TheoremProved

The degenerate case a=0

Statement

If $a=0$ , the equation $ax+b=0$ reduces to $b=0$ : it has infinitely many solutions (every real x works) when $b=0$ is a true statement, and no solution at all when $b\neq0$ .

Why is it true?

This shows the label "linear equation" secretly depends on the coefficient of x being nonzero — drop that, and the whole notion of "one unique answer" collapses into either every answer or no answer.

Proof sketch

Substitute $a=0$ directly into $ax+b=0$ : the term $ax$ becomes $0\cdot x=0$ for every real number $x$ , since any number times zero is zero. So the equation literally becomes $b=0$ , a statement about $b$ alone that no longer mentions $x$ at all.

Now there are exactly two possibilities for the fixed number $b$ . If $b=0$ , the leftover statement " $0=0$ " is true regardless of which $x$ we substituted — so every real number $x$ satisfies the original equation, giving infinitely many solutions. If instead $b\neq0$ , the leftover statement " $b=0$ " is simply false — no value of $x$ can make a false numerical statement true, so the equation has no solution whatsoever, no matter what $x$ we try.

Topics that use this theorem

Linear equations

Step-by-step proofs

No step-by-step proof yet for this theorem.