Thank you very much for this video. My lecturer who has a PhD struggled to explain this, and I'm paying them £9,000 a year. Where as you on the other hand, explained it absolutely amazingly withing 10 minutes, which I would guess was during your lunch break. You're amazing and I hope to see more videos from you!!
+Matthew Sherman The conclusion is then that you should rather pay that £9,000 to +shane molloy instead ;J Don't support bad teachers. Support good teachers. Otherwise, bad teachers will prosper, and good teachers will die out.
Please tell me.... I really need help.. Why do we need to assume that they are in the simplest form.. Rational numbers dont always happen in simolest form
All fractions can be written in their simplest form. i.e. the highest common factor of the numerator and denominator is 1. However when we do the proof we find that a and b do not have a HCF of 1 but 3. This means that if you assume that root 3 is rational you will end up with a fraction that is not in its simplest form.
I am a bit puzzled with your claim, that if a number divides a number (some square), then it divides the square root. That is not always correct. For instance 27 divides 81 three times, but 27 does certainly not divide 9 (the square root). Where's the error.
+Thomas Pulawski The error is that he misstated the theorem. The theorem actually says that if a PRIME divides a square, then it divides the square root.
+Isaacar Wolfsbane Thank you, yes that makes sense. 3 is a common divisor of 9 and 27... Great. So the proof checks out, but cannot be generalised to anything but primes... However, I recall learning that the square root of any number not a perfect square is irrational, so we could ask the author of a proof that the square root of, say, 8, also is irrational... :-)
Not the square root of 81 which is 9, but the square of 81 which is 6561. It is if p/a then p/a^2. But you have if p/a then p/d in your case "p" is 27, "a" is 81, and "d" is 9 which is the square ROOT of 'a', but it is the SQUARE of 'a', not the SQUARE ROOT of 'a' The reason is based on the fundamental theorem of arithmetic that every number is either a prime or a product of a UNIQUE set of primes. So 81 is (3x3x3x3), no other number has a product four '3' (3 is a prime). but the square of 81 which is 6561 (as said above) is (3x3x3x3, x 3x3x3x3), thus if 27 divides 81 then 27 must divide 6561 or 81^2 because 81^2 is just a repeat of 81 twice, instead of four '3's it now' has eight '3's or 3^8 instead of 3^4. So whatever is in the set of 81 is in the set of its square, just twice. 81^2 has the same unique set of numbers as 81, it is just that it is twice the amount just as if 5 divides 5 then 5 divides (5x5) or 2 divides (2x3) then 2 divides (2x3,x2x3) or the square of 2x3 or the square of 6 which is 36
yes, 27 divides 81, but 27 divides 81^2 or 6,561. 27 doesn't divide 9, since its square root could be a smaller number. It is not if a number divides some square then it divides its square root. it is if a number divides a number then it divides its square, not the same thing. you chose 81 because (9 x 9=81, and 9 hence is the square root of 81, so you say 81 is a square). If a number divides 9 its divides 81 and if a number divides 81 it divides 6561 and it divides 6561 it divides 43,046,721 or the square of 6561. 729 divides 6561, but 729 doesn't divide 81 (which the square root of 6561), 729 however divides 43,046,721 PS not because 81 is the square of a number (9), means that it applies to 9. However, if the number divides 81 (in your 27) then it will divide 81^2 or 6551
Can you now apply the same argument to √4, which is clearly rational (√4 = 2 = 2/1), to show that this argument still works in this case, correctly showing no contradiction?
