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Prove that all numbers can be expressed with 2 and 5, or give a reason why it is impossible.

Note by Bryan Lee Shi Yang
2 years, 4 months ago

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What exactly do you mean by "can be expressed with 2 and 5" ?

Prasun Biswas - 2 years, 4 months ago

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e.g. 3.5 = \(2^{2}-2^{0}+2^{-1}\), and 85.54 = \(2^{4}(2^{2}+1)+5^{1}+2^{-1}+5^{-2}\)

Bryan Lee Shi Yang - 2 years, 4 months ago

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I think that it should be that every "terminating decimal number" can be expressed with 5 and 2, because are base is 10.

Archit Boobna - 2 years, 4 months ago

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You have my algorithm

Agnishom Chattopadhyay - 2 years, 4 months ago

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I don't think \(3.5\) and \(85.54\) are integers, but I get your point.

Prasun Biswas - 2 years, 4 months ago

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Hint: \( 2 \times 5 = 10\)

Bryan Lee Shi Yang - 2 years, 4 months ago

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[This is just a mathematical explanation for the the algorithm given by Agnishom]

First, note that,

\[0=5\times 2-2\times 5\]

Now, consider any positive integer \(x\). We then have,

\[x=0+x=(5\times 2-2\times 5)+\sum_{k=1}^x 1=(5\times 2-2\times 5)+\sum_{k=1}^x (5-2\times 2)\\ \implies x=(5\times 2-2\times 5)+\underbrace{(5-2\times 2)+(5-2\times 2)+\ldots}_{x\textrm{ times}}\]

This is obviously a representation in \(2\) and \(5\), as the OP wants.

It suffices to show that there is such a representation also for \((-x)\) to complete the proof. Indeed, we just multiply the above result by \((-1)\) to get,

\[(-x)=(2\times 5-5\times 2)+\underbrace{(2\times 2-5)+(2\times 2-5)+\ldots}_{x\textrm{ times}}\]

Since there exists such representations for \(x~,~(-x)\) and \(0\) where \(x\in\Bbb{Z^+}\), the claim in the problem is proved.


This is kinda similar to an inductive proof.

Prasun Biswas - 2 years, 4 months ago

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Proof:

The following program expresses any positive integer in terms of 2 and 5

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n = int(raw_input())
buffer = ""

for i in xrange(n-1):
   buffer += "(5-(2*2)) +"

buffer += (5-(2*2))
print buffer

The proof of correctness has been left to the reader as an exercise

Agnishom Chattopadhyay - 2 years, 4 months ago

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