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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?

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    $\begingroup$ I agree with you. The book seems to be wrong. About these answers, at least. $\endgroup$ – ajotatxe Oct 11 '14 at 13:54
  • $\begingroup$ @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. $\endgroup$ – John H Oct 11 '14 at 19:03
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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 $1$?".

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  • $\begingroup$ I initially thought this, but it's only asking whether or not the point is on the line. It's not asking whether the equation the point was generated by is the same line. $\endgroup$ – John H Oct 11 '14 at 15:17
  • $\begingroup$ Perhaps I should check out the book, about its approach. A book in with mathematical logic usually has many subtleties about the definition and objects used. Is it a book in mathematical logic? $\endgroup$ – Cristhian Gz Oct 11 '14 at 15:24
  • $\begingroup$ Sorry for the delayed response, and thanks for getting back to me. The book is Core Maths for Advanced Level, which is basically covering the core sections of the UK maths curriculum that 16-18 year olds would study, so it's by no means a book on pure logic. $\endgroup$ – John H Oct 11 '14 at 19:01

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