We all here coz we’ve seen this explained to us at a 5 year olds level 50 times and still don’t get it. But we’re still trying and i think thats alright.
Let /2 be rational no /2=a/b (where a and b are co-prime integers and b is not equal to 0) Squaring Both sides 2=a^2/b^2 2b^2=a^2 a^2 is divisible by 2 (Let p be a prime number. If p divides a^2 , then p divides a, where a is a positive integer.) a is divisible by 2 let a=2c for some integer c 2b^2=4c^2 b^2=2c^2 b^2 is divisible by 2 (Let p be a prime number. If p divides a2 , then p divides a, where a is a positive integer.) b is divisible by 2 Hence our assumption is wrong as a and b has 2 as a common factor but in we have consider them co prime integer so it contradicts our statement So, /2 is irrational no hope you like it thanks you please Like if it help you
Let /2 be rational no /2=a/b (where a and b are co-prime integers and b is not equal to 0) Squaring Both sides 2=a^2/b^2 2b^2=a^2 a^2 is divisible by 2 (Let p be a prime number. If p divides a^2 , then p divides a, where a is a positive integer.) a is divisible by 2 let a=2c for some integer c 2b^2=4c^2 b^2=2c^2 b^2 is divisible by 2 (Let p be a prime number. If p divides a2 , then p divides a, where a is a positive integer.) b is divisible by 2 Hence our assumption is wrong as a and b has 2 as a common factor but in we have consider them co prime integer so it contradicts our statement So, /2 is irrational no
Is this a good way to pick lottery numbers? Just assume your guess is wrong until the last possible number which has a nearly 100% chance of being right haha
One time I broke into Michael's house and made myself a snack. I had to be rushed to the hospital to have a nail and a copper wire surgically removed from my intestines.
I once broke ChatGPT 3.5 by asking, “Are the square roots of all prime numbers irrational?” It came back with “Some prime numbers like 2 have irrational square roots, while others like 3 have rational square roots.” And no matter how much I explained things it kept bouncing back and forth between acknowledging that all prime number have an irrational square root and stating that some primes have a rational square root. 😳 I just asked 4.0 if it understood that the square roots of all primes are irrational. It said, “Yes, I understand and acknowledge that the square roots of all prime numbers are irrational. A prime number, by definition, has no divisors other than 1 and itself, and its square root cannot be expressed as a simple fraction, making it irrational.” I asked the follow up question, “Are there any prime numbers that have a rational square root?” To which it replied: “No, there are no prime numbers that have a rational square root. All prime numbers have irrational square roots.” So maybe they fixed it.
I realize this video is 4 years old now and this has been settled but i was recently struggling to understand why the second choice is not a 50/50. After all you get to chose a second time after learning where one of the zonks is making your chance to win, intuitively, a 50/50. Instead however, i found the best way to intuit why switching is the better option is think of the problem in the probability of loss rather than in the probability of winning. When you make the first choice, you have a 66.7% chance of being wrong, and that fact remains after the host has opened the door and revealed the zonk. Meaning that the probability of winning if you switch now to the other door is 66.7% and thank you for the extra 33.3%
You would have a 66.7% chance of picking a wrong door even if the host didn't know where everything is. In that case if he revealed a goat there would be no advantage in switching. So the reason there is an advantage in switching is not because you were likely in picking an incorrect door but rather it is due to the knowledge of the host.
@@klaus7443 Assuming the host doesn't know whats behind the doors, you make a selection and then he opens a door at random and it happens to reveal a zonk, you absolutely have an advantage in switching your choice. Even in the unlikely event that he randomly reveals a zonk every time over the course of 1000s of attempts, it is still in your best interest to switch your choice as the likelihood of you selecting correctly on the first attempt is only 33.3%, meaning that your chance of picking incorrectly is 66.7%. That does not change after the door is opened and the zonk revealed, which means you should switch every time. Only in the event that the host reveals the treasure would your odds change, in which case it drops to 0%. In either event, we don't rely on the knowledge of the host to statistically prove that switching is the smartest strategy, only the likelihood of a first incorrect choice determines that.
@@callsignbeaver6355 "Assuming the host doesn't know whats behind the doors, you make a selection and then he opens a door at random and it happens to reveal a zonk, you absolutely have an advantage in switching your choice." 100% Wrong. If the host doesn't know what is behind the doors then the probability of picking Goat A and the host revealing Goat B is equal to the probability of picking the Car and the host revealing Goat B. Likewise the probability of picking Goat B and the host revealing Goat A is equal to the probability of picking the Car and the host revealing Goat A.
@@callsignbeaver6355 If the host is going to randomly open a door, he can also have the player pick a door for him. This means the player is the one who choose everything. Player: I pick door 1. Player: I want door 3 to be the last door (so I open door 2). What makes you think door 3 is better than door 1?
I have come across this concept for the second time now. all of you are boggling. 1) Its always 50 50 and never a 1/3 to 2/3 because Monty is always going to reveal a goat door 2) When Monty gives the choice to switch its is a new option and its probability is independent of the previous choice.
You should have read at least basics of probability, before embarrassing yourself like this. Both doors cannot hold a car equally likely, since one door was picked randomly from player, who don't know where the car is, while the other door was carefully picked by the host, who knows where the car is. "New option" cannot be independent from previous choice since you have picked the door you are staying with in previous choice.
@@max5250 who said I am staying with the previous one. the 1/3 probability of the revealed door is distributed equally on the remaining ones making them 50 50 for my next choice which I can get using a coin flip.
@@muzammalbaig Not 50/50. There are two probabilities to consider before the host even opens a door. The host reveals the other goat when one is picked, and can reveal either goat when the car is picked.
@@muzammalbaig Who said you are staying with the previous door?! Have you even watched the video or read the problem?!!! You can either: - stay with the door you picked, which still holds a car with 33.3% chances - switch to the other door, which holds a car with the remaining 66.6% chances Revealed door holds a car with 0% chance, so even if you would want to split that probability between the other doors (which doesn’t work like that in probability calculation) you would split 0% which is still 0%. If you ask how can that door hold a car with 0% chances, well, host knows what is behind these two doors, so, for him, these two doors are holding a car with 66.6% and 0% chances, respectively. If that door had a car behind with 33.3% chance, host would reveal a car once in a while, but that never happens.
>never a 1/3 to 2/3 because Monty is always going to reveal a goat door Counter-intuitively that is actually exactly why it's 2/3. It only happens because of the conditional reveal of a goat where the host has to actively avoid revealing a car, unlike a random pick. 2/3rds of the time the host has to actively avoid revealing the car and has to leave the car as the only remaining option to switch to.
I guess It depends on if the person sees the situation as a continuous situation. Once there's a revealed door, the problem starts again : there is a car behind A or B, there is no option C. So it would still just be 50/50. continuous situation wise it's switching doors.
"Once there's a revealed door, the problem starts again" No it doesn't. The contestant picked one of the three doors and the host will open a door that has a goat behind it. So if the probabilities of those two doors are 1/3 and 0 then the only other one must be 2/3.
@@courtofcryptids7626 Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for you.