I know this request is a year old, but let's go through the argument for sqrt(4). Assume sqrt(4) = a/b, where a and b are integers (for simplicity, we will consider the principal root, so we will assume we are only dealing with nonnegative integers throughout this process), b is not 0, and gcd(a,b) = 1. Squaring both sides, we obtain 4 = a^2/b^2, so we get 4b^2 = a^2 Now, a^2 is divisible by 4. Thus, a must be divisible by 2 (this is the most we can conclude - we cannot conclude that a is divisible by 4, since there are counterexamples to this: for example, 6^2 = 36 is divisible by 4, but 6 is not. The reason why we can use that sort of argument for numbers like 2 and 3 is because they are prime. Prime numbers have the property that if a prime number divides a product of integers, then it must divide at least one of the factors; hence, if a prime divides a^2, since a is the only factor, it must also divide a. The same is not true for non-primes). Therefore, a = 2c for some integer c. Plugging this in, we obtain 4b^2 = (2c)^2 = 4c^2 Dividing both sides by 4, we get b^2 = c^2 From which we obtain that b = c. Therefore, a = 2c and b = c, so c is a common factor of a and b. This may at first seem like a contradiction to the fact that gcd(a,b) = 1, but keep in mind that c _could_ be 1. In fact, if we continue our reasoning... Since gcd(a,b) = 1, and since all common factors divide the gcd, it follows that c is a positive integer dividing 1. The only possibility is c = 1. So a = 2 and b = 1, giving that sqrt(4) = 2/1. Of course, this is not a proof that sqrt(4) = 2/1 since we are assuming that sqrt(4) can be written as a fraction, and such an argument is circular reasoning. The argument, however, does not lead to a contradiction and shows that if sqrt(4) is rational, then its principal value is 2/1 = 2.
+MolloyMaths Well, with a few minor modifications, it can work a little more generally than that too - if n is a positive integer which is divisible by some prime p but not divisible by p^2, then the proof is still very close. But Bon Bon brought up a very common question I've seen on videos like this before - what happens if you try to apply the same technique to the square root of a perfect square? And I think it's valid to go through the method then to see what happens and to show them why it doesn't work for sqrt(4), for instance. And understanding where the method breaks down for the square root of a perfect square can really help identify what makes this proof technique shine for prime numbers - the fact that whenever a prime divides a square of an integer, it must divide the integer itself, and how crucial that fact is for the method to work.
at 5:30 you said any number if it divides a squared number(a²)then it divides the number(a)...but that's false, 4 divides 196 but doesn't divide 14...this theorem is just for prime numbers, and 4 is not a prime. in the video, it worked since 3 is a prime number
Molloy, you said that since both a and b share a common factor of 3, the square root of 3 cannot be expressed as a fraction. Well the fact is, you DID get a fraction: it's a/b. The fraction a/b looks like a perfectly good fraction to me. Just because it's unsimplified, that doesn't mean it's a non-fraction. What is preventing you from reducing a/b ? If you were to reduce that fraction, then HCF(a,b) would be equal to 1 and you would no longer have a contradiction. I don't doubt that the square root of 3 really is irrational. It just wasn't demonstrated in the above proof. By the way, that cursor on Khan Academy is horribly annoying. Am I alone on this?
Molloy, although it is true that a fraction cannot have a numerator and denominator with a HCF of both 1 and 3, you never incorporated the assumption that HCF(a,b) = 1 into the body of your proof. You arbitrarily declared that HCF(a,b) = 1 at the outset, but that assumption played no role in generating your result. If "HCF(a,b) = 1" actually generated "HCF(a,b) = 3", then I would be convinced. Here is what you did prove: "If sqrt3 is in fact rational, then that fraction can be expressed as an unreduced fraction." Then again, the same could be said of any fraction. By the way, that cursor on Khan Academy is horribly annoying. Am I alone on this?
All fractions can be reduced to their simplest form i.e. HCF of numerator and denominator = 1. In the proof root 3 is assumed to be rational, therefore = a/b with HCF (a, b) = 1, but shows that a and b have a HCF of 3 not 1 or if root 3 is rational, a and b will always have a HCF of 3 and can never be simplified to a fraction where the numerator and denominator have a HCF of 1.
Okay, now that does make sense. No fraction possesses those qualities. Thank you very much. Very effective use of the adverbs "always" and "never" by the way. Incidentally, that cursor on Khan Academy is horribly annoying. Am I alone on this?
why does the proof base on the facr that a and b must be in the simplest form? because if they share a factor you can just cancel out the common factor and you will still be left with sqrt(3)=a/b. or am i completely wrong?
