Is the circle with the + inside commutative. Is the star commutative? What patterns in the table tell us this?

To determine if the operations defined by the tables (circle with a plus inside, \(\oplus\), and star, \(\star\)) are commutative, we need to verify if the order of operands affects the result. An operation is **commutative** if, for any two elements \(a\) and \(b\) in the set, the result of the operation \(a \ast b = b \ast a\) (where \(\ast\) represents the operation).

### Checking if \(\oplus\) is Commutative:

Let's analyze the table for the \(\oplus\) operation:

| \(\oplus\) | 0 | 1 | 2 | 3 |
|:---------:|:-:|:-:|:-:|:-:|
| **0**     | 0 | 1 | 2 | 3 |
| **1**     | 1 | 0 | 3 | 2 |
| **2**     | 2 | 3 | 0 | 1 |
| **3**     | 3 | 2 | 1 | 0 |

To check commutativity, we need to see if the table is symmetric across its diagonal (from top left to bottom right):

- Compare \((0 \oplus 1)\) and \((1 \oplus 0)\): Both are 1.
- Compare \((0 \oplus 2)\) and \((2 \oplus 0)\): Both are 2.
- Compare \((0 \oplus 3)\) and \((3 \oplus 0)\): Both are 3.
- Compare \((1 \oplus 2)\) and \((2 \oplus 1)\): Both are 3.
- Compare \((1 \oplus 3)\) and \((3 \oplus 1)\): Both are 2.
- Compare \((2 \oplus 3)\) and \((3 \oplus 2)\): Both are 1.

Since all corresponding pairs are equal, the operation \(\oplus\) is **commutative**.

### Checking if \(\star\) is Commutative:

Let's analyze the table for the \(\star\) operation:

| \(\star\) | 0 | 1 | 2 | 3 |
|:--------:|:-:|:-:|:-:|:-:|
| **0**    | 0 | 0 | 0 | 0 |
| **1**    | 0 | 1 | 2 | 3 |
| **2**    | 0 | 2 | 3 | 1 |
| **3**    | 0 | 3 | 1 | 2 |

To check commutativity, we again look for symmetry across the diagonal:

- Compare \((0 \star 1)\) and \((1 \star 0)\): Both are 0.
- Compare \((0 \star 2)\) and \((2 \star 0)\): Both are 0.
- Compare \((0 \star 3)\) and \((3 \star 0)\): Both are 0.
- Compare \((1 \star 2)\) and \((2 \star 1)\): Both are 2.
- Compare \((1 \star 3)\) and \((3 \star 1)\): Both are 3.
- Compare \((2 \star 3)\) and \((3 \star 2)\): Both are 1.

Since all corresponding pairs are equal, the operation \(\star\) is **commutative**.

### Conclusion:

- The \(\oplus\) operation is **commutative** because its table is symmetric across the diagonal.
- The \(\star\) operation is **commutative** for the same reason.

The **pattern** that tells us an operation is commutative is the **symmetry of the operation table** across the main diagonal. This means that for any \(a\) and \(b\), \(a \ast b = b \ast a\).

Would you like more details or have any other questions?

Here are some related questions you might find interesting:
1. What is an example of a non-commutative operation?
2. How can we check if an operation is associative?
3. What is the significance of commutativity in algebraic structures?
4. Are all operations on numbers commutative?
5. How does commutativity affect solving equations?

**Tip:** Symmetry in the operation table is a quick way to check for commutativity, but remember, an operation can be commutative without being associative. Always check both properties when analyzing algebraic structures!

Explore commutativity in mathematical operations through detailed analysis of tables for the circle with a plus inside and star operations. Learn how symmetry in operation tables indicates commutativity. Suitable for high school students and anyone interested in algebraic properties.

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