Math

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1. Say whether the following is true or false and support your answer by a proof Answer: The statement is false. There do not exist natural numbers and such that . Proof:Assume for contradiction that (where ) satisfying . Rewrite the equation as: Let and . Then . Since , we derive: Suppose and for some integer . Substituting back: . . For , must be even and a multiple of , so is a multiple of .Testing : (invalid). For : (invalid). No valid exists. Conclusion: ...

同学们大家好，欢迎来到今天的数学课堂。我是来自静安区教育学院附属学校的厉老师。前几节课，我们已经学习了因式分解的意义，以及提取公因式法、公式法和十字相乘法这三种因式分解的方法。我们知道公式法和十字相乘法主要适用于二项式和三项式。而今天，我们将重点研究如何对四项式进行因式分解。 请同学们首先思考这个问题：如何将 进行因式分解？我们一起观察分析一下这个四项式的特征。可以发现，这个四项式有四项，且没有公因式。显然，之前学过的提取公因式法、公式法和十字相乘法都不适用，因而我们遇到了一些困难。 在解决这个问题之前，我们先来看一个类似的表达式： 这个表达式存在公因式 ，我们可以使用提取公因式法来进行因式分解。同学们一定能够轻松解决这个问题。 那么请大家再观察一下这两个表达式之间的关系。你会发现，如果将第二个表达式乘展开，就得到了第一个四项式。它们本质上是相同的表达式。现在，同学们是否能够因式分解 了呢？我们可以发现，这个整式的前两项都有字母 ，而后两项都有字母 。我们可以把前两项分为一组，后两项分为一组，利用加法结合律，从第一组中提取公因式 ，得到 ；从第二组中提取公因式 ，得到 。这样一来 ...

Q1. Let Show that A has a lower bound in but no greatest lower bound in . Give all details of the proof along the lines of the proof given in the lecture that the rationals are not complete.Proof: Let . We aim to show that has a lower bound in but no greatest lower bound in . Step 1: Existence of a Lower Bound in Since all elements of are positive and satisfy , we observe that is a lower bound. For all , , so . Therefore, is a lower bound of in . Step 2: Assume the Existence of a Great ...

1.Prove that the intersection of two intervals is again an interval. Is the same true for unions?Proof: Firstly, we need to prove that . It’s vacuously true. Assume, on the contrary, .In other words, there must exist two elements x,y such that there exist at least one element z(where x<z<y) that is not in the set.Here rasies a contradction, since there are no elements in the . Hence, our assumption is invalid, so . Secondly, we come to prove that the intersection of two intervals is again ...

1. Prove or disprove the claim that there are integers m, n such that is a perfect square.Proof: Let n = 0. Then . And is a perfect square. Q.E.D 2. Prove or disprove the claim that for any positive integer m there is a positive integer n such that is a perfect square.Proof: For any positive integer m, let n=m+2, then . And is a perfect square. Q.E.D 3. Prove that there is a quadratic , with positive integers coefficients b, c, such that f(n) is composite (i.e, not prime) for all po ...

Optional ProblemsIt is a standard result about primes (known as Euclid’s Lemma) that if p is prime, then whenever p divides a product ab, p divides at least one of a, b. Prove the converse, that any natural number having this property (for any pair a, b) must be prime.

Theorem: The rational line is not complete.Proof: By contradiction. Assume the rational line is complete. Then there exists a counterexample. Let A={x| and <2}. a<2. So 2 is a upper bound of A. Now we need to show that A doesn’t have a least upper bound. Supplementary Materialshttps://home.iitk.ac.in/~psraj/mth101/lecture_notes/lecture1.pdf

1. The Euclid’s Lemma: For any integer a, b, if there exists a prime p such p divides ab, then p divides at least one of a or b.Proof: By stong induction. We need to prove that Suppose that p|ab and gcd(p,a) = 1 which means p can’t divide a. Now we have to show that p|b. Since p|ab, there is an integer q such that pq=ab.Without lose of generality, we can suppose p, q, a, and b are positive, since the divisibility relation is independent from the signs of the involved integers. For prov ...