# 查看完整版本 : A difficult MI problem [M2]

verylovediscuss 2022-10-27 19:00

Do someone know how to answer below problem?

[attach]13701633[/attach]
[attach]13701634[/attach]

[[i] 本帖最後由 verylovediscuss 於 2022-11-10 22:27 編輯 [/i]]

i.e. = that is

"Suppose the statement is true for some positive integer k, i.e. 1=2=...=k."
"Suppose the statement is true for some positive integer k" ... (*) 尚未有問題 (考慮 k=1)，

hkpal 2022-10-31 09:03

[url=https://en.wikipedia.org/wiki/All_horses_are_the_same_color]https://en.wikipedia.org/wiki/All_horses_are_the_same_color[/url]

XMing 2022-10-31 20:01

[quote]原帖由 [i]verylovediscuss[/i] 於 2022-10-27 19:00 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=553247038&ptid=30833312][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]
Do someone know how to answer below problem?

[img]https://img.discuss.com.hk/d/attachments/day_221027/20221027_2540f90bdb88d1ce7f3bRZOc6voMwfB7.jpg[/img]
[img]https://img.discuss.com.hk/d/attachments/day_221027/20221027_6405c990bee180c13ce0w44pJLEShZnt.jpg[/img] [/quote]
Mathematical induction:  Let P(n) be a statement about the natural number n.
(i) When n=1, P(1) is true
(ii) If P(n) is true then P(n+1) is true.
Then P(n) is true for all natural numbers.
Now the statement P(n) is "All first n positive integers are equal."
Now (i) is OK because P(1) is true.
But for (ii) when n=1, P(1) is true but the argument to prove 1=2 is not correct. because we do not have 0=1-1=1. Actually, it is not true or 0=1 have not proved to be true.
So, for (ii) it is not true. We cannot use this argument and Mathematical induction to prove that "All first n positive integers are equal".

[[i] 本帖最後由 XMing 於 2022-11-4 12:56 編輯 [/i]]

verylovediscuss 2022-11-1 23:26

[quote]原帖由 [i]XMing[/i] 於 2022-10-31 20:01 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=553356515&ptid=30833312][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

Mathematical induction:  Let P(n) be a statement about the natural number n.
(i) When n=1, P(1) is true
(ii) If P(n) is true then P(n+1) is true.
Then P(n) is true for all natural number.
Now th ... [/quote]
Yes, I also think that the mistake comes from the induction step. For k = 1, "k-1 = k" has not yet proved to be true.
For k=1, only 1=1 but not 0=1 was proved.

[[i] 本帖最後由 verylovediscuss 於 2022-11-2 16:35 編輯 [/i]]

verylovediscuss 2022-11-1 23:29

[quote]原帖由 [i]hkpal[/i] 於 2022-10-31 09:03 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=553341258&ptid=30833312][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]
https://en.wikipedia.org/wiki/All_horses_are_the_same_color [/quote]

verylovediscuss 2022-11-1 23:35

[quote]原帖由 [i]重重迷惑的口罩[/i] 於 2022-10-31 01:36 發表 [url=https://www.discuss.com.hk/redirect.php?goto=findpost&pid=553337600&ptid=30833312][img]https://www.discuss.com.hk/images/common/back.gif[/img][/url]

i.e. = that is

"Suppose the statement is true for some positive integer k, i.e. 1=2=...=k."
"Suppose the statement is true for some positive integer k" ... (*) 尚未有問題 (考慮 k= ... [/quote]
Thanks for your reply. I finally figure out for the case of k = 1, "1=2=...=k" should be interpreted as "1=1", but being treated as "0=1" in the given proof.

[[i] 本帖最後由 verylovediscuss 於 2022-11-2 16:36 編輯 [/i]]