Collection of Lame Jokes
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And then when the driver started talking back disrespectfully to the cop, he got another ticket for being a brat.jtbell said:I saw a report on TV this morning that a cop in Wisconsin stopped the Oscar Mayer wienermobile for a traffic violation. First he had to ketchup with the vehicle, then he grilled the driver.
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A gorilla was walking through the jungle when he saw a lion taking a drink from a lake. The lion had its back turned and was oblivious to everything except the water, so the gorilla snuck up behind and kicked the lion into the lake. Needless to say, the lion was not happy, and the gorilla took off into the jungle with the lion in hot pursuit. Soon the gorilla came to an abandoned human camp. Quick as a flash, he grabbed khakis and a pith helmet, put them on, sat down in a chair, grabbed a newspaper and hid his hairy face behind it, pretending to read.
The lion charged into the encampment a moment later. "Excuse me sir," he said, "but have you seen a gorilla come through here?"
"What," answered the gorilla, without lowering his newspaper. "You mean the gorilla who pushed the lion into the lake?"
"My goodness!" exclaimed the lion. "It's in the papers already?"
The lion charged into the encampment a moment later. "Excuse me sir," he said, "but have you seen a gorilla come through here?"
"What," answered the gorilla, without lowering his newspaper. "You mean the gorilla who pushed the lion into the lake?"
"My goodness!" exclaimed the lion. "It's in the papers already?"
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No problem in ##\mathbb{Z}_3\, : \,\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=0## so nobody has been asked..WWGD said:When asked, first half of students liked abstract Mathematics, Second half liked applied Mathematics. Third half never studied Mathematics.
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If nobody has been asked then we don't need finite fields. We got all three answers 0/2 = 0 times.fresh_42 said:No problem in ##\mathbb{Z}_3\, : \,\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=0## so nobody has been asked..
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Maybe we need another. ##\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{1}{4}##. Now what about the other three quarters?mfb said:If nobody has been asked then we don't need finite fields. We got all three answers 0/2 = 0 times.
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But if ##\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=0## then that makes no cents.WWGD said:3 quarters is 75 cents, figure out your Math!
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If I had a dollar for every time I'd heard that...jbriggs444 said:But if ##\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=0## then that makes no cents.
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##\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{1}{3}##. Someone has to solve this inflation problem.jbriggs444 said:But if ##\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=0## then that makes no cents.
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Can one of you post a wiki link (or similar) to whichever branch of mathematics you're talking about? I'm a little out of my depth here.
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I am a bit out of depth also, but I believe that it is the finite field of three elements that is being discussed. So the elements could be referred to as {0, 1, 2}. "4" would be an alias for 1. ##\frac{1}{4}=\frac{1}{1}=1##. And 1+1+1 = 0.Ibix said:Can one of you post a wiki link (or similar) to whichever branch of mathematics you're talking about? I'm a little out of my depth here.
Edit: However, this theory does not square with a claim that ##\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{1}{3}## because that would involve a division by zero. By the same logic, a finite field of four elements is also ruled out. That leaves the possibility of the finite field of five elements. In that field ##\frac{1}{4}=4##, ##4+4+4=2## and ##\frac{1}{3}=2##. So that one sounds like a winner. Plus no need to make "4" an alias.
Edit2: ##\frac{1}{4}=4## since ##4 \times 4 = 16 = 1##. Similarly ##\frac{1}{3}=2## since ##3 \times 2 = 6 = 1##
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Presumably, ## \frac {1}{4} ## is the multiplicative inverse of ##4## in the field with ##3## elements, i.e., the number ##x## with ##4x=1## (this x exists for all non-zero elements, by definition/construction). But in ## \mathbb Z_3 = \{[0],[1],[2]\}## , the field with ##3 ## elements, ##1=4=7=10=...; 2=5=8=11=... ; 0=3=6=...## ( These are classes of elements rather than just elements; you collapse each of 1=4=7=... into a single class and your elements are now classes ), so you get ## 1/4+1/4+ 1/4 =1 +1+1 =3=0 ## ( in a field with p elements, any multiple of p is the same as zero) . Hope I explained it well. This is theory of Fields, Rings in Abstract Algebra. Please ask any followup, followed by a bad joke if possible ( i.e., model any of mine ;) ).Ibix said:Can one of you post a wiki link (or similar) to whichever branch of mathematics you're talking about? I'm a little out of my depth here.
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A physicist, a biologist and a mathematician are sitting in a street café watching people entering and leaving the house on the other side of the street.
First they see two people entering the house. Time passes. After a while they notice three people leaving the house.
The physicist says, "The measurement wasn't accurate."
The biologist says, "They must have reproduced."
The mathematician says, "If one more person enters the house then it will be empty."
First they see two people entering the house. Time passes. After a while they notice three people leaving the house.
The physicist says, "The measurement wasn't accurate."
The biologist says, "They must have reproduced."
The mathematician says, "If one more person enters the house then it will be empty."
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Yeah, we really need another subforum in the Mathematics section: "Nonsensicus Absolutus".jbriggs444 said:But if ##\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=0## then that makes no cents.
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You mean a nerd corner in the nerd lounge of nerdy general discussions?strangerep said:Yeah, we really need another subforum in the Mathematics section: "Nonsensicus Absolutus".
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Why a corner in a lounge? I've yet to encounter a pure mathematician who's not like that.fresh_42 said:You mean a nerd corner in the nerd lounge of nerdy general discussions?

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Their reference #1: Parachute use to prevent death and major trauma related to gravitational challenge: systematic review of randomised controlled trials (Smith & Pell) is worth a look:mfb said:Parachute use to prevent death and major trauma when jumping from aircraft: randomized controlled trial
The parachute and the healthy cohort effect
One of the major weaknesses of observational data is the possibility of bias, including selection bias and reporting bias, which can be obviated largely by using randomised controlled trials. The relevance to parachute use is that individuals jumping from aircraft without the help of a parachute are likely to have a high prevalence of pre-existing psychiatric morbidity. Individuals who use parachutes are likely to have less psychiatric morbidity and may also differ in key demographic factors, such as income and cigarette use. It follows, therefore, that the apparent protective effect of parachutes may be merely an example of the “healthy cohort” effect. Observational studies typically use multivariate analytical approaches, using maximum likelihood based modelling methods to try to adjust estimates of relative risk for these biases. Distasteful as these statistical adjustments are for the cognoscenti of evidence based medicine, no such analyses exist for assessing the presumed effects of the parachute.
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Hmm...What do I make out of someone calling me 'Pialidotous'?BillTre said:
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Soup, a cauli, fridge, elastic and eggs?WWGD said:Hmm...What do I make out of someone calling me 'Pialidotous'?
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If it is a she, call her pistantrophobe, if it is a he, just say "Gesundheit!"WWGD said:Hmm...What do I make out of someone calling me 'Pialidotous'?
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You mean im-pert-inent? 
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