'Advanced' Functions and Mathematical Logic Help

[boyshiny]

Hello Mathhelp Community.

I'm new here. I was attempting some exercises from a previous assignment which are similar to questions in an upcoming assignment.
Please excuse me if I've posted this in the wrong place. I didn't know if this was calculus or Algebra. The class is called Algebra and Discrete maths, so I thought "Heck, algebra is algebra right?"

Anyway. I have a few equations I'm having trouble with. I'm hoping that getting help with one and learning how to do it will mean I can complete the others.
There are 3 questions that I will present which I'm having trouble with, and I hope that it's not an issue to ask for help

The first is regarding Functions.
Prove that the function f : R − {2} → R − {5} defined by f (x) = 5x+1/x−2 is a bijection.

I was able to complete the other functions questions, but completely stumped with this. In all honesty, I haven't been able to even begin this one. The function itself is confusing. R - {2} - R - {5}?!?
With some clarity of that function, I should be fine, but some tips would be helpful

The next two questions are Logic.
(p → (q → r)) and (q → (p → r)) are logically equivalent.
I have to use a Truth Table to do this.
I am under the presumption that I fill out a truth table with both equations to prove that they are equivalent?

The final question is:
"Use laws of logic (algebraic version) to show the following equivalences (clearly indicate which law you use in each step):"
((p → r) ∨ (q → r)) ≡ ((p ∧ q) → r).

From my understanding, ≡ is what is used in modulo equations, and using logic with modulo makes zero sense to me. The second thing I'm having a problem with is "which law you use in each step".

Any insight?

 

[pka]

The first is regarding Functions.
Prove that the function f : R − {2} → R − {5} defined by f (x) = 5x+1/x−2 is a bijection.

Suppose that \(\displaystyle 2 \notin \left\{ {a,b} \right\}\) then you must prove:
if \(\displaystyle 2 \notin \{a,b\} \) then \(\displaystyle f(a)=f(b)\text{ implies }a=b`.\)( ONE-TO-ONE)

show that \(\displaystyle h\in R\setminus\{5\}\text{ implies }\exists x\in R\setminus\{2\}\) such that \(\displaystyle f(x)=h\). (ONTO)

 

[HallsofIvy]

Hello Mathhelp Community.

I'm new here. I was attempting some exercises from a previous assignment which are similar to questions in an upcoming assignment.
Please excuse me if I've posted this in the wrong place. I didn't know if this was calculus or Algebra. The class is called Algebra and Discrete maths, so I thought "Heck, algebra is algebra right?"

Anyway. I have a few equations I'm having trouble with. I'm hoping that getting help with one and learning how to do it will mean I can complete the others.
There are 3 questions that I will present which I'm having trouble with, and I hope that it's not an issue to ask for help

The first is regarding Functions.
Prove that the function f : R − {2} → R − {5} defined by f (x) = 5x+1/x−2 is a bijection.

I was able to complete the other functions questions, but completely stumped with this. In all honesty, I haven't been able to even begin this one. The function itself is confusing. R - {2} - R - {5}?!?

This simply says that the domain of f is "all real numbers except 2". You can see that by noting that the denominator in f(x)= (5x+ 1)/(x- 2) would be 0 when x= 2 and division by 0 is not allowed. (PLEASE use parentheses! What you wrote would, strictly, be interpreted as "f(x)= 5x+ (1/x)- 2" but that is clearly not what you intended.)
On the other hand, if you write y= (5x+ 1)/(x- 2) and solve for x, you will get x= (2y+ 1)/(y- 5)- and now it is clear that y= 5 would not correspond to any value of x. The "range" or "co-domain" is "all real numbers except 5" which is exactly what "R- {5}" means.

Again, "f: R- {2} → R- {5}" just means that x can be any real number except 2 and the value of the function can be any value except 5.

With some clarity of that function, I should be fine, but some tips would be helpful

To prove "f is a bijection", you must prove two things:
f is an "injection" (one to one)- if f(a)= f(b) the a= b (two different values for x cannot give the same value for f(x)). To prove that you need to start from f(x)= (5a+ 1)/(a- 2)= (5b+ 1)/(b- 2)= f(b) and, algebraically, conclude that a= b.

f is a "surjection" (onto)- if y is in R- {5} (y is any real number except 5) there exist a real number, x, in R- {2}, such that y= (5x+ 1)/(x- 2). The simplest way to do that is to actually solve y= (5x+1)/(x- 2) for x.