+Simon Reeck You are right. There is a little thing which is unclear in almost every such proof, though: What if the fact that root(3) can be expressed as a fraction _depends_ on the fact that the numerator and the denominator are _not_ in their simplest form? Something in the lines of the following reasoning: root(3) = 1.732050808... = 1 + 7/10 + 3/100 + 2/1000 + 5/100000 + 8/10000000 + 8/1000000000 + ... Adding just the first two fractions would give 17/10 = 1.7, which is not root(3) yet. Adding just the first three fractions would give 173/100 = 1.73, which is also not root(3) yet. Adding just the first four fractions would give 1732/1000 = 1.732, which is still not root(3) (but closer). We would need to add infinitely many such fractions to get exactly the root(3). But we can observe the pattern in the resulting fraction: its denominator is just 1 followed by infinitely many 0s, and the numerator is all the significant digits (infinitely many of them again!) of the decimal expansion of root(3). Something in the lines of: 1732050808... / 1000000000... So technically we _could_ express the root(3) as a ratio of two natural numbers. The problem is that these numbers would have to be both infinitely big. And the bigger they are, the more there is the possibility that they have some common factors. So what this proof above does, is not contradicting that root(3) cannot be expressed as _a/b_ at all, but that _a_ and _b_ were _not_ in the simplest form _yet_ - we could still divide both of them by 3, to get _k_ and some other number (let's say _m_). And in fact we can repeat the entire argument again with _k/m_, which means that they weren't in the lowest terms too. And in fact, _no_ two numbers will ever be in the simplest terms in this argument - we can divide them still and still more by another 3. Which is pretty much equivalent to the fact that the numerator and the denominator must be infinitely divisible by 3, and therefore must be infinitely big themselves. Which is exactly the case in our example with decimal expansion! Therefore I would rather say that this proof doesn't mean that we _cannot_ express root(3) as _a/b_ at all - it just means that _a_ and _b_ would have to be both _infinitely big_ to be infinitely divisible by 3, and to fulfil our requirements.
Simon Reeck because we proved a/b must have a common factor 2, even we simpiflied it it will still have a common factor of 2, but no number can keep be divided by two but keep as an integer
What bothers me about this sort of proof is that it "works" with any integer. You can show that sqrt 4 is also irrational by this method. What I actually mean by this, precisely, is that the "3" didn't really take a special part of the demonstration.
The principal square root of 4 is 2, which is a rational number so Proof by contradiction is irrelevant here. This proof is a specific example of the general proof that the square root of prime numbers are irrational.
@@molloymaths1092 But what you're actually saying is "Well, it's irrational -- unless it's rational" LOL. Seriously, I'm well away of the proof, but it's only now that I recognize that it isn't particularly "proofy." Again, the reason is that the "3" doesn't really play a part in the "proof." It's just a place holder, really.
@@GetMeThere1 However if you use 4, the first line would be "assume root 4 is rational, then root 4 = a/b" but root 4 is 2 which CAN be written as a/b so the proof ends here. Anyway - thanks for the comment. Always good to hear different points of view on these proofs.
@@molloymaths1092 I have a card trick for you: You pick a card from a deck, and I guarantee it's the queen of clubs -- unless it's not. I have to check first, and if it's not then we'll try another card.
@@GetMeThere1 Ok so, you're missing the point really hard and its super cringe. The square roots of certain numbers are irrational, and the square roots of others are rational. IF YOU PICK ONE, LIKE 4, THEN IT WILL BE RATIONAL, IF YOU PICK ONE LIKE 2, IT WILL BE IRRATIONAL. U STOOPID. you're not "trying other cards" you're just checking if the one specific card is rational or not by contradiction. you would not pass the turing test
You can prove that the square root of any non-square integer is irrational by the same proof. However, it doesn’t. work for perfect squares because then you can’t assume that the perfect square divides the integers a and b themselves. Where has there been any proof that any number that divides a perfect square but not its square root must itself be a perfect square?
MolloyMaths Certainly. This is the same classical proof that is used to prove that sqrt2 is irrational. in each case, we assume that since a^2 is divisible by 3 or 2, respectively, that number must also divide a. But it doesn’t work for sqrt4, because 4 | a^2 does not imply that 4 | a. I suppose that my supposition was false anyway because it doesn’t work for n = 8. 8 | 16 = 4 ^ 2, but 8 !| 4. That is because sqrt8 is not a reduced square root; that would be 2 * sqrt2. However, your proof does work for all reduced square roots.
