多伦多大学 MAT 137 课业解析
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2019-09-23 13:03:41
MA 多伦多大学 MAT 137 课业解析

题意：

完成三道计算题

解析：

第三题：. For which positive integers n ≥ 1 does 2^n > n^2 hold? Prove your claim by induction.

证明：

n>=5

（1）当 n=5 时，2^5=32 > 5^2=25，不等式成立

（2）假设 n=k (k>5) 时，2^k > k^2;

则 n = k+1 时，2^(k+1)=22^k > 2(k^2)=(k-1)^2-2+(k+1)^2 当 k >5 时，(k-1)^2-2>0 所以 2^(k+1)>(k+1)^2 即 n>5 时，假设成立由数学归纳法可知，V n>=5，2^n>n^2。

涉及知识点：

数学归纳法，集合

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MAT 137

Problem Set #1

Due on Thursday September 26, 2019 by 11:59 pm

Submit via Crowdmark

Instructions

• You will need to submit your solutions electronically. For instructions, see the

MAT137 Crowdmark help page. Make sure you understand how to submit and

that you try the system ahead of time. If you leave it for the last minute and you

run into technical problems, you will be late. There are no extensions for any reason.

• You will need to submit your answer to each question separately.

• You may submit jointly written answers in groups of up to two people. Your partner

can be anyone in MAT137 from any lecture section. You can also submit jointly

written answers with a different person for each problem set.

• If you do not jointly write your solutions with someone else then you must submit

your answers individually.

• This problem set is about the introduction to logic, notation, quantifiers, conditionals, definitions, and proofs (Playlist 1).

Problems

1. Read Notes on Collaboration on the course website. Copy out the following sentence

and sign below it, to certify that you have read the Notes on Collaboration”.

I have read and understood the notes on collaboration for this course, as

explained in the course website.”

If submitting as a group of two, both people must sign and submit.

1. Negate the following statement without using any negative words (no”, not”,

none”, zero”, etc.):

All students at a university in Canada are enrolled in an odd-numbered course

that is taught by a professor whose last name starts with a letter alphabetically

before Q and who lectures only on weekdays.”

1. In this problem we will only consider (real-value

多伦多大学 MAT 137 课业解析

题意：

完成三道计算题

解析：

第三题：. For which positive integers n ≥ 1 does 2^n > n^2 hold? Prove your claim by induction.

证明：

n>=5

（1）当 n=5 时，2^5=32 > 5^2=25，不等式成立

（2）假设 n=k (k>5) 时，2^k > k^2;

则 n = k+1 时，2^(k+1)=22^k > 2(k^2)=(k-1)^2-2+(k+1)^2 当 k >5 时，(k-1)^2-2>0 所以 2^(k+1)>(k+1)^2 即 n>5 时，假设成立 由数学归纳法可知，V n>=5，2^n>n^2。

涉及知识点：

数学归纳法，集合

更多可加微信号：alexa_au