1. The problem asks which statement about the order relation $\leq$ on the natural numbers $\mathbb{N}$ is not true. 2. Recall the properties of the usual order $\leq$ on $\mathbb{N}$: - Reflexivity: For all $a \in \mathbb{N}$, $a \leq a$. - Transitivity: For all $a,b,c \in \mathbb{N}$, if $a \leq b$ and $b \leq c$, then $a \leq c$. - Antisymmetry: For all $a,b \in \mathbb{N}$, if $a \leq b$ and $b \leq a$, then $a = b$. - Well-ordering: Every non-empty subset of $\mathbb{N}$ has a least element. 3. Analyze each option: - Option A: "$\leq$ is reflexive" is true by definition. - Option B: "For all $a,b,c \in \mathbb{N}$, $a \leq b$ and $b \leq c$ implies $a \leq c$" is the transitivity property, which is true. - Option C: "For all $a,b \in \mathbb{N}$, $a \leq b$ and $b \leq a$ implies $a = b$" is antisymmetry, which is true. - Option D: "Every non-empty subset of $\mathbb{N}$ admits a largest element" is false because $\mathbb{N}$ is well-ordered by least element, but infinite subsets do not have a largest element. 4. Therefore, the statement that is not true is Option D. Final answer: Option D is not true concerning order in $\mathbb{N}$.