I could talk about nipples in the context of algebraic topology, would that work for you?It would be nice to have a few pints and scan thru the Weather thread… but Tom screwed us!!
Ring, or mental circle jerk?More specifically, it can be a direct sum of copies of Z_2. In particular, we have to have x+x=0 always. Here’s why. If x^2=x then 1+x=(1+x)^2=1+2x+x^2=1+2x+x and you cancel out 1+x from both sides, get 0=2x. You also have x(1+x)=x+x^2=x+x=0, so you have a whole bunch of zero divisors and only one unit, 1. In the first sentence I said it can be a direct sum of copies of Z_2, and I’m trying to figure out if this is always the case. If the ring is finite I am positive that it has to be the case (I need a proof) but what if it’s infinite? I’m not trying to be a show off here, it’s just that I find this stuff gloriously fascinating.
Ring. You have both multiplication and addition. When you have just addition or just multiplication, it’s called a group. When you have a mental circle jerk, it’s a PSU BOT meeting.Ring, or mental circle jerk?
I’m balancing workouts with rest and am also hitting the carbs hard at the training table. Otherwise, I’m watching a lot of nd game film and feel confident with what I want to do on offense and defense. To be clear, I will go both ways on Thursday night.
You?
Pics please!I could talk about nipples in the context of algebraic topology, would that work for you?
Sometimes the jokes just write themselvesWorking on Boolean rings. These are commutative rings where x^2=x always. And now I’m going to shovel snow, what a pain in the butt, got six inches out there, just now stopped snowing.
x^2=xMore specifically, it can be a direct sum of copies of Z_2. In particular, we have to have x+x=0 always. Here’s why. If x^2=x then 1+x=(1+x)^2=1+2x+x^2=1+2x+x and you cancel out 1+x from both sides, get 0=2x. You also have x(1+x)=x+x^2=x+x=0, so you have a whole bunch of zero divisors and only one unit, 1. In the first sentence I said it can be a direct sum of copies of Z_2, and I’m trying to figure out if this is always the case. If the ring is finite I am positive that it has to be the case (I need a proof) but what if it’s infinite? I’m not trying to be a show off here, it’s just that I find this stuff gloriously fascinating.
This is true for the real numbers. The real numbers are a field (you can cancel) so anytime you multiply two numbers to get zero, one of them has to be zero. This isn’t true in general rings. As a simple example of this, suppose that you’re working in ZxZ, which is a ring consisting of all ordered pairs (x,y) where x and y are integers. Simple idea, you have elements like (3,2) and (-4,7) and you multiply and add component wise so that (5,7)+(4,2)=(9,9) and (2,3)(5,4)=(10,12). So you have (0,5)(2,0)=(0,0), the zero element in ZxZ, you have two non-zero elements multiplying to the zero element. Same thing with, say, Z_6. When you divide a number by 6, you get a reminder, right, and this reminder will be 0,1,2,3,4,5, right? So you just focus on the remainders and do the math from there. So you get 2+3=5 and 5+2=1 because 5+2=7 and when you divide 7 by 6 your remainder is 1. In the same way you get 2x5=4 (because the remainder you get when you divide 10 by 6 is 4) and 3x5=3 (because the remainder you get when you divide 15 by 6 is 3). In particular you have 3=3x5=3x1, and you cannot cancel the 3. Same way, you get 3x4=0 (because 3x4=12 and 12 is evenly divided by 6, the remainder is 0). You have two non-zero elements multiplying to zero. What you wrote is true in a field. The ring of real numbers is a field. Boolean rings are, in general, not fields.x^2=x
x^2 - x = 0
x(x-1) =0
x=0 or x=1
If one of my students gave me that supposed proof, I give them a ZERO>
If you’re working in ZxZxZxZ, thenThis is true only for the real numbers. The real numbers are a field (you can cancel) so anytime you multiply two numbers to get zero, one of them has to be zero. This isn’t true in general rings. As a simple example of this, suppose that you’re working in ZxZ, which is a ring consisting of all ordered pairs (x,y) where x and y are integers. Simple idea, you have elements like (3,2) and (-4,7) and you multiply and add component wise so that (5,7)+(4,2)=(9,9) and (2,3)(5,4)=(10,12). So you have (0,5)(2,0)=(0,0), the zero element in ZxZ, you have two non-zero elements multiplying to the zero element. Same thing with, say, Z_6. When you divide a number by 6, you get a reminder, right, and this reminder will be 0,1,2,3,4,5, right? So you just focus on the remainders and do the math from there. So you get 2+3=5 and 5+2=1 because 5+2=7 and when you divide 7 by 6 your remainder is 1. In the same way you get 2x5=4 (because the remainder you get when you divide 10 by 6 is 4) and 3x5=3 (because the remainder you get when you divide 15 by 6 is 3). In particular you have 3=3x5=3x1, and you cannot cancel the 3. Same way, you get 3x4=0 (because 3x4=12 and 12 is evenly divided by 6, the remainder is 0). You have two non-zero elements multiplying to zero. What you wrote is true in a field. The ring of real numbers is a field. Boolean rings are, in general, not fields.
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Field (mathematics) - Wikipedia
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Lock the f— in, or LTFI for short, has become a theme for Penn State’s Playoff run. It embodies Franklin’s 1-0 mantra but with a less corny and more edgy delivery. LTFI has become a hashtag used by Penn State fans on social media when mentioning the next game. Like their head coach, they’re not looking past any opponent. It’s also a message that’s been printed on merchandise with Penn State’s NIL collective, Happy Valley United.---Audrey Snyder The Athletic
You've only seen one Superbowl run so I understand it's still new to you.I don’t understand that. Win 2? Wut?
That’s cold.You've only seen one Superbowl run so I understand it's still new to you.
hehe
Fitting that melon into a football helmet had to be work yo .
More specifically, it can be a direct sum of copies of Z_2. In particular, we have to have x+x=0 always. Here’s why. If x^2=x then 1+x=(1+x)^2=1+2x+x^2=1+2x+x and you cancel out 1+x from both sides, get 0=2x. You also have x(1+x)=x+x^2=x+x=0, so you have a whole bunch of zero divisors and only one unit, 1. In the first sentence I said it can be a direct sum of copies of Z_2, and I’m trying to figure out if this is always the case. If the ring is finite I am positive that it has to be the case (I need a proof) but what if it’s infinite? I’m not trying to be a show off here, it’s just that I find this stuff gloriously fascinating.
I just scouted ND, not a single TD earned. We just dominated 2 with lane discipline. If we PUNISH 13 like we did 2, ND got nothing for us. I guarantee a lions win.
I just scouted ND, not a single TD earned. We just dominated 2 with lane discipline. If we PUNISH 13 like we did 2, ND got nothing for us. I guarantee a lions win.
Your guarantee is the "kiss of death."I just scouted ND, not a single TD earned. We just dominated 2 with lane discipline. If we PUNISH 13 like we did 2, ND got nothing for us. I guarantee a lions win.
In fairness, Rick predicted way early that we’d be playing in the semifinals.Your guarantee is the "kiss of death."
Balloons come in handy!Practicing throwing things at my television. Got to make sure I have enough items on the couch.
Unfortunately they usually do not fly well enough.Balloons come in handy!
they do if you wizz in them firstUnfortunately they usually do not fly well enough.
No sex day of game - it’s bad luck. Now that I think about it we could schedule a game with an hour’s notice and still be safe.