1. Let the personal domain be an integer set, and the true value of the following formula is (a).
(1)? x？ y(x+y=0)(B)？ y？ x(x+y=0)
(3)? x？ y(x+y=0)(D)┓？ x？ y(x+y=0)
2. let a = {a, b, c} and r = {
(a) reflexivity (b) reflexivity (c) antisymmetry (d) equivalence
3. Let the function f:R→R, f (a) = 2a+1; G:R→R, g(a)=a2, then (c) has an inverse function.
(A) g？ f (B)f？ Britain, France, Germany and Britain
4. In the following algebraic system (a, *), there are unitary elements (a unitary 0, c unitary 1, d unitary 1).
(a), A=R, r is a real number, a*b=a+b-ab.
(b), A=R, r is a real number, and A * B = B
(c), A=N, n is a natural number, and a*b=ab.
(d), A=N, n is a natural number, and a*b=gcd(a, b), where gcd () is the greatest common divisor of a and b.
5. The following algebraic system can be divided into (a, d)
(a) One-dimensional real coefficient polynomial set P(x) (including 0 polynomials), and the operation * is the addition of polynomials.
(b) One-dimensional polynomial set with real coefficients P(x) (including 0 polynomials), and the operation * is the multiplication of polynomials.
(c) Positive real number set R+, about the division operation of numbers.
(d) Let Q+ be a positive rational number and the operation * be a common subtraction.
8. Try to determine whether the following functions are surjective, injective and bijective.
① let A={a, b, c}, B={ 1, 2}, function f: a → b, f = {
② Let n be a set of natural numbers, and the function s:N→N, s(n)=n+ 1, injective.
③ Let z be an integer set, e an even integer set, and the function g:Z→E, bijection.
9. Prove the problem and give it uniqueness
( 1)(a^- 1)^- 1=a，
A- 1 ○ A = E, a-1= e, so the inverse of (a-1) is a, (A- 1)- 1 = A,
(2)a○b has an inverse element, (A ○ b)-1= b-1○ a-1.
(b^- 1○a^- 1)○( a○b)= b^- 1○a^- 1○a○b = b^- 1○b=e
(a○b)○(b^- 1○a^- 1)=a○b○b^- 1○a^- 1 =a○a^- 1 = e
Therefore, (a ○ b)-1= b-1○ a-1.
10. An algebraic system (z,) in which to operate? For what? a，b∈Z，a？ B=a+b-2, which proves that algebraic system (z,) is a group.
①(a？ b)？ c=(a+b-2)？ c =a+b-2+c-2=a+b+c-4，a？ (b？ c)= a？ (b+c-2) =a+b+c-2-2=a+b+c-4，
That is (a? b)？ c=a？ (b？ C), so the combination is established.
②? a∈Z，a？ 2=a+2-2=a，2？ A=2+a-2=a, so what's 2?
③? A∈Z, a+(4-a)=a+4-a-2=2, so the inverse of A is 4-a,
It can be seen from ① ② ③ that algebraic system (z,) is a group.