√2 's Last Decimal Digit!

I am starting to post a few discussions on irrational numbers, where you comment to find the flaw in the argument, comment your opinion, or more things related to the discussion. This is the first discussion. I will soon update this into a good Latex-ed discussion.

You can check this proof where it says that if √2 is written as the quotient of 2 integers aa and bb where aa is divided by bb, then aa and bb would then always have a common factor of 2 or i.e they are even and will be infinite integers as they cannot be reducible as they will always have a common factor of 2.

√2 can be written as 1.41421356237309504880168872420969807856967187537694... and what if we write it as a fraction like this,

141421356237309504880168872420969807856967187537694...1000000000000000000000000000000000000000000000000000...\LARGE \frac { 141421356237309504880168872420969807856967187537694... }{ 1000000000000000000000000000000000000000000000000000... }.

It is interesting to note in the fraction that these numbers will be never-ending or infinite integers which satisfy the condition of the proof as I mentioned above that they will have to be infinite integers so that they cannot be never reducible. But also, these numbers would also satisfy another condition of the proof that both the numbers would be even. Obviously, 10000000000... in the denominator would be even, but also 141421356237309504880168872420969807856967187537694... the last digit would also be even, which if again written in the decimal form means √2 's last decimal digit would be even i.e 0, 2, 4, 6, or 8.

Note by Siddharth Chakravarty
5 days, 9 hours ago

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I don't understand why you wrote:

But also, as these numbers are integers, they would also satisfy another condition of the proof that both the numbers would be even.

Are all integers even?

Vinayak Srivastava - 5 days, 9 hours ago

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No, of the proof, in the proof the numbers would have to be never-ending and even.

Siddharth Chakravarty - 5 days, 9 hours ago

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Mention people as I can't IDK why?

Siddharth Chakravarty - 5 days, 9 hours ago

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For your mentioning I mention some people, @David Vreken, @Mahdi Raza, @Aryan Sanghi, @Kriti Kamal

Vinayak Srivastava - 5 days, 9 hours ago

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Mention Zakir Hussain, Alak Bhattacharya, Pi Han Gogh, Alice Smith, Chris Lewis and others whose names I cannot remember for now.

Siddharth Chakravarty - 5 days, 9 hours ago

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@Zakir Husain, @Alak Bhattacharya, @Pi Han Goh

Vinayak Srivastava - 5 days, 9 hours ago

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other names are not coming in the list, and I have to sleep.

Vinayak Srivastava - 5 days, 9 hours ago

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Ok, Thanks a lot!

Siddharth Chakravarty - 5 days, 9 hours ago

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@Alice Smith

Vinayak Srivastava - 4 days, 23 hours ago

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I think you misunderstood the proof, it is the proof that 2\sqrt{2} is irrational, and so it can't have any pattern in its decimal expansion, which itself is never-ending.

Vinayak Srivastava - 4 days, 20 hours ago

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The proof suggests that if it is written in the form of a/b then that numbers would always be reducible as they would always have a common factor of 2 or to say the numbers will be even. And this is possible only of the numbers never end i.e infinite. So if write it as I showed as with 10000000... in the denominator both the numerator and denominator are integers and plus they will be never ending, and would have to be evn as per the proof. So it suggests that 141... should have even number in its end so that the whole number is even. It does not suggest any pattern in between.

Siddharth Chakravarty - 4 days, 20 hours ago

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Why? All powers of 10 have factors of 5 as well, which can also reduce. How can you say that there has to be an even factor?

Vinayak Srivastava - 4 days, 20 hours ago

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@Vinayak Srivastava You are correct, but here it means that the numbers will obviously have a common factor of 2, although they could have other common and distinct factors. As having a common factor of 2 means even, the number will be even.

Siddharth Chakravarty - 4 days, 19 hours ago

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@Siddharth Chakravarty Sorry, its not obviuos to me at all.

Vinayak Srivastava - 4 days, 19 hours ago

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@Vinayak Srivastava By obvious, I meant the proof suggests. You should first, better understand the proof and then try to think about it.

