问题1.证明由f(x)=(1 / x)定义的函数f： R * ⇢R *是一对一的，而R *是所有非零实数的集合。如果将域R *替换为N，且共域与R *相同，则结果是否正确?
解决方案：
One-one:
f(x)=f(y)
⇒1/x =1/y
⇒x=y
Therefore, f is one-one.
Onto:
It is clear that for y∈ R* there exists x=(1/y)∈ R* (exists as y ≠ 0) such that f(x)=1/(1/y)=y
Therefore, f is onto.
Thus, consider function g: N⇢R* defined by g(x)=1/x
We have, f(x1)=g(x2)⇒1/x1=1/x2⇒x1=x2
Therefore, g is one-one.
Further, it is clear that g is not onto as for 1.2∈ R* there does not exist any x in N such that g(x)=1/(1.2)
Hence, function g is one-one but not onto.
问题2.检查以下函数的内射性和外射性：
(i)f：由f(x)= x 2给出的N⇢N
解决方案：
It is seen that for x, y ∈ N, f(x)=f(y) ⇒x2=y2⇒x=y
Therefore, f is injective.
Now, 2 ∈ N but there does not exist any x in N such that f(x)=x2=2.
Therefore, f is not surjective.
(ii)f：由f(x)= x 2给出的Z given Z
解决方案：
It is seen that f(-1)=f(1), but -1 ≠1. Therefore, f is not injective.
-2 ∈ Z. But, there does not exist any x in Z such that f(x)= x2=-2.Therefore, f is not surjective
(ⅲ)F：R⇢ř用f(x)给出= X 2
解决方案：
It is seen that f(-1)=f(1), but -1 ≠1. Therefore, f is not injective.
-2 ∈ R. But, there does not exist any x in R such that f(x)= x2=-2.Therefore, f is not surjective.
(ⅳ)的F：N⇢N用f(x)给出=×3
解决方案：
It is seen that for x, y ∈ N, f(x)=f(y)⇒x3=y3⇒x=y. Therefore, f is injective.
2∈ N. But, there does not exist any element x in domain N such that f(x)=x3=2. Therefore, f is not surjective.
(ⅴ)的F位：Z⇢ž用f(x)给出=×3
解决方案：
It is seen that for x, y ∈Z, f(x)=f(y)⇒x3=y3⇒x=y. Therefore, f is injective.
2∈Z. But, there does not exist any element x in domain Z such that f(x)=x3=2. Therefore, f is not surjective.
问题3.证明f(x)= [x]给出的最大整数函数f： R⇢R既不为一，也不为on ，其中[x]表示小于或等于x的最大整数。
解决方案：
It is seen that f(1.2)=[1.2]=1, f(1.9)=[1.9]=1.
f(1.2)=f(1.9), but 1.2≠1.9. Therefore, f is not one-one.
Consider 0.7∈R. It is known that f(x)=[x] is always an integer. Thus, there does not exist any element x ∈R such that f(x)=0.7. Therefore, f is not onto.
Hence, the greatest integer function is neither one-one nor onto.
问题4.证明f(x)= | x |给出的模函数f： R⇢R既不是一，也不是在| x |上。如果x为正数或0且| x |为x如果x为负，则为-x。
解决方案：
It is seen that f(-1)=|-1|=1, f(1)=|1|=1.
f(-1)=f(1), but -1≠1. Therefore, f is not one-one.
Consider, -1∈R. It is known that f(x)=|x| is always non-negative. Thus, there does not exist any element x in domain R such that f(x)=|x|=-1. Therefore, f is not onto.
Hence, the modulus function is neither one-one nor onto.
问题5.证明符号函数f： R⇢R由f(x)= {(如果x> 0则为1，(如果x = 0则为0)，(-1如果x <0))给定既不是一个人也不是一个人。
解决方案：
It is seen that f(1)=f(2)=1, but 1≠2. Therefore, f is not one-one.
As f(x) takes only 3 values (1, 0, or -1) for the element -2 in co-domain R, there does not exist any x in domain R such that f(x)=-2. Therefore, f is not onto.
Hence, the signum function is neither one-one nor onto.
问题6。令A = {1、2、3}，B = {4、5、6、7}，令f = {(1,4)，(2,5)，(3,6)}为a函数从A到B。证明f为一。
解决方案：
It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f:A⇢B is defined as f={(1,4), (2,5), (3,6)}
Therefore, f(1)=4, f(2)=5, f(3)=6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.
