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Prove that $$S=\{I_0,I_1,I_2,I_3,I_4\}$$

is a partition of $\mathbb{Z}$. Where

$$I_k=\{x \in \mathbb{Z}:x\text{ has remainder $k$ when divided by $5$}\}$$

I"m not sure how to approach this I remember a partition also forms an equivalence relation but not sure how to use that.

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    How to approach this is to look up the definition of a partition. (added). How would you show that $I_0$, $I_1$, $I_2$, $I_3$, and $I_4$ satisfy it? – 2017-01-18
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    I'm not sure how to show the union of the sets covers the integers the other two conditions seem simple enough to state. – 2017-01-18
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    Every integer has *some* remainder when divided by 5, which is between $0$ and $4$. – 2017-01-18

2 Answers 2

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Yes, you remember correctly: The equivalence relation here is $a \sim b \iff a \equiv b\pmod 5$

Another ways of saying this is that $a\sim b \iff a-b \equiv 0 \pmod 5$.

Every integer $n\in \mathbb Z$, when divided by $5$, has one and only one of the following remainders: $0, 1, 2, 3, 4$. Those elements that have a remainder of $0$ form one class (are mutually related); those elements that have a remainder of $2$ form another class; $\cdots$; and all those elements that have a remainder of $4$ form the final class.

That is, every integer belongs to one and only one of the equivalence classes: $[0] = I_0, [1] = I_1, [2] = I_2, [3] = I_3, [4]= I_4,$ such that the classes are mutually disjoint.

So, since every integer is an element of and only one equivalence class, we have a partition of $$\mathbb Z = \{I_0, I_1, I_2, I_3, I_4\}$$

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Any integer has exactly one remainder $k$ when divided by 5 (0, 1, 2, 3, or 4) therefore it belongs to exactly one of the sets you defined. This also implies the union of the sets is equal to $\mathbb Z$ which means they form a partition.