Siddharth Chakravarty - 4 days, 19 hours ago

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@Siddharth Chakravarty It is assumed at first that 2\sqrt{2} is rational, and as we arrive at a contradiction, it is irrational. This means that none of the steps are true, since the initial assumption itself is wrong.

Vinayak Srivastava - 4 days, 19 hours ago

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@Vinayak Srivastava You arrive at the contradiction because you assume that one point the numbers won't be reducible at all, i.e they will be be coprime and this is only possible if the integers are infinite or never-ending numbers, hence the precise way to define a rational number is that the integers p and q would be finitely expressed and not never-ending numbers, which still holds that root2 is irrational. So the integers are now infinite which means they have to satisfy the another condition of the proof that they will have to be even. :)

Siddharth Chakravarty - 4 days, 19 hours ago

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@Siddharth Chakravarty I don't agree with the last line. Why are you assuming that any statement in between the proof has to be true? It doesn't have to be.

Vinayak Srivastava - 4 days, 19 hours ago

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@Vinayak Srivastava Vinayak In the proof we contradicted that the numbers won't be reducible ever but for a number to be rational the integers would have to be in their reducible form by definition, and thus root2 does not follow this definition and hence is irrational and also for the integers to be never reducible they would have to be infinite, and we have cleary written it as a fraction of two infinite numbers so that 2 numbers would have to be even then by the proof. You have to understand what you contradicted and what you didn't.

Siddharth Chakravarty - 4 days, 19 hours ago

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@Siddharth Chakravarty Ok, but there do not exist a pair of integers (a,b)(a,b) for which ab=2\dfrac{a}{b}=\sqrt{2} , so even if you prove both are even if they existed, that is not useful.

Vinayak Srivastava - 4 days, 19 hours ago

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@Vinayak Srivastava They exist but are infinite, in rational number's definition we consider the integers as finite but I showed you a=141... and b=1000... but infinite numbers l So in a way it is useful as we found out the last decimal digit of root2 so suppose if it was 2 then root2 would 141.....2 or whatever the even number is out of 0, 2, 4, 6 or 8 at last 141.....E

Siddharth Chakravarty - 4 days, 19 hours ago

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@Siddharth Chakravarty Maybe I am not understanding, but I can't digest the concept "Last digit of an irrational number".

Vinayak Srivastava - 4 days, 18 hours ago

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@Vinayak Srivastava Some things are weird but true, and thus I opened this discussion if everybody agrees or somebody can disprove it, but as far as I see it is correct. Root2 is then a number which has infinity in its middle.

Siddharth Chakravarty - 4 days, 18 hours ago

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@Siddharth Chakravarty I will be against the argument, whatever the case. I will try to disprove it, but I have my intuition that it is wrong.

Vinayak Srivastava - 4 days, 18 hours ago

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@Vinayak Srivastava Will wait for you to disprove :) I'm on the opposite side for now as everything seems fine :)

Siddharth Chakravarty - 4 days, 18 hours ago

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@Siddharth Chakravarty Any statement inside a proof by contradiction does not need to be true.

Vinayak Srivastava - 4 days, 19 hours ago

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I really don't understand this note at all.

577408\frac{577}{408} is a good approximation for 2\sqrt2. Neither of the numerator nor denominator is a power of 10.

Pi Han Goh - 4 days, 17 hours ago

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@Pi Han Goh It is a good approximation, not root2.

Siddharth Chakravarty - 4 days, 17 hours ago

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@Pi Han Goh What do you not understand? The proof says if a and b are the integers and we express root2 as a/b then a/b would never be reducible as they would always be even or have a common factor of 2. This is only possible if a and b are never-ending numbers. In the note, we can write root2= 141..../100... where both the numerator and denominator have become infinite integers, so it satisfies the 1st condition and thus both the numbers would even as per the proof, so 141.... will be an even number and thus its last digit would be 0, 2, 4, 6 or 8.

