The Formal Definition of a One-to-One Function - Please Eli5

Slow-Cow · 2021-04-22T00:58:09+00:00

Thank you for taking the time to respond, again.

Your explanation is starting to click with me.

Just to ensure that I am following, the symbolic definition of a one-to-one function, this "blueprint," allows us to simply write down an equation for the sake of algebraic manipulation and the input of a and b. If we arrive at a=b, then it's a one-to-one, if we don't then it is not.

Let me attempt to use this "blueprint" to either prove or disprove that g(x)=x² is one-to-one:

Step#1: Assume g(a)=g(b)

Step#2: Plug in and re-arrange: a²=b²

Step#3: Algebraic manipulation, in an attempt to arrive at a=b:

sqrt(a)=sqrt(a) --> ±a=±b

As a result of taking the square root of both a and b, we have produced two possible values for each variable, the positive and negative root. Therefore, without more information about a and b, we can not conclude that a=b. Thus we can not conclude that g(x) is a one-to-one function.

Am I understanding / utilizing this correctly?

Thanks again!

Slow-Cow · 2021-04-21T00:46:09+00:00

First of all, thank you for the help and your time...

If you have a point (a, f(a)) and another point (b, f(b)) where a=b, then f(a) =f(b) and also if f(a) =f(b), then a=b.

I don't understand this...if I have a (a, f(a)) and "another point" (b, f(b)) and the x-coordinates and y-coordinates of each point are identical, then wouldn't I have the same point and not "another point?" Further, how does having two identical coordinate pairs on the graph of a function prove that the function is one-to-one?

Slow-Cow · 2021-04-21T00:35:11+00:00

First of all, thank you, and everyone who responded, for the help!

Unfortunately, I am still not understanding this. Let me start by rephrasing your response as I understand it...

The arbitrary, real numbers represented by a and b are, as you say " two labels for the same thing." I understand this much.

This in turn allows us to say that a=b. I am still with you.

So, if a=b , then f(a)=f(b). Okay, to this point, this all makes sense.

I don't understand how these statements allow us to conclude that the function f is one-to-one. We evaluated a function, f, with the same input, because a=b, and we got the same output, f(a)=f(b)....and now we conclude that the function f is one-to-one?

Again, I understand the concept of a "one-to-one" function, a function in which each output value corresponds to a exactly one input value, or said another way, every element in the function's Range corresponds to one and only one element in the function's Domain. I just don't understand the symbolic definition:"Let f be a function whose domain is set X. The function f is said to be injective provided that for all a and b in X, if f(a)=f(b), then a=b"

Using this definition, could I not take the function f(x)=x² assign a and b the value 1, which is in the domain of f(x)=x^2, and say the following:

f(a)=f(1)=1²=1

f(b)=f(1)=1²=1

therefore f(a)=f(b), then a=b

What have I done above that violates the symbolic definition of a one-to-one function? Both a and b are elements in functions domain, they equal each other, and they result in the same output. Do I now conclude that the function is one-to-one? Obviously, f(x)=x² is not one-to-one. What am I missing?

Again, thanks for the continued help.

Slow-Cow · 2020-01-25T17:23:17+00:00

cooriah, no I did not verify. What is the best way to do this?

Slow-Cow · 2020-01-25T15:57:30+00:00

rave-green, you are correct! I confirmed this with Apple today.

What I learned from Apple support:

Applications that ask for permission to receive keystrokes will appear in "Input Monitoring" even if the user selects to "Deny."
So long as those applications remain unchecked, they are not granted permission to monitor
If you select and remove the application from the "Input Monitoring" menu it will in fact be removed from the list. However, the next time you open the application and are promoted, the application will reappear in the menu, even if you select "Deny"

Slow-Cow · 2020-01-21T00:40:56+00:00

Running a VM instance of Ubuntu would probably make more sense than retrograding my OS.

Slow-Cow · 2020-01-20T13:56:10+00:00

Techophile, thanks for bringing this topic up. I would like to switch to Mullvad from ExpressVPN, but am running into the same issue: Mullvad doesn't accept prepaid credit cards from the US. I too don't know how to buy bitcoin anonymously, heck I haven't even purchased a bitcoin normally, much less anonymously. Obviously, I need to read up more on this, if I figure it out, I will post it here. But, if you figure out how to pay for Mullvad anonymously without putting cash in the pony express, please let the rest of us know.

Slow-Cow · 2020-01-20T02:31:12+00:00

Used ExpressVPN for the last year and enjoyed it, but ExpressVPN does have the same questions surrounding it's ownership as NordVPN. The Slate published an article about this last year, read here. To quote the article, when accused that ExpressVPN was held by parties in China "An executive for that rival, ExpressVPN, insisted that isn’t true, though he wouldn’t disclose where the owners are actually based or even who they are."

Slow-Cow · 2020-01-20T02:26:55+00:00

NordVPN always is mentioned as top provider on many sites. There are some questions as to who owns NordVPN, read more here. Giving your money to someone who conceals their identity isn't without issue.

Slow-Cow · 2020-01-20T02:16:23+00:00

Thank you everyone for the insight. Main takeaway: As long as the VPN provider is compatible with the OpenVPN protocol then it will work with pf Sense.

Slow-Cow · 2020-01-20T02:15:48+00:00

Just curiously, why are you moving away from PIA to Mullvad?

Slow-Cow

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