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Dave
2004-01-17 09:37:51 |
究竟是 True or False?
Determine the truth value of each statement, assuming that x, y, and z are real numbers.
(1) For all x and all y, there exists at least one z such that z > y implies that z > x + y
(2) For all x there exists at least one y such that for all z, z > y implies that z > x + y
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andycool
2004-01-17 13:14:33 |
Dave
1.False
2都係false,但如果改為
For all "y "there exists at least one "x" such that for all z, z > y implies that z > x + y咁就"arm".
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*圓
2004-01-17 13:27:30 |
同意. (無乜說服力地)
哎 ... 乜呢個forum 星期六冇人 ...
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2004-01-18 16:08:59 |
Re : Dave 2004-01-17 09:37:51
// True or False?
Determine the truth value of each statement, assuming that x, y, and z are real numbers.
(1) For all x and all y, there exists at least one z such that z > y implies that z > x + y
(2) For all x there exists at least one y such that for all z, z > y implies that z > x + y //
>>>>>>>>>>>>>>>>>>
As more than 24 hours have lapsed and there are no further discussions on Dave 2004-01-17 09:37:51 quoted above, it seems not inappropriate for me to give some comments on the issues concerned ---- just to entertain the “thinkological* / logical / mathematical guys” in this Sp(Spiritual) Webgarden.
[ 1 ]
Statement (1), expressible in mathematical logic as
ΠxΠyΣz(z > y → z > x + y),
is true. This can be proved by Reductio ad Absurdum(RAA):
__Proof__
<1> ~ΠxΠyΣz(z > y → z > x + y) [Assumption]
<2> ΣxΣyΠz~( z > y → z > x + y) [Q-transformation]
<3> ΣxΣyΠz( z > y & ~(z > x + y)) [P-transformation]
<4> ΣyΠz( z > y & ~(z > a + y)) [EE on x]
<5> Πz( z > b & ~(z > a + b)) [EE on y]
<6> b > b & ~(b > a + b) [UE on z]
<7> ΠxΠyΣz(z > y → z > x + y) [RAA, with <1> discharged]
Q.E.D.^_^
[ 2 ]
Statement ( 2 ), expressible in mathematical logic as
ΠxΣyΠz(z > y → z > x + y),
is false. This can be established by proving the negation of ( 2 ).
__Proof__
Let R be the set of real numbers and λεR be positive. It is obvious that for any yεR there is a zεR such that z>y but NOT z>λ+y, since {r : y < r < λ+y} =/= Φ. Hence in R as the domain under consideration, we have
<1> ΠyΣz(z > y & ~(z > λ+y)) [Proved]
<2> ΣxΠyΣz(z > y & ~(z > x+y)) [EI on λ]
<3> ΣxΠyΣz~(z > y → z > x+y) [P-transformation]
<4> ~ΠxΣyΠz(z > y → z > x + y) [Q-transformation]
Q.E.D.^_^
------------------------
*“Thinkology” is here coined for “思方學”.
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andycool
2004-01-18 17:55:18 |
依,衰左tim.閃.
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採菊
2004-01-18 18:03:26 |
打渾
雖然我己經好俾心機睇，但都係睇唔明到底點解在一大堆ｘｙｚ和很多奇怪的符號之後，會有如此一個笑臉^_^？
真是太神秘了！
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Alicia
2004-01-18 18:38:17 |
andycool
^^ 你睇李生幾俾面，prove晒出黎。第二個冇咁好招呼。佢出唔到新書，都係因為你。
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長老
2004-01-18 18:47:18 |
andycool
oh.....你因住做咗罪人
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andycool
2004-01-18 19:14:31 |
依.個問題唔係我問架喎.
XDDDDDD
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Alicia
2004-01-18 19:15:46 |
但係你答錯。XDDDDDDDD
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andycool
2004-01-18 19:16:34 |
依....好衰架.
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Joejones
2004-01-18 20:52:37 |
E... 李生好衰架...
本來我己收心養性不留言, 都引我在SP. webgarden 再留言...
唉...
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圓
2004-01-18 20:56:19 |
er.... 關唔關我事 ... 係咪阻左出書時間 ... 同李生同大家講聲唔好意思先.
乜咁鬼複雜, 我都係玩下執屋搵到果本陳年智力體操算了.
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圓
2004-01-18 20:59:41 |
JoeJones ~
你姣婆守唔到寡就別賴噢
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Joejones
2004-01-18 21:15:53 |
>你姣婆守唔到寡就別賴噢
哈!
從這句話的語氣, 我猜它的構思者是S.C. 但以"圓"之名出文的概然性很高
:-目
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圓
2004-01-18 21:53:58 |
True or False?
/哈!
從這句話的語氣, 我猜它的構思者是S.C. 但以"圓"之名出文的概然性很高
:-目 /
False.
^++++++^
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S.C.
2004-01-18 22:01:50 |
Joejones
不是。大概是人同此心﹐心同此理。
來還是不來﹖愛還是不愛﹖再還是不再﹖待還是不待﹖來如春夢不多時﹐去似朝雲無覓處。去既然不要理由﹐來當然也不要原因。True or False?今爾往矣﹐楊柳不依依﹐False or true?及時宜行樂﹐歲月不待人。Very true.
歡迎歸來。Very cool.
(S.C.以上這種貼文才是我會說的。)
2004-01-23 23:34:38
.
(羅素+(李白+峨默)/2)/2
^_^
.
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Joejones
2004-01-19 00:34:01 |
收到!
thx S.C.
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黐線B
2004-01-19 01:12:13 |
嘩.......
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Dave
2004-01-19 17:23:11 |
感恩......
非常感謝李先生的說明,使我明白
亦要謝謝andycool 和*圓的回應
我從 ~@@ 到 ^_^ 都花了一段時間 _0_
加多三題練習題: True or False?
(3) ΠxΣyΠz(x + y = z)
(4) ΠxΠyΣz(x + y = z)
(5) ΣxΠyΣz(xz = xy)
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