Siddharth Chakravarty - 4 days, 17 hours ago

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@Siddharth Chakravarty The whole point of writing it as a fraction of integers a/b is to have a and b to be finite integers, what you have done is saying that a and b are "infinite integers", which is basically saying that a/b = infty/infty. Which is just absurd

An "infinite integer" doesn't have a last digit because it is not a finite number to begin with!

Pi Han Goh - 4 days, 17 hours ago

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@Pi Han Goh Infinity can arise in the middle also and yes infinity is absurd, and thus it does not say root2 is rational just because it can be written as the fraction as shown in the note because for the number to be rational the numbers would have to be finite integers. But if you look at the proof, a and b could be infinite integers also and as we proceed further we see they both will be even, so 141....E would be an even number with the last digit as E and infinite numbers in between, which causes it to be infinite.

Siddharth Chakravarty - 4 days, 17 hours ago

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@Siddharth Chakravarty What do you mean "infinity can arise in the middle?"

But if you look at the proof, a and b could be infinite integers also and as we proceed further we see they both will be even

I already told you this proof made no sense. There's no such thing as "infinite integers."

If you're saying that a and b can be (finite) integers to begin with, then sqrt2 is not rational, which is false!

Pi Han Goh - 4 days, 16 hours ago

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@Pi Han Goh The definition of rational numbers is that they are the quotient of finite integers a and b, with remainder 0. However, in the proof performing arithmetic with 'a' and 'b' is possible with infinitely expanded integers, so in a way, it suggests 'a' and 'b' should have infinite expansion.

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty

However, in the proof performing arithmetic with 'a' and 'b' is possible with infinitely expanded integers, so in a way, it suggests 'a' and 'b' should have infinite expansion.

So you're telling me a and b becomes unboundedly large? And so a/b = infty/infty again.

Pi Han Goh - 4 days, 16 hours ago

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@Pi Han Goh There is a difference between infinity and infinitely expanded numbers, infinity is something which is always far to a number no matter how big the number is.

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty "Infinity is not a number, its a concept. You can't perform simple arithmetic using infinity" -Some YouTube video I saw.

Vinayak Srivastava - 4 days, 16 hours ago

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@Vinayak Srivastava Yes, when did we perform arithmetic with infinity? Maybe you misinterpreted something.

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty How can you convert it to a fraction???

Vinayak Srivastava - 4 days, 16 hours ago

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@Vinayak Srivastava What did we convert into a fraction?

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty You converted a never-ending number to a fraction, which means you got ×2=\dfrac{\infty\times \sqrt{2}}{\infty}=\dfrac{\infty}{\infty}

Vinayak Srivastava - 4 days, 16 hours ago

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@Vinayak Srivastava So you mean 14142135....=1000... because they both are infinitely expanded and thus they both are infinity. Vinayak infinity is not a number, but rather a limit or concept.

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty Ok, I ask you just one thing- can you write out the numerator of your fraction completely? If not, then it tends to infinity, since it does not have any ending.

Vinayak Srivastava - 4 days, 16 hours ago

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@Vinayak Srivastava Therefore, it is called an infinitely expanded fraction, it does not tend to infinity, and how can it also, it is not like 0.99999... tends to 1 or =1. Any number whether finitely expanded or infinitely expanded is close to 0, so you can think that infinity is never possible to reach. @Pi Han Goh always just uses approximation and then mix infinity with anything. I would appreciate if what he/she Idk has a proper reasoning.

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty

infinitely expanded fraction, it does not tend to infinity

You said it's infinite, then you said it's not infinite. Which one is it?

Pi Han Goh - 4 days, 16 hours ago

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@Pi Han Goh Infinitely expanded means you could write the number forever, and infinity means the largest thing ever. Writing a nested radicaly is infinitely expanding but it is not infinity.

Siddharth Chakravarty - 4 days, 16 hours ago

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@Pi Han Goh @Siddharth Chakravarty, not all nested radicals also converge.
See this

Vinayak Srivastava - 4 days, 16 hours ago

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@Vinayak Srivastava I was talking about the different ones like 666...\sqrt { 6\sqrt { 6\sqrt { 6\sqrt { ... } } } }

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty We've been through this already. I already said your proof made no sense, you created some math definitions that aren't sound to justify your proof.

