## UnkleRhaukus Group Title $(\forall n\in \mathbb Z)[n|n^2]$ 2 years ago 2 years ago

1. UnkleRhaukus

t/f

2. ParthKohli

What does "|" mean in logic?

3. joemath314159

divides

4. ParthKohli

The division thing? Yes.

5. ParthKohli

Except for 0.

6. swissgirl

True

7. ParthKohli

What if $$\rm n = 0$$?

8. swissgirl

n^2=n*n so (n*n)/n=n

9. ParthKohli

False.

10. swissgirl

Truesomeness

11. ParthKohli

Though the statement is true for $$\mathbb{Z}^+$$

12. joemath314159

Dont read :$n\mid n^2$as division. By definition, $a\mid b \iff b=ak ,k\in \mathbb{Z}$there is no division taking place. By this definition, 0 does divide 0^2 since 0^2=0*0

13. ParthKohli

I've always heard, zero cannot divide anything.

14. joemath314159

You are thinking of fractions. Notice the definition doesnt contain anything about fractions.

15. joemath314159

its a statement only about multiplication.

16. ParthKohli

Oh.

17. UnkleRhaukus

$a|b \iff \exists q[b=aq],a\neq 0$

18. joemath314159

ah, then i am mistaken. if a cant be zero then what i posted is wrong.

19. ParthKohli

So, in fact, we do have fractions in the definition.

20. joemath314159

hmmm, i still wouldnt say there are fractions. This is generally the way they talk about division in Rings, where only multiplication and addition are defined. But yes, a cannot be zero. http://primes.utm.edu/glossary/xpage/divides.html

21. UnkleRhaukus

$$a|b$$ denotes $$b$$ is divisible by $$a$$