he let a=3k since "a" is divisible by 3 given that it was FOUND that a^2 is divisible by 3 so since "a" now is 3k or can be expressed as 3 times a number k, then "a^2"= 3k x 3k or 9k^2. so he moves from 3b^2= a^2 to 3b^2= 9k^2 (here the goal is to also show that 3 also divides b). so now both sides of the equation are divided by 3 to get b^2=3k^2 hence b^2 is also divisible by 3, and if b^2 is divisible by 3 (or 3/b^2) then 3 also divides 'b' or 'b' is divisible by 3 due to the fundamental theorem of arithmetic and so the set of 'b' consists of a unique product of prime, and therefore 'b^2' will also consist of those unique primes BUT only twice. So what if 'b' is '6' then 'b' is unique (2x3) no other number but '6' has these 2 primes with the product of 2x3 or 3x2. therefore 'b^2' also has the same unique set but just twice (2x3,2x3) hence if 3 divides 6, then 3 divides 36 or 6squared. But 'a' and 'b' are irreducible hence sqrt 3 is an irrational number, the proof, since 3 actually divides, 'a' and 'b'
If you're confused as to why 3|a^2 makes it true that 3|a, it's because of the fundamental theorem of arithmatic. if p|ab then p|a or p|b
wouldnt this rule not hold as 9|3^2 but 9 doesnt divide into 3?
@@ethis2091 9 is not prime
@@ethis2091 remember that a and b are coprime, meaning that none of their prime factors can have the same prime number
Thank you very much for this video. My lecturer who has a PhD struggled to explain this, and I'm paying them £9,000 a year. Where as you on the other hand, explained it absolutely amazingly withing 10 minutes, which I would guess was during your lunch break. You're amazing and I hope to see more videos from you!!
+Matthew Sherman
Thanks for your comment. Glad it makes sense. Any videos I've done are on www.molloymaths.com
+Matthew Sherman The conclusion is then that you should rather pay that £9,000 to +shane molloy instead ;J
Don't support bad teachers. Support good teachers. Otherwise, bad teachers will prosper, and good teachers will die out.
lol same
TO those who are lazy to watch : Its basically like proof of irrationailt of root 2 but the common factor of the "fraction" is 3 instead of 2
Please tell me.... I really need help.. Why do we need to assume that they are in the simplest form.. Rational numbers dont always happen in simolest form
All fractions can be written in their simplest form. i.e. the highest common factor of the numerator and denominator is 1. However when we do the proof we find that a and b do not have a HCF of 1 but 3. This means that if you assume that root 3 is rational you will end up with a fraction that is not in its simplest form.
I am a bit puzzled with your claim, that if a number divides a number (some square), then it divides the square root. That is not always correct. For instance 27 divides 81 three times, but 27 does certainly not divide 9 (the square root). Where's the error.
+Thomas Pulawski The error is that he misstated the theorem. The theorem actually says that if a PRIME divides a square, then it divides the square root.
+Isaacar Wolfsbane Thank you, yes that makes sense. 3 is a common divisor of 9 and 27... Great. So the proof checks out, but cannot be generalised to anything but primes... However, I recall learning that the square root of any number not a perfect square is irrational, so we could ask the author of a proof that the square root of, say, 8, also is irrational... :-)
@@pulawski Root 8 = 2 root 2
Not the square root of 81 which is 9, but the square of 81 which is 6561. It is if p/a then p/a^2. But you have if p/a then p/d in your case "p" is 27, "a" is 81, and "d" is 9 which is the square ROOT of 'a', but it is the SQUARE of 'a', not the SQUARE ROOT of 'a'
The reason is based on the fundamental theorem of arithmetic that every number is either a prime or a product of a UNIQUE set of primes. So 81 is (3x3x3x3), no other number has a product four '3' (3 is a prime).
but the square of 81 which is 6561 (as said above) is (3x3x3x3, x 3x3x3x3), thus if 27 divides 81
then 27 must divide 6561 or 81^2 because 81^2 is just a repeat of 81 twice, instead of four '3's it now' has eight '3's or 3^8 instead of 3^4. So whatever is in the set of 81 is in the set of its square, just twice.
81^2 has the same unique set of numbers as 81, it is just that it is twice the amount just as if 5 divides 5 then 5 divides (5x5) or 2 divides (2x3) then 2 divides (2x3,x2x3) or
the square of 2x3 or the square of 6 which is 36
yes, 27 divides 81, but 27 divides 81^2 or 6,561. 27 doesn't divide 9, since its square root could be a smaller number. It is not if a number divides some square then it divides its square root. it is if a number divides a number then it divides its square, not the same thing. you chose 81 because (9 x 9=81, and 9 hence is the square root of 81, so you say 81 is a square). If a number divides 9 its divides 81 and if a number divides 81 it divides 6561 and it divides 6561 it divides 43,046,721 or the square of 6561.
