# Logic

#### JeffM

##### Elite Member
That is quite incoherent.

Letters are used to represent different things.

You particularly need to understand the difference between an unknown and a variable.

An unknown represents one from a finite set of numbers, each of which satisfies an equation.

$$\displaystyle x^2 - 9 = 0 \implies x = 3 \text { or } x = -\ 3.$$

So cannot be 100 or 19 or anything but one of those two numbers.

A variable stands for any number in a given set, which may be an infinite set, without any further limitation.

$$\displaystyle f(x) = \dfrac{1}{2x + 3}, \text { where } x \in \mathbb R \text { and } x \ne -\ 1.5.$$

We are saying there that the number specified by f(x) cannot be determined until one of the numbers in its domain is specified, any of the numbers in the domain can be specified.

$$\displaystyle a,\ b \in \mathbb R \implies a * b = b * a.$$

We are saying that no matter which two real numbers you specify, the order in which they are multiplied has no effect on the resulting product.

The symbols chosen to represent unknowns and variables are arbitrary. They have no implicit meaning except as defined for a particular purpose.

$$\displaystyle u,\ v \in \mathbb R \implies u * v = v * u$$ means exactly the same thing as $$\displaystyle a,\ b \in \mathbb R \implies a * b = b * a.$$

Finally, it was explained to you that x = y is true only under certain conditions. It is not in and of itself a true statement.

$$\displaystyle f(x) = \dfrac{1}{2x + 3} \text { and } f(x) = \dfrac{1}{6y + 3} \implies x = 3y.$$

$$\displaystyle f(x) \equiv \dfrac{1}{2x + 3} \implies f(x) \ne \dfrac{1}{3x + 3}.$$

It has been explained now to you that your questions are better posed in the context of specific texts or specific problems. As it is you jump around from idea to idea and confuse yourself.

##### Full Member
We have to guess a lot in your threads because your rambling-style of writing combined with poor English is very difficult to understand.

That's not the meaning here. Moving from f(x) to f(2x) means the input has changed. (It does not mean the inputs are equal.)

2x means the input has doubled. What does the output look like, when the input is doubled?

Given:

f(
INPUT ) = 1 ÷ [ 2 ∙ INPUT + 4 ]

Now, double the input.

f( 2 INPUT ) = 1 ÷ [ 2 ∙ 2 INPUT + 4 ]

When the input is doubled, the output is 1/(4 INPUT + 4)

If the input were halved, then the output would be 1/(INPUT + 4)

f(x/2) = 1/[(2)(x/2) + 4] = 1/[x + 4] Sorry I'm still confused !
but the first equation is given for input as "INPUT" and not as "2*INPUT" ! ..so how I assign 2*INPUT in the function at all ?! and why am I replacing 2*INPUT instead of INPUT ! really weird !

#### JeffM

##### Elite Member
$$\displaystyle f(x) = \dfrac{1}{2x + 4}$$

MEANS

that whatever is inside the parentheses after f is to replace x in the formula to the right of the equal sign. It is an instruction on how to complete a formula.

#### Otis

##### Senior Member
… how [can] I assign 2*INPUT in the function at all …
We may input any expression we like, as long as it represents a value in the function's domain.

f(√3) $$\quad$$ f(5z+4) $$\quad$$ f(cos(k∙π/7)) $$\quad$$ f(g(x))

… why am I [using] 2*INPUT instead of INPUT ! really weird !
I don't know why you started a thread about f(x) and f(2x). What were you looking at?

Given f(x), a textbook might ask students to find f(2x) just to practice symbolic reasoning. In that case, there's probably no real-world meaning. But, there are many situations in the real world where people would want to see what happens to a function's value, when its input is doubled.

Can you explain why you think doubling the input is really weird? #### Ryan$##### Full Member Is that book written in English? (Maybe that's part of your issue.) If you and I have exactly$1 in our pocket, and somebody else says to you, "I have exactly $1 in my pocket, just like you", then why think the$1 in my pocket would suddenly change. Magic? I'm totally with you, but in the second part who said that it didn't change?! that's my problem. if I didn't touch your pocket, then is it definitely still the same?!

#### Ryan$##### Full Member Is that book written in English? (Maybe that's part of your issue.) If you and I have exactly$1 in our pocket, and somebody else says to you, "I have exactly $1 in my pocket, just like you", then why think the$1 in my pocket would suddenly change. Magic? I'm not saying magic, because there's no magic in math. what I'm thinking that none tells that you didn't change what's in your pocket after a while (after we read and finish the second part of your sentence..maybe the first case(your pocket) would be changed ! because a time have passed since we have read your first statement), so how I know that's still $1 in your pocket?!! #### Denis ##### Senior Member I'm not saying magic, because there's no magic in math. what I'm thinking that none tells that you didn't change what's in your pocket after a while (after we read and finish the second part of your sentence..maybe the first case(your pocket) would be changed ! because a time have passed since we have read your first statement), so how I know that's still$1 in your pocket?!!
Ryan, if I was a moderator, I'd "close" such ridiculous threads....

##### Full Member
Ryan, if I was a moderator, I'd "close" such ridiculous threads....
and instead of just bothering, it could be more than appreciated and honor to help your OP no more nothing else, yeah it might be a silly question, but I face a problem with it which related to me it's magnificent problem, thanks alot

#### Otis

##### Senior Member
… who said that [given information] didn't change?! that's my problem …
… maybe the [amount of money in your pocket] changed ! because a time have passed since we have read your first statement …
If any of the dollar amounts (in my example) had changed, then you would have been told, already. So, don't worry about given information changing. Just accept the information in exercises as described.

In your exercise, it's given that x=y and y=z, so those relationships are fixed. Nothing about them is going to change, in that exercise.

If you worry that given information might change, then you won't be able to believe anything! An exercise statement will tell you, if anything changes.

In summary: Do not change given information. • Ryan$#### Ryan$

##### Full Member
If any of the dollar amounts (in my example) had changed, then you would have been told, already. So, don't worry about given information changing. Just accept the information in exercises as described.

In your exercise, it's given that x=y and y=z, so those relationships are fixed. Nothing about them is going to change, in that exercise.

If you worry that given information might change, then you won't be able to believe anything! An exercise statement will tell you, if anything changes.

In summary: Do not change given information. thanks alot ! convinced me much appreciated!

##### Full Member
HEY Ryan, if 3 + 2 = 5, then 2 + 3 = ? Hi , is that relevant to my question tho?

5

#### Denis

##### Senior Member
Thing is, we never know what YOUR questions are!

• • topsquark and MarkFL

#### Ryan\$

##### Full Member
If any of the dollar amounts (in my example) had changed, then you would have been told, already. So, don't worry about given information changing. Just accept the information in exercises as described.

In your exercise, it's given that x=y and y=z, so those relationships are fixed. Nothing about them is going to change, in that exercise.

If you worry that given information might change, then you won't be able to believe anything! An exercise statement will tell you, if anything changes.

In summary: Do not change given information. 