The next two questions are Logic.
(p → (q → r)) and (q → (p → r)) are logically equivalent.
I have to use a Truth Table to do this.
I am under the presumption that I fill out a truth table with both equations to prove that they are equivalent?

Yes, that is the case. I would start with two columns, labeled "p" and "q" with the four "T" "F" combinations then, in one table, columns for "q → p" and "p → (q → r)" and, in the other table, columns for "p → r" and "q → (p → r)".

The final question is:
"Use laws of logic (algebraic version) to show the following equivalences (clearly indicate which law you use in each step):"
((p → r) ∨ (q → r)) ≡ ((p ∧ q) → r).

From my understanding, ≡ is what is used in modulo equations, and using logic with modulo makes zero sense to me. The second thing I'm having a problem with is "which law you use in each step".

Any insight?

Various mathematical symbols, such as ≡, can be used in a variety of different ways. No, this NOT "modulo". It is "the two sides are equivalent". As for "which law you use in each step", what laws do you know for symbolic logic? What I would do is look for a law that will allow me to change the left side into something a little more like the right side, then look for another that will change it to a little more like the right side, etc.
(And, since this is an equivalence, you will also have to show that the right side can be changed into the left side.)


(I am concerned about you attempting problems like this without knowing the basic symbols used. Surely your text book either has a paragraph at the beginning of the chapter explaining them, or, perhaps, tables at the beginning or end of thebook?)

 

[boyshiny]

This simply says that the domain of f is "all real numbers except 2". You can see that by noting that the denominator in f(x)= (5x+ 1)/(x- 2) would be 0 when x= 2 and division by 0 is not allowed. (PLEASE use parentheses! What you wrote would, strictly, be interpreted as "f(x)= 5x+ (1/x)- 2" but that is clearly not what you intended.)
On the other hand, if you write y= (5x+ 1)/(x- 2) and solve for x, you will get x= (2y+ 1)/(y- 5)- and now it is clear that y= 5 would not correspond to any value of x. The "range" or "co-domain" is "all real numbers except 5" which is exactly what "R- {5}" means.

Again, "f: R- {2} → R- {5}" just means that x can be any real number except 2 and the value of the function can be any value except 5.


To prove "f is a bijection", you must prove two things:
f is an "injection" (one to one)- if f(a)= f(b) the a= b (two different values for x cannot give the same value for f(x)). To prove that you need to start from f(x)= (5a+ 1)/(a- 2)= (5b+ 1)/(b- 2)= f(b) and, algebraically, conclude that a= b.

f is a "surjection" (onto)- if y is in R- {5} (y is any real number except 5) there exist a real number, x, in R- {2}, such that y= (5x+ 1)/(x- 2). The simplest way to do that is to actually solve y= (5x+1)/(x- 2) for x.


Yes, that is the case. I would start with two columns, labeled "p" and "q" with the four "T" "F" combinations then, in one table, columns for "q → p" and "p → (q → r)" and, in the other table, columns for "p → r" and "q → (p → r)".


Various mathematical symbols, such as ≡, can be used in a variety of different ways. No, this NOT "modulo". It is "the two sides are equivalent". As for "which law you use in each step", what laws do you know for symbolic logic? What I would do is look for a law that will allow me to change the left side into something a little more like the right side, then look for another that will change it to a little more like the right side, etc.
(And, since this is an equivalence, you will also have to show that the right side can be changed into the left side.)


(I am concerned about you attempting problems like this without knowing the basic symbols used. Surely your text book either has a paragraph at the beginning of the chapter explaining them, or, perhaps, tables at the beginning or end of thebook?)

Thank you so much for the explanation.

You'd think so, but my lecturer didn't give us a textbook. He just gave us lecture slides with a brief explanation. I ended up learning what I know through YouTube and Khan Academy, lecturer has taught me naught.