729 divides 6561, but 729 doesn't divide 81 (which the square root of 6561), 729 however divides 43,046,721
PS not because 81 is the square of a number (9), means that it applies to 9. However, if the number divides 81 (in your 27) then it will divide 81^2 or 6551
Can you now apply the same argument to √4, which is clearly rational (√4 = 2 = 2/1), to show that this argument still works in this case, correctly showing no contradiction?
I know this request is a year old, but let's go through the argument for sqrt(4).
Assume sqrt(4) = a/b, where a and b are integers (for simplicity, we will consider the principal root, so we will assume we are only dealing with nonnegative integers throughout this process), b is not 0, and gcd(a,b) = 1.
Squaring both sides, we obtain 4 = a^2/b^2, so we get
4b^2 = a^2
Now, a^2 is divisible by 4. Thus, a must be divisible by 2 (this is the most we can conclude - we cannot conclude that a is divisible by 4, since there are counterexamples to this: for example, 6^2 = 36 is divisible by 4, but 6 is not. The reason why we can use that sort of argument for numbers like 2 and 3 is because they are prime. Prime numbers have the property that if a prime number divides a product of integers, then it must divide at least one of the factors; hence, if a prime divides a^2, since a is the only factor, it must also divide a. The same is not true for non-primes).
Therefore, a = 2c for some integer c.
Plugging this in, we obtain
4b^2 = (2c)^2 = 4c^2
Dividing both sides by 4, we get
b^2 = c^2
From which we obtain that b = c.
Therefore, a = 2c and b = c, so c is a common factor of a and b. This may at first seem like a contradiction to the fact that gcd(a,b) = 1, but keep in mind that c _could_ be 1. In fact, if we continue our reasoning...
Since gcd(a,b) = 1, and since all common factors divide the gcd, it follows that c is a positive integer dividing 1. The only possibility is c = 1. So a = 2 and b = 1, giving that sqrt(4) = 2/1.
Of course, this is not a proof that sqrt(4) = 2/1 since we are assuming that sqrt(4) can be written as a fraction, and such an argument is circular reasoning. The argument, however, does not lead to a contradiction and shows that if sqrt(4) is rational, then its principal value is 2/1 = 2.
This proof only applies to the square root of prime numbers.
+MolloyMaths
Well, with a few minor modifications, it can work a little more generally than that too - if n is a positive integer which is divisible by some prime p but not divisible by p^2, then the proof is still very close.
But Bon Bon brought up a very common question I've seen on videos like this before - what happens if you try to apply the same technique to the square root of a perfect square? And I think it's valid to go through the method then to see what happens and to show them why it doesn't work for sqrt(4), for instance. And understanding where the method breaks down for the square root of a perfect square can really help identify what makes this proof technique shine for prime numbers - the fact that whenever a prime divides a square of an integer, it must divide the integer itself, and how crucial that fact is for the method to work.
@@MuffinsAPlenty THANK YOU. I was searching for over an hour why this prove doesnt work for square roots of 4, 9...
This video has been re-recorded as "Proof by contradiction that the square root of 3 is irrational"
Where did you come up with the assumption that if 3 divides a^2 then 3 also divides a
See the following video: th-cam.com/video/GD6VYrr78ic/w-d-xo.html
Thank you for taking time out of your day to make this video
+Marcus Hines
No problem. Hope it makes sense.
Thank you so much! You're the only teacher that I understood!
Your welcome. Glad it helps.
Thank you, legend, my math exam is coming up and I would be quite cooked without this knowledge.
Best of luck!
at 5:30 you said any number if it divides a squared number(a²)then it divides the number(a)...but that's false, 4 divides 196 but doesn't divide 14...this theorem is just for prime numbers, and 4 is not a prime.
in the video, it worked since 3 is a prime number
Yes. I didn't explain this in detail since I had already done this theorem in a previous video. See th-cam.com/video/GD6VYrr78ic/w-d-xo.html
Awesome video!
Thanks!
Molloy, you said that since both a and b share a common factor of 3, the square root of 3 cannot be expressed as a fraction. Well the fact is, you DID get a fraction: it's a/b. The fraction a/b looks like a perfectly good fraction to me. Just because it's unsimplified, that doesn't mean it's a non-fraction. What is preventing you from reducing a/b ? If you were to reduce that fraction, then HCF(a,b) would be equal to 1 and you would no longer have a contradiction. I don't doubt that the square root of 3 really is irrational. It just wasn't demonstrated in the above proof. By the way, that cursor on Khan Academy is horribly annoying. Am I alone on this?
