MHB Cantor Diagonalization | Find a Number Not on the List

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Consider the following list of numbers. Using Cantor's diagonalization argument, find a number not on the list (use 2 and 4 when applying Cantor' argument). Give a brief explanation of the process.

0.123456876…

0.254896487…

0.143256876…

0.758468126…

0.534157162…
 
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It's not a good idea to ask other people to do something trivial, which you can easily do yourself. For example, suppose you are told to compute $1\oplus 1$. You say, "I don't know, I need help with this". The person who gave you this problem asks, "What kind of help? What exactly don't you understand?" You say, "I don't know what $\oplus$ denotes". "Well, to calculate $x\oplus y$ you need to add $x$ and $y$ and then take the remainder when the sum is divided by 2. So the answer is 0 if the sum is even and 1 if the sum is odd. Can you solve the problem now?" "Of course, the answer is 0".

So it is with this problem. If you understood what the Cantor's diagonalization argument is, solving the problem would be trivial. But if you don't understand the argument, you should ask questions about the argument itself, not about how to use it. Otherwise you leave open a possibility that you know and understand the Cantor's argument but cannot be bothered to do a trivial computation. So get a good textbook and tell us what is the first sentence in the description of the Cantor's argument that you don't understand and why.
 
Do you not know what "Cantor diagonalization" is or do you just want some one to do the work for you?

The idea is that we can create a new number, not on a given list, by changing the first digit of the first number, the second digit of the second number, the third digit of the third number, etc.

The first digit of the first number is "1" so write any digit except 1. Since the instructions say " use 2 and 4 when applying Cantor' argument", write "2"instead. The second digit of the second number is "5" so write "2" instead. The third digit of the third word is "3" so write "2" instead. So far that gives 222...

Keep using "2" until the number you want to replace IS "2" and then use "4" instead.

Now, can you explain why this guarantees, even though we have an infinite list of numbers, that this number is not any where on the list.
 
Since this has been here a while:
The first number is
0.123456876…
Cantor's method would replace that "1" by any other digit. Since here we have been told to "use 2 and 4", I will replace it by "2" so my number starts "0.2"

The second number is
0.254896487…
We want to replace the second digit, "5", by any other digit. I will choose "4" so now we have "0.24".

The third number is 0.143256876…
We want to repace the third digit, "3", by any other digit. I will choose "2" so now we have "0.242".

The fourth number is
0.758468126…
We want to replace the fourth digit, "4", by any other digit. I will have to use "2" since I am supposed to use either "2" or "4" and I cannot use "4". Now we have "0.2422".

The fifth number is 0.534157162…
We want to replace the fifth digit, "5", by any other digit. I choose "4" so now we have 0.24224.

It is trivial to see that the number 0.24224... is not on this list. The point is that we can continue doing this "infinitely" so that, even if we had an infinite list of numbers we could create another number that is NOT on that list- the set of all real numbers is NOT "countable".
 
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Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

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