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#1
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Discrete math anyone
Anyone ever take a discrete math course? I'm stumped beyond infinity.
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#2
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is it linear? groups?
effin' hate that 5hit! hey... wait a second... you can't have infinity in discrete math... |
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#3
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Nope. Soon as the numbers go above ten my brain shuts down.
- Peter.
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2021 Chevrolet Spark Formerly... 2000 GMC Sonoma 1981 240D 4spd stick. 347000 miles. Deceased Feb 14 2021 ![]() 2002 Kia Rio. Worst crap on four wheels 1981 240D 4spd stick. 389000 miles. 1984 123 200 1979 116 280S 1972 Cadillac Sedan DeVille 1971 108 280S |
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#4
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Peter, take your shoes off & you will be good for 20 !!!
Don, infinity is just an 8 that fell over !!
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Grumpy Old Diesel Owners Club group I no longer question authority, I annoy authority. More effect, less effort.... ![]() 1967 230-6 auto parts car. rust bucket. 1980 300D now parts car 800k miles 1984 300D 500k miles 1987 250td 160k miles English import ![]() 2001 jeep turbo diesel 130k miles ![]() 1998 jeep tdi ~ followed me home. Needs a turbo. 1968 Ford F750 truck. 6-354 diesel conversion. Other toys ~J.D.,Cat & GM ~ mainly earth moving |
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#5
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Yes, I took one. It was alright, it was mainly logic, and proofs of various items. Only math course I've ever gotten anything other than an A in..
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Cruise Control not working? Send me PM or email ([email protected]). I might be able to help out. Check here for compatibility, diagnostics, and availability! (4/11/2020: Hi Everyone! I am still taking orders and replying to emails/PMs/etc, I appreciate your patience in these crazy times. Stay safe and healthy!) 82 300SD 145k 89 420SEL 210k 89 560SEL 118k 90 300SE 262k RIP 5/25/2010 90 560SEL 154k 91 300D 2.5 Turbo. 241k 93 190E 3.0 235k 93 300E 195k |
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#6
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No thanks, I prefer my math to be blatant.
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1980 300TD-China Blue/Blue MBTex-2nd Owner, 107K (Alt Blau) OBK #15 '06 Chevy Tahoe Z71 (for the wife & 4 kids, current mule) '03 Honda Odyssey (son #1's ride, reluctantly) '99 GMC Suburban (255K+ miles, semi-retired mule) 21' SeaRay Seville (summer escape pod) |
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#7
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Quote:
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"The law, in its majestic equality, forbids the rich as well as the poor to sleep under bridges, to beg in the streets, and to steal bread." |
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#8
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its killing me. I get the idea of whats going on but dont understand how to do the proofs
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#9
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Which book are you using for logic?
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You don't need a weatherman to know which way the wind blows - Robert A. Zimmerman |
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#10
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Discrete Mathmatics and Its Applications sixth edition Kenneth H. Rosen
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#11
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Hmm... never used the system. What kind of rules are you allowed so far, are you using quantifier logic, and are you allowed to use reductio ad absurdium?
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You don't need a weatherman to know which way the wind blows - Robert A. Zimmerman |
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#12
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quantifiers, laws of disjuntion/conjunction
here is an in class example Prove (p → r) ∨ (q → r) ≡ (p ∧ q) → r using logical equivalence. Please the name of the laws you use. I've got no idea how to start. |
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#13
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Don, I actually recognize that from over 35 years ago!! There are a series of laws, if you start with the RHS you can use the laws to change it a couple of times, you will then be able to make it look like the LHS & so proved
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Grumpy Old Diesel Owners Club group I no longer question authority, I annoy authority. More effect, less effort.... ![]() 1967 230-6 auto parts car. rust bucket. 1980 300D now parts car 800k miles 1984 300D 500k miles 1987 250td 160k miles English import ![]() 2001 jeep turbo diesel 130k miles ![]() 1998 jeep tdi ~ followed me home. Needs a turbo. 1968 Ford F750 truck. 6-354 diesel conversion. Other toys ~J.D.,Cat & GM ~ mainly earth moving |
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#14
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yeah.. I dont follow how to actually do that. it doesnt click
so I ignore the left hand side all together and make the right look like the left.. somehow? |
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#15
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Quote:
Since you are proving equivalence you need to prove on direction then prove the other. First prove [(p → r) ∨ (q → r)] -> [(p ∧ q) → r ] Assume one side of the conditional 1) (p → r) ∨ (q → r) (assumption) Now you need to prove the other side which is a conditional also. So you assume the left side of the conditional as well. Indent one time. I can't seem to do it on the forum. 2) (p ∧ q) (assumption) Now you need to prove r. 3) -r (assumption) . . . Once you've prove the above conditional you need to prove the conditional the the other way. (since p ≡ q is p -> q and q -> p right?)
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You don't need a weatherman to know which way the wind blows - Robert A. Zimmerman |
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