A fraction cannot have a numerator and denominator with a HCF of both 1 and 3. That's the contradiction. Thanks for your comment.
Molloy, although it is true that a fraction cannot have a numerator and denominator with a HCF of both 1 and 3, you never incorporated the assumption that HCF(a,b) = 1 into the body of your proof. You arbitrarily declared that HCF(a,b) = 1 at the outset, but that assumption played no role in generating your result. If "HCF(a,b) = 1" actually generated "HCF(a,b) = 3", then I would be convinced. Here is what you did prove: "If sqrt3 is in fact rational, then that fraction can be expressed as an unreduced fraction." Then again, the same could be said of any fraction.
By the way, that cursor on Khan Academy is horribly annoying. Am I alone on this?
All fractions can be reduced to their simplest form i.e. HCF of numerator and denominator = 1. In the proof root 3 is assumed to be rational, therefore = a/b with HCF (a, b) = 1, but shows that a and b have a HCF of 3 not 1 or if root 3 is rational, a and b will always have a HCF of 3 and can never be simplified to a fraction where the numerator and denominator have a HCF of 1.
Okay, now that does make sense. No fraction possesses those qualities. Thank you very much. Very effective use of the adverbs "always" and "never" by the way.
Incidentally, that cursor on Khan Academy is horribly annoying. Am I alone on this?
why does the proof base on the facr that a and b must be in the simplest form? because if they share a factor you can just cancel out the common factor and you will still be left with sqrt(3)=a/b. or am i completely wrong?
+Simon Reeck You are right.
There is a little thing which is unclear in almost every such proof, though:
What if the fact that root(3) can be expressed as a fraction _depends_ on the fact that the numerator and the denominator are _not_ in their simplest form? Something in the lines of the following reasoning:
root(3) = 1.732050808... = 1 + 7/10 + 3/100 + 2/1000 + 5/100000 + 8/10000000 + 8/1000000000 + ...
Adding just the first two fractions would give 17/10 = 1.7, which is not root(3) yet.
Adding just the first three fractions would give 173/100 = 1.73, which is also not root(3) yet.
Adding just the first four fractions would give 1732/1000 = 1.732, which is still not root(3) (but closer).
We would need to add infinitely many such fractions to get exactly the root(3).
But we can observe the pattern in the resulting fraction: its denominator is just 1 followed by infinitely many 0s, and the numerator is all the significant digits (infinitely many of them again!) of the decimal expansion of root(3). Something in the lines of: 1732050808... / 1000000000...
So technically we _could_ express the root(3) as a ratio of two natural numbers. The problem is that these numbers would have to be both infinitely big. And the bigger they are, the more there is the possibility that they have some common factors.
So what this proof above does, is not contradicting that root(3) cannot be expressed as _a/b_ at all, but that _a_ and _b_ were _not_ in the simplest form _yet_ - we could still divide both of them by 3, to get _k_ and some other number (let's say _m_). And in fact we can repeat the entire argument again with _k/m_, which means that they weren't in the lowest terms too. And in fact, _no_ two numbers will ever be in the simplest terms in this argument - we can divide them still and still more by another 3. Which is pretty much equivalent to the fact that the numerator and the denominator must be infinitely divisible by 3, and therefore must be infinitely big themselves. Which is exactly the case in our example with decimal expansion!
Therefore I would rather say that this proof doesn't mean that we _cannot_ express root(3) as _a/b_ at all - it just means that _a_ and _b_ would have to be both _infinitely big_ to be infinitely divisible by 3, and to fulfil our requirements.
+Bon Bon oh man. thank you very much for your explanation!
Simon Reeck I'm not a man, I'm a pony ;) I accept your thanks, though ;)
Simon Reeck because we proved a/b must have a common factor 2, even we simpiflied it it will still have a common factor of 2, but no number can keep be divided by two but keep as an integer
What happen if we change "3" to "16", will the proof change? But we do know square root of "16" is rational and can be written as 8/4, or 4/1.
3 is a prime number. 16 is not. This is a proof for the square root of prime numbers.