I'm not continuing this conversation any longer.

Pi Han Goh - 4 days, 16 hours ago

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@Pi Han Goh What does not make sense, you always have to discontinue the conversations? Provide a formal proof or reasoning, you are mixing everything and then you are telling it doesn't make sense. You are simply trying to tell then 100000....=141..... because they both are infinitely expanded and both are thus the same or infinity.

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty I already said:

a and b must be finite integers such that a/b is finite.

What you did is:

a and b are "infinite integers" with "infinite expansion" (which doesn't make sense)

because this means that a and b are unboundedly large, and so a/b = infty/infty = does not exist.

Pi Han Goh - 4 days, 16 hours ago

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@Pi Han Goh No, infinitely expanded means they could be written forever, but that doesn't mean arithmetic can't be applied to them. Infinity is not a number, but a limit so a number cannot approach infinity

Siddharth Chakravarty - 4 days, 16 hours ago

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@Pi Han Goh Also, both the numerator and denominator are not the same thing, so you cannot write infty/infty the similarity is that they have an infinite expansion. Infinity can arise in the middle, for example the denominator 1000.... would obviously have the last digit as 0, which 10000....0 so which means there is infinity in the middle or there are infinite zeroes in between of 10000....0

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty

so which means there is infinity in the middle or there are infinite zeroes in between of 10000....0

So you're basically saying b = 10^infinity = infinity, which is plain absurd. You can't perform arithmetic operation on infinity.

Pi Han Goh - 4 days, 16 hours ago

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@Pi Han Goh No, it doesn't suggest that it doesn't suggest infinity, it suggests infinite expansion. We are performing arithmetic with numbers of infinite expansion, not with infinity :)

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty How can you perform :

arithmetic with numbers of infinite expansion, not with infinity

?????

Vinayak Srivastava - 4 days, 16 hours ago

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@Siddharth Chakravarty I'm sorry, but it still doesn't make sense. "Infinite expansion" isn't sound either.

I feel that you're trying to use some unfounded reason to justify why your proof works, which in fact it doesn't.

Pi Han Goh - 4 days, 16 hours ago

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@Pi Han Goh Infinite expansion, in the sense, for example, adding 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...., You obviously know if I have suppose 1234567... and 10000... and divide both 1234567.../100000... and the infinite expansion of 100000.... is accordingly done so that there will be the decimal always after 1 then we know it would be 1.234567...

Siddharth Chakravarty - 4 days, 16 hours ago

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Really good post! I can't even see the flaw! @Siddharth Chakravarty

Yajat Shamji - 4 days, 18 hours ago

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Thanks, but still I would appreciate if you find the flaw or anybody :) I will soon post the another part.

Siddharth Chakravarty - 4 days, 18 hours ago

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I think that since a,ba, b are even, it contradicts that ab\frac{a}{b} is irreducible, therefore, Q.E.D - 2\sqrt{2} is irrational.

@Siddharth Chakravarty

Yajat Shamji - 4 days, 18 hours ago

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@Yajat Shamji The discussion suggests that root2 is irrational and that its last digit would be even.

Siddharth Chakravarty - 4 days, 18 hours ago

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@Siddharth Chakravarty Well...

Yajat Shamji - 4 days, 18 hours ago

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What did I miss?

Yajat Shamji - 4 days, 16 hours ago

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@Siddharth Chakravarty, watch this

Not all series of any numbers converge, some tend to infinity.

Vinayak Srivastava - 4 days, 16 hours ago

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But here it does not :)

Siddharth Chakravarty - 4 days, 16 hours ago

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Prove it! :)

Vinayak Srivastava - 4 days, 16 hours ago

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@Vinayak Srivastava You prove it that it tends to infinity. If infinity is defined as a number which is the largest of all, rather a never-ending limit. Then taking an infinitely expanded number approaching to infinity would then be greater than infinity which is not possible.

