Determine if the function \( y=x^{2}-x \) is one-to-one. Select an answer \( v \)

**Step 1.** Recall that a function is one-to-one if for any \( a \) and \( b \), \( f(a)=f(b) \) implies \( a=b \). **Step 2.** For the function \( f(x)=x^2-x \), assume that for some \( a \) and \( b \) we have \[ a^2 - a = b^2 - b. \] **Step 3.** Rearranging the equation, we write: \[ a^2 - a - b^2 + b = 0. \] Factor by grouping: \[ (a^2 - b^2) - (a - b) = 0. \] **Step 4.** Notice that \( a^2 - b^2 \) factors as \( (a-b)(a+b) \): \[ (a-b)(a+b) - (a-b) = (a-b)(a+b-1) = 0. \] **Step 5.** The equation \( (a-b)(a+b-1) = 0 \) implies that either \[ a - b = 0 \quad \text{or} \quad a + b - 1 = 0. \] - The first case, \( a - b = 0 \), gives \( a=b \), which is acceptable for one-to-one functions. - The second case, \( a+b=1 \), can yield \( a \neq b \). For example, \( a=0 \) and \( b=1 \) satisfy \( a+b=1 \) and \[ f(0)=0^2-0=0 \quad \text{and} \quad f(1)=1^2-1=0. \] **Conclusion.** Since there exist distinct values \( a \) and \( b \) (such as \( 0 \) and \( 1 \)) with \( f(a)=f(b) \), the function is not one-to-one. Select the answer: \( v = \text{No} \).