If you "hate writing the word divisible" there is an operator for that. It's much easier to write. It's basically the "pipe" operator--> |, So 2|5 is false, but 2|6 is true.
@@NeilDoesMaths Even if it doesn't explicitly appear at A-level, is there any problem introducing the notation as a short aside? It is simple, speeds up writing the proof and is used fairly widely elsewhere, so mentioning it allows a student to read beyond the A-level spec. I could see the argument "extra notation might confuse an examiner" - but at A-level my understanding is that scripts are marked by people who know maths, which is not necessarily the case at GCSE, so would either already know the notation or recognise it as a notation and use 10 seconds to google it and check it is used correctly. The only sound reason I've heard for not using it is due to potential confusion arriving from it meaning a variety of different things in different contexts: divisibility, evaluating a derivative at a point, shorthand for "such that", conditional probability, restricting domains.
@plasmacarrot6863 yeah there is markers are inconsistent and their knowledge is limited, a level maths is very write it like this or you don't get marks
Its actually easier to prove that ✓n is irrational, where n is not a perfect square. Assume: ✓n=p/q. => n=p^2/q^2. => nq^2=p^2 This means that n is a factor in p^2. But if n is not a perfect square, at least one of its prime factors will have an odd exponent in the factorized form. However p^2 is a perfect square and therfore all prime factors have an even exponent in the factorized form. Therefore we have a contradiction. And thus the square root of all non perfect squares are irrational.
@@frs2512312 is also not a square. But it isnt Prime either. The point is that that line of reasoning only removes squares from the picture. Whereas we need to remove all the non square composites as well. As a formal proof its incomplete is all. Its the same every square is a rhombus but not every rhombus is a square kind of thing.
@@alexicon2006Wdym, i showed that all the NON-squares to be irrational. So the root of 2,3,5,6,7,8,10,11,12,13,14,15,17,18,.... And therefore all the primes as well, since a perfect square isnt prime.
cool video/proof! i did it just before watching and was glad to see we used the same approach one question: are we allowed to divide by k (when showing that (k^2)(n^2) = kq^2 implies k | q^2)? it makes sense since k ≠ 0 and we're working within the rational numbers so division makes sense, but whenever doing number theory-esque proofs like this i've been told that division / multiplicative inverses in the rationals aren't appropriate. i went back into mine and used another contradiction to show it but i was curious about your thoughts enjoying the content!
can you prove using rational root theorem ? so for the function x2 - k = 0 with solution sqrk if we assume its a rational number it should be found by some factor of the constant divided by the coefficient of the highest power so therefore the possible rational roots would be + or - k and + or - 1 and sqrk does not belong to the set so it cannot be a rational number
kq^2 cant be a square since k is a prime by hypothesis. Not square * square = not square . Not as watertight as your explanation but a bit of a shortcut if you want intuition without longing out the proof
can we not just deduce that q must be 1 on account of the primes being a subset of the naturals. thus leading to the statement k=p^2 which is a contradiction to the statement that k is prime?
Why can't we just: Let √k = u, where u is natural number greater than 1. k = u² => factors of k are 1, u(=√k), u²(=k) This contradicts the fact that prime numbers have only two factors, 1 and the number itself, which implies √k must be irrational
We don't need k to be a perfect square, just for its square root to be rational. So for example, sqrt(9/4) = 3/2 is rational, so maybe there's a chance that the square root of some integer could be rational too? So that's what we need to disprove. Checking that k is not square just ensures the square root won't be an integer, which is still useful, but not as powerful as the full result.
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If you "hate writing the word divisible" there is an operator for that. It's much easier to write. It's basically the "pipe" operator--> |, So 2|5 is false, but 2|6 is true.
We don't use notation like that at A-Levels.
@@NeilDoesMathsSo you don't use the notation that mathematicians use 😂
@@NeilDoesMaths further pure 2?
@@NeilDoesMaths Even if it doesn't explicitly appear at A-level, is there any problem introducing the notation as a short aside? It is simple, speeds up writing the proof and is used fairly widely elsewhere, so mentioning it allows a student to read beyond the A-level spec.
I could see the argument "extra notation might confuse an examiner" - but at A-level my understanding is that scripts are marked by people who know maths, which is not necessarily the case at GCSE, so would either already know the notation or recognise it as a notation and use 10 seconds to google it and check it is used correctly.
The only sound reason I've heard for not using it is due to potential confusion arriving from it meaning a variety of different things in different contexts: divisibility, evaluating a derivative at a point, shorthand for "such that", conditional probability, restricting domains.