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty I don't need to, if an expression doesn't converge, it certainly tends to infinity. I want your proof, its your note. :)

Vinayak Srivastava - 4 days, 16 hours ago

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@Vinayak Srivastava Simple, suppose 45/10 which is 4.5 so similarly if the numerator has n digits then to obtain the decimal point after the first digit of the number after dividing, the power of 10 should have n-1 zeroes. So 141... in the numerator would have a 1 digit more than 10000... in the denominator no matter how much infinitely it is expanded, and then if you say they both approach infinity is wrong, and also there might be numbers greater than 141.... which doesn't make sense at all. Simply put they would be different always.

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty Ok, now since you talked about limits, how does the limit exist when it has so many different answers?

Vinayak Srivastava - 4 days, 14 hours ago

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@Vinayak Srivastava What limit? It depends on what problem we are talking about. Simple thing is any number is smaller than infinity. Like one if just expands 141..., It just increases in value it doesn't tend to infinity. Otherwise you could take other number, say it will also tend to infinity, equate both and then say 1=10 which is wrong.

Siddharth Chakravarty - 4 days, 13 hours ago

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@Brilliant Mathematics, can you resolve, @Pi Han Goh, @Vinayak Srivastava and @Siddharth Chakravarty's arguement over the reasoning of this note?

Yajat Shamji - 4 days, 16 hours ago

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I had talked to some other people in my college, they are also a bit philosophically confused but say it is correct.

Siddharth Chakravarty - 4 days, 16 hours ago

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I know but I have been reading all of your arguments for 3030 minutes and I think that the staff should give the final verdict.

Yajat Shamji - 4 days, 16 hours ago

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@Yajat Shamji We aren't fighting a case, we are discussing math, "verdict" should not be used :)

Vinayak Srivastava - 4 days, 16 hours ago

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@Vinayak Srivastava Yes, I know. But, this?

Definitely the staff should be involved.

Also: https://brilliant.org/problems/euler-totient-function-equations-problem-1/

Yajat Shamji - 4 days, 16 hours ago

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@Yajat Shamji Remember the Staff are also people, so they also have their range of knowledge and ideas,

Siddharth Chakravarty - 4 days, 16 hours ago

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@Siddharth Chakravarty I do understand. But I believe that the staff should help you!

Yajat Shamji - 4 days, 16 hours ago

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All I can say right now is:

"Infinite integers" and "infinite integer expansion" are not sound terms.

If you could come up with a similar looking proof for 3\sqrt3 or equivalent with these terms, then maybe I can convince you what went wrong. Otherwise, all conversations are circular that leads us nowhere.

Pi Han Goh - 4 days, 7 hours ago

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@Pi Han Goh I agree, this won't lead us nowhere if nobody will do the rebuttal logically. I had talked to Kriti Kamal, who claimed that he will prove the last digit will be odd by today. After he posts his proof, I will sooon upload a similar looking proof for root3.

Edit: Root3's proof can't be done similarly.

Siddharth Chakravarty - 3 days, 22 hours ago

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@Siddharth Chakravarty Well, I hope I can convince you this time:

You have made a blunder in your logic here. We all know that 2\sqrt2 is irrational, so it can't be stated in terms of a ratio of two integers.

Now you said something about "infinite integers" and "infinite integer expansion." I've read through all our discussion in detail again, and it all boils down to you saying:

"Well, the number of digits of the numerator and denominator are identical, and they both have many many digits, infinitely many digits in between."

Which is basically saying that both the (values of) numerator and denominator diverge to infinity. Which doesn't make sense, because if they both diverge, then their ratio is infty/infty. Which is obviously undefined.

And even if you can somehow justify your way out of this, you're basically saying that you've found two integers whose ratio is 2\sqrt2. Which is again, contradicting the irrationality of 2\sqrt2.

I'm afraid, you really wanted to create some observation regarding the "last digit of the numerator of 2\sqrt2." Unfortunately, that doesn't work.

Also, "infinite integers" and "infinite integer expansion" are not (known) math terms. So it definitely makes it harder for readers to understand what you're trying to say.

