Is every integer derived from 1?

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find me an integer that isn't divisible by 1.
 
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fresh_42 said:
This is a contradiction in itself. ##1## is a unit.
what do you mean?
 
donglepuss said:
Is every integer derived from 1?
No. Why would you think it is?

Every even integer is divisible by 2. So what? Do you think that means that all even integers are derived from 2?
 
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I don't think anyone here understands what you mean by "derived from".
Are you referring to specific operations on the integers, like addition, multiplication, etc. Perhaps you could give us some examples of things that are "derived from" other things.
So, 3 = 1+1+1, then is 3 "derived from" 1?

Also, every integer is divisible by one, that is sort of the definition of "1". This is the basis of the "1 is a unit" comment. I would call it the multiplicative identity for numbers.
 
donglepuss said:
what do you mean?
The statement that "1 is a unit" comes from a generalization of the notions of addition and multiplication into abstract "rings" The notions of divisibility and of being a "prime number" can apply to such structures. The notion of a "unit" is also definable.

In grade school, we classified the positive integers as "prime", "composite" and "one". In the more general context, the classification is "prime", "composite" and "unit". [@fresh_42 would likely be quick to point out that we need not classify 0 since it is not a member of the multiplicative group]

One way to define "unit" is "any element which can be multiplied by another element to obtain 1 as a result". Using this definition and considering the signed integers, -1 is a unit since -1 * -1 = 1. Of course, 1 itself is always a unit since 1 * 1 = 1.
 
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What is the connection between title question and the OP?
 
Thread title: Is every integer derived from 1?
If by "derived by" you mean "can we get any integer by repeatedly adding 1?" -- Yes

OP question: find me an integer that isn't divisible by 1.
Answer: There aren't any. Besides being a unit, the number 1 is the multiplicative identity. For any real number r (which includes the integers), ##1 \times r = r##. This clearly shows that 1 is a factor of r, hence r is divisible by 1.

Since the question has been asked and answered, I'm closing the thread.
 
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