# Negation of converse of a theorem related to linear independence

• Theorem $$T:\quad$$ Let $$S = \{v_1, v_2,..., v_p\}$$ be a set of vectors in $$\mathbb R^m$$. If $$p>m$$, then this set is linearly dependent.

Theorem $$T$$ can be proved to be true.

• Converse $$T_c$$ of theorem $$T:\quad$$ If the set $$S$$ is linearly dependent then $$p > m$$.

The converse of theorem $$T$$ is not a true statement. Does it mean that its negation is true?

• Negation $$T_{nc}$$ of $$T_c:\quad$$ The set $$S$$ is linearly dependent and $$p \le m$$.

In simple words, what does $$T_{nc}$$ mean, and how to prove it?

• Does this answer your question? Is a counterexample considered a rigorous proof that a property is not true? The answers to that post can be considered as "abstract duplicates", so I shall delete my answer below, and just copy-paste it as a comment. Mar 27 at 7:16
• The "converse" of the theorem T: For all $S=(v_1,\dots,v_p),$ if $p>m$ then $S$ is dependent is the (wrong) statement Tc: For all $S=(v_1,\dots,v_p),$ if $S$ is dependent then $p>m.$ The negation of this "converse" is the (true) statement Tnc: There exists $S=(v_1,\dots,v_p)$ such that $S$ is dependent and $p\le m.$ A proof that Tc is false (or Tnc is true) is a "counterexample", i.e. an example of such an $S.$ E.g. $S=(0).$ Mar 27 at 7:20
• @AnneBauval The quantifiers introduced by you for the explanation is really valuable. Then the negation is more clear. Mar 27 at 9:16

The converse of theorem $$T$$ is not a true statement. Does it mean that its negation is true?

Negating a statement logically flips its truth value; this means that even if you abstractly redefine the phrases (e.g., 'linearly dependent') and non-logical symbols (e.g., the 'less than' symbol) within it, the statement and its negation continues to have opposite truth values. So, the negation of a false statement has to be true.

• Theorem $$T:\quad$$ Let $$S = \{v_1, v_2,..., v_p\}$$ be a set of vectors in $$\mathbb R^m$$. If $$p>m$$, then this set is linearly dependent.
• Converse $$T_c$$ of theorem $$T:\quad$$ If $$S$$ is linearly dependent then $$p > m$$.

You have not fully stated the converse $$T_c:$$ the part about $$S$$ being an arbitrary set of $$p$$ vectors in $$\mathbb R^m$$ is important, because in full: $$\forall p{\in}\mathbb Z_0^+\quad\forall m{\in}\mathbb Z^+\\\forall S{\in}\{\text{sets of p vectors in }\mathbb R^m\}\;(S\text{ is linearly dependent}\implies p>m).\tag1$$

• Negation $$T_{nc}$$ of $$T_c:\quad$$ $$S$$ is linearly dependent and $$p \le m$$.

Consequently, your negation is incomplete; compare it with the full negation $$\exists p{\in}\mathbb Z_0^+\quad\exists m{\in}\mathbb Z^+\\\exists S{\in}\{\text{sets of p vectors in }\mathbb R^m\}\;(S\text{ is linearly dependent and } p\le m).\tag{1n}$$

how to prove $$T_{nc}$$ ?

As suggested by Anne Bauval, just exhibit $$(p,m,S)=\left\{1,1,\{0\}\right\}.$$

In my second paragraph above, the point is that the "Let $$S = \{v_1, v_2,..., v_p\}$$ be a set of vectors in $$\mathbb R^m$$" part of theorem $$T$$ is understood to mean that $$S$$ is an arbitrary set of $$p$$ vectors in $$\mathbb R^m,$$ and that this stipulation is equivalent to the sequence of universal quantifiers that I made explicit in statement $$(1).$$