What bothers me about this sort of proof is that it "works" with any integer. You can show that sqrt 4 is also irrational by this method. What I actually mean by this, precisely, is that the "3" didn't really take a special part of the demonstration.
The principal square root of 4 is 2, which is a rational number so Proof by contradiction is irrelevant here.
This proof is a specific example of the general proof that the square root of prime numbers are irrational.
@@molloymaths1092 But what you're actually saying is "Well, it's irrational -- unless it's rational" LOL. Seriously, I'm well away of the proof, but it's only now that I recognize that it isn't particularly "proofy." Again, the reason is that the "3" doesn't really play a part in the "proof." It's just a place holder, really.
@@GetMeThere1 However if you use 4, the first line would be "assume root 4 is rational, then root 4 = a/b" but root 4 is 2 which CAN be written as a/b so the proof ends here.
Anyway - thanks for the comment. Always good to hear different points of view on these proofs.
@@molloymaths1092 I have a card trick for you: You pick a card from a deck, and I guarantee it's the queen of clubs -- unless it's not. I have to check first, and if it's not then we'll try another card.
@@GetMeThere1 Ok so, you're missing the point really hard and its super cringe. The square roots of certain numbers are irrational, and the square roots of others are rational. IF YOU PICK ONE, LIKE 4, THEN IT WILL BE RATIONAL, IF YOU PICK ONE LIKE 2, IT WILL BE IRRATIONAL. U STOOPID. you're not "trying other cards" you're just checking if the one specific card is rational or not by contradiction. you would not pass the turing test
You can prove that the square root of any non-square integer is irrational by the same proof. However, it doesn’t. work for perfect squares because then you can’t assume that the perfect square divides the integers a and b themselves. Where has there been any proof that any number that divides a perfect square but not its square root must itself be a perfect square?
Can you give me an example?
MolloyMaths Certainly. This is the same classical proof that is used to prove that sqrt2 is irrational. in each case, we assume that since a^2 is divisible by 3 or 2, respectively, that number must also divide a. But it doesn’t work for sqrt4, because 4 | a^2 does not imply that 4 | a. I suppose that my supposition was false anyway because it doesn’t work for n = 8. 8 | 16 = 4 ^ 2, but 8 !| 4. That is because sqrt8 is not a reduced square root; that would be 2 * sqrt2. However, your proof does work for all reduced square roots.
Your hand writting is not clear
Thank you! This video was very helpful!
Glad it was helpful!
How do you go from "3k squared = 3b squared" to "9k squared = 3b squared"?
nvm I get it
he let a=3k since "a" is divisible by 3 given that it was FOUND that a^2 is divisible by 3
so since "a" now is 3k or can be expressed as 3 times a number k, then "a^2"= 3k x 3k or 9k^2. so he moves from 3b^2= a^2 to 3b^2= 9k^2
(here the goal is to also show that 3 also divides b). so now both sides of the equation are divided by 3 to get b^2=3k^2 hence b^2 is also divisible by 3, and if b^2
is divisible by 3 (or 3/b^2) then 3 also divides 'b' or 'b' is divisible by 3 due to the fundamental theorem of arithmetic and so the set of 'b' consists of a unique product of prime, and therefore 'b^2' will also consist of those unique primes BUT only twice. So what if 'b' is '6' then 'b' is unique
(2x3) no other number but '6' has these 2 primes with the product of 2x3 or 3x2. therefore 'b^2' also has the same unique set but just twice (2x3,2x3) hence if 3 divides 6, then 3 divides 36 or 6squared.
But 'a' and 'b' are irreducible hence sqrt 3 is an irrational number, the proof, since 3 actually divides, 'a' and 'b'
Very good explanation, thanks so much
You are welcome!
Mr molloy u the goat u carry me all year
Thanks. Glad to hear it.
so damn helpful thanks a bunch
No problem. Glad it helps.
Found this very helpful thankyou
Glad it was helpful!
i dont get if A and B have a HCF greater than one, how is root 3 irrational (where root 3 is equal to ratio of those numbers)
can anyone help +MolloyMaths
Can you expand on your question. I don't quite understand what you are asking
Thank you !!
No problem
Danke 🤗
Thank you Sir.
No problem.
thnks
Your welcome
why?
??
who are you?
hello?
Good
Thanks
Can3=tangpi/3
Phần hơi. DatlangsoN
You're too slow
Not everyone is as fast and intelligent as you!!