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# Is it.......

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, 7 months ago

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

- 2 years, 7 months ago

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

- 2 years, 7 months ago

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

- 2 years, 7 months ago

You have my algorithm

- 2 years, 7 months ago

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

- 2 years, 7 months ago

Hint: $$2 \times 5 = 10$$

- 2 years, 7 months ago

[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.

- 2 years, 7 months ago

Proof:

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

 1 2 3 4 5 6 7 8 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

- 2 years, 7 months ago