Exercise · ZW-INJECTIVE-FUNCTION-007 · ★★☆ Intermediate
Let \(f\colon\mathbb{Z}\to\mathbb{Z}\), \(f(n)=2n\). Is \(f\) injective?
Choose one option.
Show answer
- B. Yes; \(2a=2b\) implies \(a=b\)
Explanation
If \(2a=2b\), dividing by \(2\) gives \(a=b\). The other options are false: \(f(0)=0\) and \(f(1)=2\) are distinct, and so are \(f(1)=2\) and \(f(-1)=-2\). Odd numbers missing from the image concerns surjectivity, not injectivity.
Review the concept: Injective function
Function \(f\colon A\to B\) in which distinct elements of \(A\) have distinct images in \(B\).
\[\forall a_1,a_2\in A\,\bigl(f(a_1)=f(a_2)\Rightarrow a_1=a_2\bigr)\]
More practice problems: injective function
- ZW-INJECTIVE-FUNCTION-001 What does it mean for fcolon A→ B to be injective?
- ZW-INJECTIVE-FUNCTION-002 Which of these functions from ℝ to ℝ is injective?
- ZW-INJECTIVE-FUNCTION-003 What is the logical form of the definition of an injective function?
- ZW-INJECTIVE-FUNCTION-004 If fcolon A→ B is not injective, what holds?
- ZW-INJECTIVE-FUNCTION-005 Can there be an injective function fcolon\1,2,3\→\a,b\?
- ZW-INJECTIVE-FUNCTION-006 If f is injective and a_1ne a_2, what happens to f(a_1) and f(a_2)?
- ZW-INJECTIVE-FUNCTION-008 If fcolon A→ B and gcolon B→ C are injective, is g∘ f?
- ZW-INJECTIVE-FUNCTION-009 Is every injective function also surjective?