Problem 6 let x be any positive real number if y1 y2 0 are...

Problem 6. Let $x$ be any positive real number. If $y_1$, $y_2 > 0$ are any two real solutions of the equation $y^2 = x$, show that $y_1 = y_2$. Problem 7 Prove that if $(a_n)$ and $(b_n)$ are Cauchy sequences in $\mathbb{R}$, then $(a_n - b_n)$ is a

Transcript

00:01 Okay, for the first of the question, i mean, in the first question, we are given some real number x, y.

00:08 We assume x is strictly less than y.

00:11 I mean, on the real line r, we want to show there exists some m and n, which are both natural numbers, such that x plus 1 over m is strictly less than y minus 1 over n.

00:32 Okay, how do we do that? to prove this strict inequality, we want to use the property for this sequence.

00:43 It's easy for us to see this sequence converges to zero, right? it's a very simple fact.

00:51 That means for any epsilon greater than zero, there exists some very large m, such that for any little n greater or equal to n, we have one over n is strictly less than epsilon.

01:09 Of course, this is a positive sequence, so it must be greater than zero for sure.

01:16 Okay, here, let's just, as x is assumed to be strictly less than y, we know y minus x is some number strictly positive.

01:27 So, it does 1 minus x over 2.

01:31 So here, let's just use our epsilon.

01:33 Take our epsilon to be 1, y minus x over 2.

01:38 And then there is some capital m such that for any n greater than 2n, we have this number is less, strictly less than y minus x over 2.

01:49 Okay, so take m and n to be strictly greater than m.

01:56 We know 1 over m and 1 over m will be both bounded by 0 and y minus x over 2.

02:06 Now let's consider what is x plus 1 over m.

02:16 Here is x, here is y, this is a middle point, as m is assumed to be strictly less than this middle point, so x plus 1 over m will be here.

02:28 And then 1 over n is also assumed to be strictly less than this guy, so 1.

02:35 This is our x plus 1 over n.

02:39 Here we have y minus 1 over n.

02:42 So you can see by our construction this number is in fact strictly less than the later one.

02:52 Okay, then for the second question, we're given the convergence sequence xn.

02:59 Let's assume xn converges to m in r.

03:04 And we have another sequence yn such that for any epsilon greater than zero, there exists a capital k, which is a natural number.

03:18 Maybe even r sequence y satisfies.

03:30 For any epsilon greater than zero, there exists a capital k such that for any n greater or equal to this capital k, okay, we have xn minus yn less than epsilon.

03:46 Okay, here we want to show our code is to show yn is convergent.

03:58 I mean, we want to show yn also converges to this n.

04:05 How do we, let's use the epsilon delta language to prove it, epsilon k language to prove it.

04:11 For n given epsilon greater than zero by this fact...

Question

Problem 6. Let $x$ be any positive real number. If $y_1$, $y_2 > 0$ are any two real solutions of the equation $y^2 = x$, show that $y_1 = y_2$. Problem 7 Prove that if $(a_n)$ and $(b_n)$ are Cauchy sequences in $\mathbb{R}$, then $(a_n - b_n)$ is a

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