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Maths query

As NF points out
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Therefore .9 recurring is 3 thirds. Which IS exactly equal to 1.
 
Code:
double d = 0.99999999999999999;

if( d == 1.0 )
{
	std::cout << "Yes";
}
else
{
        std::cout << "No";
}

It says Yes. One less 9 and it says No.
 
Therefore .9 recurring is 3 thirds. Which IS exactly equal to 1.

That's conjecture. Develop a theory and prove it.

There's no evidence that .999 to infinity will equal 1.

Honestly Euclid would be turning in his grave. 2000 years later and humans still cannot grasp infinity.

Understand his thought experiment on infinite primes and you'll understand why .999 recurring will always be that.

Euclid's Proof of the Infinitude of Primes
 
Last edited: 09:48

I like Khan academy, a valuable resource.

She said '0.999 recurring' is not a real number. Infinity is not a real number. That's blasphemy that is.

Of course if I take 0.999 and stop three places after the decimal point I too can also prove that is equal to 1 algebraically. But that does not answer the question.

From that video the answer is, I cannot divide 0.99 recurring anymore than I could divide 0
 
I like Khan academy, a valuable resource.

She said '0.999 recurring' is not a real number. Infinity is not a real number. That's blasphemy that is.

Of course if I take 0.999 and stop three places after the decimal point I too can also prove that is equal to 1 algebraically. But that does not answer the question.

From that video the answer is, I cannot divide 0.99 recurring anymore than I could divide 0

You can divide 0 :eyes
 

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