, not that all lack one.
“每个存储文件都有哈希值”为假，意味着至少有一个文件没有，而不是所有文件都没有。
f inverse of C can mean a preimage set without an inverse function.
f⁻¹(C) 可以表示原像集合，而不需要反函数存在。
A reflexive, symmetric relation may still fail transitivity.
自反且对称的关系仍可能不传递。
Negate the whole statement before simplifying its parts.
先否定整个命题，再化简各部分。
If y is f of x with x in both sets, then y lies in both images.
若 y = f(x) 且 x 在两个集合中，则 y 在两个像中。
Mapping two distinct points to the same c makes the inclusion strict.
把两个不同的点映到同一个 c，就使包含关系严格。
If f is injective, a shared output forces a shared input, giving equality; and the singleton example shows equality for all pairs forces injectivity.
若 f 是单射，相同的输出迫使输入相同，从而相等；单点集的例子说明，对所有集合对都相等就迫使 f 为单射。
Reflexivity uses a minus a equals 0; symmetry, the negative of a multiple of four; transitivity, a sum of two such multiples.
自反性用 a − a = 0；对称性用 4 的倍数的相反数；传递性用两个这样的倍数之和。
The classes are the four residues modulo four.
等价类是模 4 的四个剩余类。
The negation says: some integer n has every larger m failing P.
否定是：存在某个整数 n，使得每个比它大的 m 都不满足 P。
Check yourself.
自我检测。
Images preserve unions always, and intersections when f is injective.
像总是保持并集，单射时也保持交集。
x minus y at most 1 is not transitive.
|x − y| ≤ 1 不传递。
Not every becomes some does not.
“并非所有”变成“存在某个不”。
And congruence modulo 4 has four classes.
模 4 同余有四个类。