Pi Han Goh - 3 days, 19 hours ago

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@Pi Han Goh I agree with the majority of your things, but partially or not fully with this following lines and the few last lines:

so it can't be stated in terms of a ratio of two integers.

Which is basically saying that both the (values of) numerator and denominator diverge to infinity. Which doesn't make sense, because if they both diverge, then their ratio is infty/infty. Which is obviously undefined.

  • (I agree that infty/infty is absurd)

The first line I mentioned, I partially agree because the full definition for a number to be rational is that the integers have to be finite, and not infinite. Because then any number could be expressed as rational. So 141421356237309504880168872420969807856967187537694...1000000000000000000000000000000000000000000000000000...\frac { 141421356237309504880168872420969807856967187537694... }{ 1000000000000000000000000000000000000000000000000000... } is a fraction but with never-ending integers which satisfy that √2 is irrational.

Now your second line, both do not diverge to infinity, writing 10000.... forever is just like the value is getting BIG and BIG but it never should approach infinity, unboundedly large, because infinity is basically saying no matter how BIG and BIG a number could ever approach infinity, infinity is something not comprehensible because it is just weird and strange to do things with it.

Well, basically simply put, the proof says if you try to write √2 as a/b then a/b will never be reducible as they will always have the common factor of 2, and never reducible is only possible when the numbers never end, not infinity otherwise you could prove absurd things like 1=0. So writing √2 as 141421356237309504880168872420969807856967187537694...1000000000000000000000000000000000000000000000000000...\frac { 141421356237309504880168872420969807856967187537694... }{ 1000000000000000000000000000000000000000000000000000... } satisfies the properties of 'a' and 'b'.

Just think infinity is bigger than anything, no matter how never-ending a number is, it is always close to 0 because 10000.... as a numerical value whereas infinity does not. You have to understand the difference. At first, I thought even the same but it is not. Approaching Infinity is different and being infinity is different, otherwise, you would have no numbers in between, which actually is possible. I had to revise all the details and messages for this.

Siddharth Chakravarty - 3 days, 18 hours ago

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@Siddharth Chakravarty I still want somebody who could disprove this with a proper-defined logic, I have told some other people who have a good knowledge of mathematic, who are on their way trying to disprove it. I am myself trying to, thus I opened the discussion.

Siddharth Chakravarty - 3 days, 18 hours ago

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@Siddharth Chakravarty

The first line I mentioned, I partially agree because the full definition for a number to be rational is that the integers have to be finite, and not infinite. Because then any number could be expressed as rational. So 141421356237309504880168872420969807856967187537694...1000000000000000000000000000000000000000000000000000...\frac { 141421356237309504880168872420969807856967187537694... }{ 1000000000000000000000000000000000000000000000000000... } is a fraction but with never-ending integers which satisfy that √2 is irrational.

Let us break this down. We know that 1,1410,141100,14141000,1414210000, 1, \frac{14}{10}, \frac{141}{100}, \frac{1414}{1000}, \frac{14142}{10000}, \ldots are all rational approximation of 2\sqrt2. But you made the leap to say that your large fraction above is EQUAL to 2\sqrt2.

You shouldn't use "never-ending integers", it's not a correct term. It should be called "infinitely many digits". And by infinitely many digits, they DIVERGE to infinity.

Take this number(?) for example, 11111111\ldots. You call it "never-ending integers", when it should be called "infinitely many digits."

So using your logic, we can say that 19=1199=111999=11119999=1111199999 \frac19 = \frac{11}{99} = \frac{111}{999} = \frac{1111}{9999} = \frac{11111\ldots}{99999\ldots} . Right? Does it mean that the ratio of these two integers(?) 1111÷99991111\ldots \div 9999\ldots is equal to 19 \frac19?

You can't! That's because you're making the implicit assumption that the digits in the numerator and the denominator are equal to each other!

Note that there's a distinction between these 3 expressions:

  • 11111÷99999 {11111\ldots} \div {99999\ldots} ,

  • limn(11111n digits)÷(99999n digits) \lim \limits_{n\to\infty} \left( \underbrace{11111\ldots}_{n \text{ digits}} \right) \div \left( \underbrace{99999\ldots}_{n \text{ digits}} \right) .

