Defining New Operations: Practice

Evaluate the inner parentheses first.

$2 \star 3 = 2(2) + 3 - 1 = 6$

Then use $6$ as the first input of the outer star.

$6 \star 4 = 2(6) + 4 - 1 = 15$

Set $a \otimes e = a$ and solve for $e$.

$a + e - 3 = a \;\Longrightarrow\; e - 3 = 0 \;\Longrightarrow\; e = 3$

Check: $a \otimes 3 = a + 3 - 3 = a$, so $3$ leaves every input unchanged.

Substitute to get a quadratic.

$x^2 - 7 = 18 \;\Longrightarrow\; x^2 = 25 \;\Longrightarrow\; x = 5 \text{ or } x = -5$

Both values square to $25$, so both are solutions.

Set up the identity condition.

$2a + e - 1 = a \;\Longrightarrow\; e = 1 - a$

The required $e$ depends on $a$, so no fixed number works for every input. The star has no identity element.

Evaluate the inner parentheses first.

$1 \oplus 2 = 1 + 2 + 2 = 5$

Then apply $\oplus$ with $3$.

$5 \oplus 3 = 5 + 3 + 15 = 23$

The identity leaves every input unchanged.

$5 \oplus 0 = 5 + 0 + (5)(0) = 5$

This is exactly what makes $0$ the identity for $\oplus$.

Evaluate the inner star first.

$(-1) \star 2 = 2(-1) + 2 - 1 = -1$

Then apply the star with $-3$.

$(-1) \star (-3) = 2(-1) + (-3) - 1 = -6$

Put $x$ in both slots.

$x \oplus x = x + x + x^2 = x^2 + 2x$

Set it equal to $15$ and factor.

$x^2 + 2x - 15 = 0 \;\Longrightarrow\; (x + 5)(x - 3) = 0 \;\Longrightarrow\; x = 3 \text{ or } x = -5$

Left grouping: $2 \triangle 1 = 6 - 2 = 4$, then $4 \triangle 1 = 12 - 2 = 10$. Right grouping: $1 \triangle 1 = 3 - 2 = 1$, then $2 \triangle 1 = 6 - 2 = 4$.

$(2 \triangle 1) \triangle 1 = 10, \qquad 2 \triangle (1 \triangle 1) = 4$

Since $10 \ne 4$, the grouping matters, so $\triangle$ is not associative.

Set $a \otimes e = a$ and solve for $e$.

$a + e + 2 = a \;\Longrightarrow\; e + 2 = 0 \;\Longrightarrow\; e = -2$

Check: $a \otimes (-2) = a - 2 + 2 = a$.

Write each side with the rule.

$2 \star x = 4 + x - 1 = x + 3, \qquad x \star 2 = 2x + 2 - 1 = 2x + 1$

Setting them equal gives $x + 3 = 2x + 1$, so $x = 2$. The two orders agree only at that one input.

Evaluate the inner parentheses first. Since $0$ is the identity, $0 \oplus 5 = 5$.

$0 \oplus 5 = 0 + 5 + 0 = 5$

Then $2 \oplus 5 = 2 + 5 + 10 = 17$.