# Why are these logical statements not deemed to be equivalent?

I'm working through a book on my own which has just introduced the ideas of $A \Rightarrow B, B \Leftarrow A$ and $A \Leftrightarrow B$. It then gave 20 exercise questions to answer. I've correctly answered all of them except for two, and I don't understand why my answers are incorrect. The two questions are as follows:

Insert the symbol $\Rightarrow$, $\Leftarrow$ or $\Leftrightarrow$ which fully represents the link between the two statements.

Question 1:

Statement A: $\frac{x}{x + 1} = 0$

Statement B: $x = 0$

My answer: $A \Leftrightarrow B$

Book's answer: $A \Rightarrow B$

For this one, it seems simple enough to see that $A \Rightarrow B$, but I don't understand why $A \Leftarrow B$ is false. It seems no different to expressing $x = 0$ as something like $x + 1 - 1 = 0$. Wouldn't $x = 0$ imply $x + 1 - 1 = 0$?

Question 2:

Statement A: $(a, b)$ is a point on the line $y = 2x-1$

Statement B: $b = 2a - 1$

My answer: $A \Leftrightarrow B$

Book's answer: $A \Rightarrow B$

Similar situation with this one. If $(a, b)$ is a point on the line $y = 2x - 1$, that should imply $b = 2a - 1$. Then surely, if $b = 2a - 1$, $(a, b)$ is going to be a point on the line $y = 2x - 1$. So, again, I don't understand why $A \Leftarrow B$ is false.

Is my logic flawed?

• I agree with you. The book seems to be wrong. About these answers, at least. – ajotatxe Oct 11 '14 at 13:54
• @ajotatxe Thanks for the comment. I'm inclined to agree, but I'd like some more feedback as the answer below has left me uncertain. – John H Oct 11 '14 at 19:03

In the second question. Clearly $A \implies B$. But if you have a point $(a,b)$ such that $b = 2a -1$, it is not sufficient to conclude $y = 2x - 1$. Think about the line $y = -5x -1$. The point $(0, -1)$ is a inteserction between them, but only from either $-1 = 2 \times 0 - 1$ or $-1 = -5 \times 0 - 1$ you cannot conclude the line equation. Remember, in geometry to evaluate the line equation you need two points.
In your first question. The rational $a/b$ provide a aditional information: $b \ne 0$. So you know $x + 1 \ne 0$, because of this you can multiply the both sides by $x + 1$ and so conclude $A \implies B$. But from $x = 0$ you only know there is a variable $x$ which can be equal to $0$. Think about "If $x^2 = 4$, then $x = 2$", it can be false if $x$ is equal to $-2$. Note, a variable is a unkonwn value. Maybe you need to find if the book use the concepts $:=$ (definition) and $=$ (equality evaluation) with the same symbol $=$. The first, $:=$, is useful to define a value, e.g., $x := 1$ means $x$ is an alias for $1$. By other hand, $=$ is a evaluation of truth, e.g., $x = 1$ means "$x$ is equal to $1$?".