  • [limn(11111n digits)]÷[limn(99999n digits)] \left [ \lim \limits_{n\to\infty} \left( \underbrace{11111\ldots}_{n \text{ digits}} \right) \right ] \div \left [ \lim \limits_{n\to\infty} \left( \underbrace{99999\ldots}_{n \text{ digits}} \right) \right ] .

Can you tell which one is finite and which one does not exist?

See this and this one too.

Now your second line, both do not diverge to infinity, writing 10000.... forever is just like the value is getting BIG and BIG but it never should approach infinity, unboundedly large, because infinity is basically saying no matter how BIG and BIG a number could ever approach infinity, infinity is something not comprehensible because it is just weird and strange to do things with it.

When you add \ldots into the end your expression, it means that it goes on forever. So limn10n=1000=\lim \limits_{n\to\infty} 10^n = 1000\ldots = \infty, which diverge!

Well, basically simply put, the proof says if you try to write √2 as a/b then a/b will never be reducible as they will always have the common factor of 2, and never reducible is only possible when the numbers never end, not infinity otherwise you could prove absurd things like 1=0.

You're wrong here again. The whole point of the (classical) proof of the irrationality of 2\sqrt2 is by starting with the assumption of coprime integers a,ba,b and somehow realizing that this assumption is wrong.

And when you say "numbers never end", it means both a and b DIVERGE TO INFINITY!!!!


I totally understand where you're coming from. You have basically made a bunch of misconceptions regarding the concept of infinity.

Your proof always skirts around the concept of infinity by replacing them with "never-ending digits" or other seemingly innocuous terms, which at the end of the day, are synonymous to the concept of infinity.

Try using more formal math terms and you will see what went wrong.

Pi Han Goh - 3 days, 17 hours ago

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@Pi Han Goh Excellent explanation!!!!!! :)

Vinayak Srivastava - 3 days, 17 hours ago

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@Pi Han Goh Neither do I claim for now, and for sure root2's last digit will be even because even I am also trying to see where is the flaw. I've tried to do it with more mathematical terms and yes I had mentioned it earlier it would have the digits accordingly, and when you talk about infinity there is no effect of adding 1 or 2. I think you have misunderstood the concept of infinity, certain things are not the same when they reach the exact value. If we do not do things properly, there are a lot of crazy things going to happen. We all are good at maths, but including infinity can several times cause different routes for us to follow the truth conventionally. For now, I will study and research a bit more keeping your antithesis because I assume your point also to be considered. I will still wait for others, till then I will do the synthesis on some thesis and some other anthesis including yours. Thanks for giving your time, though I am not fully convinced, for now, I will see how things go. And never-ending integers or integers with infinitely many digits is the same thing. I would not love to distinguish somebody here like a fool or genius, because there are some professors whom I showed your replies who don't agree with you while some do agree. For now, I can only perform some more studies with proper logic and synthesize :)

Siddharth Chakravarty - 3 days, 17 hours ago

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@Siddharth Chakravarty That's fine.

What I would recommend for you to do is figure out the distinction between the three expressions above. And understand what makes one (or more?) of them finite, and the others diverge.

Pi Han Goh - 3 days, 16 hours ago

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@Pi Han Goh Yes for sure, you can continue the discussion if you find something more convincing or logical as the definitions of infinity and certain terms here are debatable among us. I will look further into the matter or infinity and the weird things being proved or disproved. I will also reply if I see something else which is also convincing/logical. We will wait for some other Brilliant member also to post his views :)

Siddharth Chakravarty - 3 days, 16 hours ago

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@Siddharth Chakravarty It is not debatable to me. I'm pretty sure I'm right here.

Pi Han Goh - 3 days, 15 hours ago

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@Pi Han Goh No for root3 it is not possible because factors of 3 contain all the digits from 0 to 9 at the last.

Siddharth Chakravarty - 3 days, 21 hours ago

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