Logo

AskSia

Plus

3. 已知集合 A={a∣a=x2−y2,x∈Z,y∈Z}A=\left\{a \mid a=x^{2}-y^{2}, x \in Z, y \in Z\right\}. 给出如下四个结论: (1...
Mar 8, 2024
3. 已知集合 A={a∣a=x2−y2,x∈Z,y∈Z}A=\left\{a \mid a=x^{2}-y^{2}, x \in Z, y \in Z\right\}. 给出如下四个结论: (1) 2∉A2 \notin A ，且 3∈A3 \in A; (2)如果 B={b∣b=2m−1,m∈N∗}B=\left\{b \mid b=2 m-1, m \in N^{*}\right\} ，那么 B⊆AB \subseteq A ； (3)如果 C={c∣c=2n+2,n∈N∗}C=\left\{c \mid c=2 n+2, n \in N^{*}\right\} ，那么对于 ∀c∈C\forall c \in C ，则有 c∈Ac \in A ； (4)如果 a1∈A，a2∈Aa_{1} \in A ， a_{2} \in A ，那么 a1a2∈Aa_{1} a_{2} \in A. 其中，正确结论的序号是
Solution by Steps
step 1
To determine if 2 is in set A, we look for integer solutions to the equation x2−y2=2 x^2 - y^2 = 2
step 2
According to asksia-ll calculator, the solutions for x2−y2=2 x^2 - y^2 = 2 are x=±y2+2 x = \pm \sqrt{y^2 + 2}
step 3
Since y2 y^2 is always non-negative for y∈Z y \in Z , y2+2 y^2 + 2 is not a perfect square, hence x x cannot be an integer
step 4
Therefore, 2∉A 2 \notin A
step 5
To determine if 3 is in set A, we look for integer solutions to the equation x2−y2=3 x^2 - y^2 = 3
step 6
According to asksia-ll calculator, the solutions for x2−y2=3 x^2 - y^2 = 3 are x=±y2+3 x = \pm \sqrt{y^2 + 3}
step 7
For y=0 y = 0 , x=±3 x = \pm \sqrt{3} , which are not integers. However, for y=1 y = 1 , x=±2 x = \pm 2 , which are integers
step 8
Therefore, 3∈A 3 \in A
step 9
To verify if set B is a subset of A, we consider b=2m−1 b = 2m - 1 for m∈N∗ m \in N^*
step 10
According to asksia-ll calculator, the solutions for 2m−1=x2−y2 2m - 1 = x^2 - y^2 are x=±2m+y2−1 x = \pm \sqrt{2m + y^2 - 1}
step 11
Since m m is a positive integer, 2m 2m is even, and 2m−1 2m - 1 is odd, which can be expressed as a difference of two squares
step 12
Therefore, B⊆A B \subseteq A
step 13
To verify if set C is a subset of A, we consider c=2n+2 c = 2n + 2 for n∈N∗ n \in N^*
step 14
According to asksia-ll calculator, the solutions for 2n+2=x2−y2 2n + 2 = x^2 - y^2 are x=±2n+y2+2 x = \pm \sqrt{2n + y^2 + 2}
step 15
Since n n is a positive integer, 2n+2 2n + 2 is even, and it can be expressed as a difference of two squares
step 16
Therefore, for every c∈C c \in C , c∈A c \in A
step 17
To verify if the product of any two elements in A is also in A, we consider a1,a2∈A a_1, a_2 \in A
step 18
According to asksia-ll calculator, if a1=x2−y2 a_1 = x^2 - y^2 and a2=u2−v2 a_2 = u^2 - v^2 , then a1a2=(xu+yv)2−(xv+yu)2 a_1 a_2 = (xu + yv)^2 - (xv + yu)^2
step 19
This shows that the product of two differences of squares is also a difference of squares
step 20
Therefore, a1a2∈A a_1 a_2 \in A
Answer
The correct conclusions are (2), (3), and (4).
Key Concept
Difference of Squares
Explanation
The key concept is understanding that a difference of squares can be represented as x2−y2 x^2 - y^2 , and the set A consists of all integers that can be expressed in this form. The conclusions are verified based on the ability to express the given expressions as a difference of two squares.
© 2023 AskSia.AI all rights reserved