Interpreting a statement in first order logic

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The discussion revolves around rewriting two mathematical statements in symbolic form related to the solutions of the equation ax + b = 0. The proposed symbolic representations for both statements are presented, with the first indicating the existence of a solution and the second asserting the uniqueness of that solution. Participants suggest improvements, such as adding the condition y ∈ ℝ for clarity and mentioning the use of the symbol ∃! to denote uniqueness, despite it not being an official notation. There is a consensus that while the initial attempts are correct, they could be simplified. Overall, the conversation highlights the complexities and nuances of expressing logical statements in symbolic form.
Mr Davis 97
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Homework Statement


Rewrite the following statements in symbolic form:

a) If ##a## and ##b## are real numbers with ##a \ne 0##, then ##ax+b=0## has a solution.
b) If ##a## and ##b## are real numbers with ##a \ne 0##, then ##ax+b=0## has a unique solution.

Homework Equations

The Attempt at a Solution



Attempts at solution:

Let ##P(x,a,b)## be the statement that ##ax+b=0## is true.
a) ##\forall a \in \mathbb{R} - \{0\} \forall b \in \mathbb{R} \exists x \in \mathbb{R} P(x,a,b)##
b) ##\forall a \in \mathbb{R} - \{0\} \forall b \in \mathbb{R} \exists x \in \mathbb{R} (P(x,a,b) \wedge \forall y (P(y,a,b) \implies y=x))##

Is that at all right? Is there an easier way? It all seems very cumbersome.
 
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Mr Davis 97 said:

Homework Statement


Rewrite the following statements in symbolic form:

a) If ##a## and ##b## are real numbers with ##a \ne 0##, then ##ax+b=0## has a solution.
b) If ##a## and ##b## are real numbers with ##a \ne 0##, then ##ax+b=0## has a unique solution.

Homework Equations

The Attempt at a Solution



Attempts at solution:

Let ##P(x,a,b)## be the statement that ##ax+b=0## is true.
a) ##\forall a \in \mathbb{R} - \{0\} \forall b \in \mathbb{R} \exists x \in \mathbb{R} P(x,a,b)##
b) ##\forall a \in \mathbb{R} - \{0\} \forall b \in \mathbb{R} \exists x \in \mathbb{R} (P(x,a,b) \wedge \forall y (P(y,a,b) \implies y=x))##

Is that at all right? Is there an easier way? It all seems very cumbersome.
Looks good to me, although I'm no logic expert, and I would add an ##\in \mathbb{R}## to the ##y##.
To make it less cumbersome a professor of mine used ##\exists !## to express uniqueness, but this isn't an official symbol. Some even used a symbol for "without loss of generality".
 
fresh_42 said:
Looks good to me, although I'm no logic expert, and I would add an ##\in \mathbb{R}## to the ##y##.
To make it less cumbersome a professor of mine used ##\exists !## to express uniqueness, but this isn't an official symbol. Some even used a symbol for "without loss of generality".

Didn't know that this isn't an official symbol. I use it all the time.
 
Math_QED said:
Didn't know that this isn't an official symbol. I use it all the time.
I don't know for certain, I only checked it in Wikipedia. But it makes sense, as it combines two statements in one. It is convenient, but not pure in a logical sense.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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