问题7.在以下每种情况下，请说明函数是一对一的，对对的或双射的。证明你的答案。
(i)f：由f(x)= 3-4x定义的R⇢R
解决方案：
Let x1, x2 ∈R such that f(x1)=f(x2)
⇒3-4x1=3-4x2
⇒-4x1=-4x2
⇒x1=x2
Therefore, f is one-one.
For any real number (y) in R, there exists {(3-y)/4} in R such that f((3-y)/4)=3-4((3-y)/4)=y.
Therefore, f is onto
Hence, f is bijective.
(ii)f： R⇢R定义bf(x)= 1 + x 2
解决方案：
Let x1, x2 ∈ R such that f(x1)=f(x2)
⇒1+x12=1+x22
⇒x12=x22
⇒x1=±x2
Therefore, f(x1)=f(x2) does not imply that x1=x2
For instance, f(1)=f(-1)=2
Therefore, f is not one-one.
Consider, an element -2 in co-domain R.
It is seen that f(x)=1+x2 is positive for all x ∈ R.
Thus, there does not exist any x in domain R such that f(x)=-1.
Therefore, f is not onto.
Hence, f is neither one-one nor onto.
问题8：设A和B。证明f：A x B⇢Bx A使得(a，b)=(b，a)是双射函数。
解决方案：
Let (a1, b1), (a2, b2) ∈ A x b such that f(a1, b1)=f(a2, b2)
⇒(b1, a1)=(b2, a2)
⇒b1=b2 and a1=a2
⇒(a1, b1)=(a2, b2)
Therefore, f is one-one.
Let (b,a) ∈ B x A such that f(a, b)=(b,a).
Therefore, f is onto.
Hence, f is bijective.
问题9.设f：N⇢N乘F(N)= {定义(第(n + 1)/ 2，如果n是奇数)，(N / 2，当n为偶数)对于所有的n∈N.国家是否函数f是双射的。证明你的答案。
解决方案：
It can be observed that:
f(1)=(1+1)/2=1 and f(2)=2/2=1
So, f(1)=f(2), where, 1≠2
Therefore, f is not one-one.
Therefore, it is not bijective. (Since, it needs to be both one-one and onto to be bijective).
问题10：让A = R- {1}。考虑函数f：由f(x)=(x-2)/(x-3)定义的A⇢B。 f是一对一的吗?证明你的答案。
解决方案：
Let x, y ∈ A such that f(x)=f(y)
⇒ (x-2)/(x-3)=(y-2)/(y-3)
⇒(x-2)(y-3)=(y-2)(x-3)
⇒ xy-3x-2y+6=xy-3y-2x+6
⇒ -3x-2y=-3y-2x
⇒ 3x-2x=3y-2y
⇒ x=y
Therefore, f is one-one.
Let, y ∈ B= R-{1}. Then y≠1.
The function f is onto if there exists x ∈ A such that f(x)=y
Now,
f(x)=y
⇒ (x-2)/(x-3)=y
⇒ x-2=xy-3y
⇒ x(1-y)=-3y+2
⇒ x=(2-3y)/(1-y) ∈ A
Thus, for any y ∈ B, there exists (2-3y)/(1-y) ∈ A such that f((2-3y)/(1-y))={((2-3y)/(1-y))-2}/{((2-3y)/(1-y))-3}=(2-3y-2+2y)/(2-3y-3+3y)=(-y)/(-1)=y
Therefore, f is onto.
Hence, function f is one-one and onto.
问题11：令f： R⇢R定义为f(x)= x 4 。选择正确的答案：
(A)f是一对(B)f是多个(C)f是一对但不是(D)f既不是一对也不是
解决方案：
Let x, y ∈ R such that f(x)=f(y)
⇒ x4=y4
⇒ x=±y
Therefore, f(x1)=f(x2) does not imply that x1=x2
For instance, f(1)=f(-1)=1
Therefore, f(1)=f(-1)=1
Therefore, f is not one-one
Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x)=2
Therefore, f is not onto.
The correct answer is D.
问题12设f，R⇢R上定义为F(X)= 3次。选择正确的答案：
(A)f是一对(B)f是多个(C)f是一对但不是(D)f既不是一对也不是
解决方案：
Let x, y ∈ R such that f(x)=f(y)
⇒ 3x = 3y
⇒ x=y
Therefore, f is one-one.
Also, for any real number (y) in co-domain R, there exists y/3 in R such that f(y/3) = 3(y/3) = y
Therefore, f is onto.
Hence, the correct answer is A.