@plasmacarrot6863 yeah there is markers are inconsistent and their knowledge is limited, a level maths is very write it like this or you don't get marks
Love how you highlight useful vocabulary.
i love how this guy is simultaneously a roadman and a smart math guy.
roadman maths cmon
Its actually easier to prove that ✓n is irrational, where n is not a perfect square.
Assume: ✓n=p/q.
=> n=p^2/q^2.
=> nq^2=p^2
This means that n is a factor in p^2. But if n is not a perfect square, at least one of its prime factors will have an odd exponent in the factorized form. However p^2 is a perfect square and therfore all prime factors have an even exponent in the factorized form. Therefore we have a contradiction. And thus the square root of all non perfect squares are irrational.
answer the question g, you are making up new questions
@NeilDoesMaths Whats up with the tone? Every child in the kindergarten understands that if a number is a perfect square it isn't prime.
@@NeilDoesMaths😂
@@frs2512312 is also not a square. But it isnt Prime either. The point is that that line of reasoning only removes squares from the picture. Whereas we need to remove all the non square composites as well. As a formal proof its incomplete is all.
Its the same every square is a rhombus but not every rhombus is a square kind of thing.
@@alexicon2006Wdym, i showed that all the NON-squares to be irrational. So the root of 2,3,5,6,7,8,10,11,12,13,14,15,17,18,.... And therefore all the primes as well, since a perfect square isnt prime.
cool video/proof! i did it just before watching and was glad to see we used the same approach
one question: are we allowed to divide by k (when showing that (k^2)(n^2) = kq^2 implies k | q^2)? it makes sense since k ≠ 0 and we're working within the rational numbers so division makes sense, but whenever doing number theory-esque proofs like this i've been told that division / multiplicative inverses in the rationals aren't appropriate. i went back into mine and used another contradiction to show it but i was curious about your thoughts
enjoying the content!
Thanks for your question! Yeah in this context it is fine because k represents a prime number, and zero is not prime. The proof is still complete
can you prove using rational root theorem ? so for the function x2 - k = 0 with solution sqrk if we assume its a rational number it should be found by some factor of the constant divided by the coefficient of the highest power so therefore the possible rational roots would be + or - k and + or - 1 and sqrk does not belong to the set so it cannot be a rational number
What do u do for triceps boss
5 reps of solving the riemann hypothesis
This is just a generalisation of the classic proof that the square root of 2 is irrational.
kq^2 cant be a square since k is a prime by hypothesis. Not square * square = not square . Not as watertight as your explanation but a bit of a shortcut if you want intuition without longing out the proof
can we not just deduce that q must be 1 on account of the primes being a subset of the naturals. thus leading to the statement k=p^2 which is a contradiction to the statement that k is prime?
We, as a student of 10th class, get this question at 2 marks 😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂
we can generalise this even more and say where k is not a perfect square sqrt(k) is irrational
Why can't we just:
Let √k = u, where u is natural number greater than 1.
k = u² => factors of k are 1, u(=√k), u²(=k)
This contradicts the fact that prime numbers have only two factors, 1 and the number itself, which implies √k must be irrational
√k may not be an integer, but still a fraction. Factors are integers that divide the number.
Why couldnt you just say that in order for k to be a perfect square, there must be some n s.t. n×n=k, which contradicts the fact that k must be prime?
We don't need k to be a perfect square, just for its square root to be rational. So for example, sqrt(9/4) = 3/2 is rational, so maybe there's a chance that the square root of some integer could be rational too? So that's what we need to disprove.
Checking that k is not square just ensures the square root won't be an integer, which is still useful, but not as powerful as the full result.
calm luh question
calmaaa suiiii
Hey man love your videos just wanna point out though it is one of the easier questions here in India (no bragging)
Thanks ! Send me your hardest question
@@NeilDoesMathssure
How's this hard? It's one of the first proofs you learn.
two fifths
Lets try JEE advanced question 😂
This is like saying ok now do Cambridge step 3 ? Youre so basic
Lets try putnam questions 😂😂😂😂
You called this hardest problem but in INDIA class 9th student solve this questions within seconds even his eyes are close 😂
do you want a medal or something? Here you go I guess... 🥉 bronze for you!
@ give this to all the students of class 9th not me
hope that helps your ego mate
Bro that was so simple even in 8th grade we got 10 times harder tasks to do.
Here is your medal 🥇
Also here is your cap 🧢
@@NeilDoesMaths💀 lun gang on top
this guy boogin
I don't understand that conclusion
@@onaman04 the fact that p and q share a prime factor contradicts their irreducibility condition imposed by the assumed rationality of sqrt p