Prove \(f(x)=x^2+x+1\) is One-One but Not Onto
📺 Video Explanation
📝 Question
Prove that the function
\[ f:\mathbb{N}\to\mathbb{N}, \quad f(x)=x^2+x+1 \]
is one-one but not onto.
✅ Solution
🔹 Step 1: Prove One-One (Injective)
Assume:
\[ f(x_1)=f(x_2) \]
Then:
\[ x_1^2+x_1+1=x_2^2+x_2+1 \]
Cancel 1:
\[ x_1^2+x_1=x_2^2+x_2 \]
Rearrange:
\[ x_1^2-x_2^2+x_1-x_2=0 \]
Factor:
\[ (x_1-x_2)(x_1+x_2+1)=0 \]
Since:
\[ x_1+x_2+1>0 \]
Therefore:
\[ x_1-x_2=0 \]
So:
\[ x_1=x_2 \]
✔ Hence, \(f\) is one-one.
🔹 Step 2: Prove Not Onto (Not Surjective)
For onto, every natural number must have a pre-image.
But:
\[ 2\in\mathbb{N} \]
Check if there exists \(x\in\mathbb{N}\) such that:
\[ x^2+x+1=2 \]
This gives:
\[ x^2+x-1=0 \]
This has no natural number solution.
So:
\[ 2 \] is not in the range.
❌ Hence, function is not onto.
🎯 Final Answer
\[ \boxed{f(x)=x^2+x+1 \text{ is one-one but not onto}} \]
🚀 Exam Shortcut
- To prove one-one: assume equal outputs and compare
- Use factorization carefully
- To prove not onto: find one missing value in codomain