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Can fg ever equal gf?by Kakateo
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#1
Jan1314, 01:54 PM

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I was wondering that if told the functions for f(x) and g(x) are different, can fg ever equal gf?
My take on this was that they can never equal each other but some of my friends said they can sometimes equal each other because they plugged in x=0 for f(x) = 2x and g(x) = 3x. I was told that I couldn't put any random coordinates in. Another instance I just thought of is 1^x and 1^2x, as both will always equal 1, holding the equality statement true. However, wouldn't those two equations be the same as writing y=1 therefore having the same function? I'm not really sure so I ask for guidance please :) 


#2
Jan1314, 02:59 PM

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I'm not sure why you want to avoid doing algebra?
Let fg = gf Then f  g + f = g f + f = g + g 2f = 2g f = g So yes, they are the same if f  g = g  f, clearly. 


#3
Jan1314, 03:01 PM

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Hi Kakateo!
(try using the X^{2} button just above the Reply box ) For two functions to be equal, they have to be the same … I can't think of another way of putting it! If two functions of x are equal at (eg) x = 0, that doesn't make them equal. 


#4
Jan1314, 03:33 PM

P: 4

Can fg ever equal gf?
Thank you for the help :)
One last question: would fg=gf be always true, sometimes true, or never true? Once again, this is said with f(x) =/= g(x) 


#5
Jan1314, 03:38 PM

P: 1,287

Always true when f(x) = g(x) Again, simple algebra makes quick work of the question. Another way to do it: Let f  g = g  f Then f  g = (f  g) Then f  g = 0 [where 0 is the zero function] And so f = g 


#6
Jan1314, 03:55 PM

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Part of the difficulty here seems to me to be that the term "equal" is ambiguous, as is the symbol we use, =.
If we let f(x) = x and g(x) = 4  x, then f(x) = g(x) if x = 2, hence f(x)  g(x) = g(x)  f(x) = 0 when x = 2. "Equal" here is conditional, with the condition being that x = 2. OTOH, if "equal" means "identically equal" then f and g are not identically equal, as their graphs are different. In this case, f(x)  g(x) is not in general equal to either g(x)  f(x) or zero. 


#7
Jan1314, 03:59 PM

P: 1,287

That's why I like the notation ##f: X \rightarrow Y##. It's hard to interpret that as being some element of the range of f. It's clearly talking about the function. If we say ##f: X \rightarrow Y = g: X \rightarrow Y##, we all know what is meant. I would argue that the only legitimate interpretation of ##f(x) = g(x)## where ##f(x)## and ##g(x)## are to be viewed as functions, is that they are identically equal. Presenting interpretations where they are referring to elements of the range is just going to cause confusion, and is likely not the OP's question (because then there is no reason to phrase this question in terms of functions anyway). 


#8
Jan1314, 04:39 PM

P: 4

So then in this case would you be able to plug in 0 as a point and get fg = gf?



#9
Jan1314, 05:08 PM

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1. (x + 1)^{2} = x^{2} + 2x + 1 2. (x + 1)^{2} = 4 The first equation is identically true. The second equation is true only for x = 1 or x = 3. If you write f(x) = g(x) versus f(x) ##\equiv## g(x), it's clear what the difference is. 


#10
Jan1314, 05:54 PM

P: 1,287

The equals sign means the same thing in both cases. By saying "then f(x) = g(x) if x = 2" you are saying that "the range element f(x) is equal to the range element g(x) whenever x = 2," you are not saying that the functions ##f(x)## and ##g(x)## are equal whenever x = 2, you are saying that the functions map the domain element 2 to the same range element, 2. In other words, you are using f(x) and g(x) to refer to a specific member of the range corresponding to a specific member of the domain, x. They are not being used to represent the functions f and g in this usage. In other words, when I say "when x = 2, f(x) = 5" I am using the symbol "f(x)" to refer to an element of the range of the function f, I am not using it to refer to the function itself, I am not asserting that the function f is equal to the number 5. Similarly, if I say "when x = 2, f(x) = 5 and g(x) = 5, so f(x) = g(x) when x = 2" I am using f(x) and g(x) to refer to elements of the ranges of f and g respectively. I am not saying that the function f(x) is equal to the function g(x) whenever x = 2. The notation with the arrow provides clarity because ##f: X \rightarrow Y## never refers to some element of the codomain, it always means the function itself. ##f(x)## may mean both, depending on the context. Let ##f: X \rightarrow Y## be a function. Then for any ##x \in X, f(x) \in Y##. I am not saying that the function f itself is a member of its own codomain.. "##f(x)##" does not refer to the function ##f## in this usage. Have you ever seen anyone write something like: Let ##f(x)## be a function. Then for any ##x \in X, ( f: X \rightarrow Y ) \in Y##. Because I sure haven't. If I say: ##f: X \rightarrow Y = g: X \rightarrow Y## there is absolutely no way to interpret that as "conditional equality of two functions." It means the functions are equal, always. 


#11
Jan1314, 06:23 PM

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Everybody is assuming "function" means something similar to ##f:\mathbb R \to \mathbb R##.
In general, the OP's statement can be true. For example, take addition modulo 2, and f(x) = x, g(x) = 1x. f(x)g(x) and g(x)f(x) are both equal to 1, for all (i.e. both) values of x. 1MileCrash's proof in #2 fails, because here 1+1=0. 


#12
Jan1314, 06:26 PM

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#13
Jan1314, 06:38 PM

P: 1,287

(x + 1)2 = 4 to be a statement about two functions, it is a false statement, always. If you say "it is true whenever x = 1 or x = 3," it may be true statement, but it is not a true statement about the functions (x+1)2 and 4, it is a true statement about the numbers (1+1)2, (3 + 1)2, and 4. You could say "the functions are equal at those points" but it is never correct to write f(x) = g(x) with those regarded as functions even if that is the case. But I digress, and this becomes a debate about nothing in particular, because whether or not we perceive ambiguity in what f(x) means or what "=" means, clarification of one clarifies the other. My only point is that in the statement "f(x) = g(x)" I would immediately consider this to be a statement about functions themselves, and thus there is no ambiguity in =. 


#14
Jan1314, 07:06 PM

P: 1,287

f(x)  g(x) is defined to be the difference of f(x) and g(x) for any x. That means that ##f(x)  g(x) = (x (mod 2))  ((1 x) mod 2)## which is completely different from ##(x  (1  x)) (mod 2) = (2x  1) (mod 2)## With your f(x) and g(x), I assert that ##(f  g)(3) = 1## and ##(g  f)(3) = 1## and so f  g =/= g  f You have defined your functions themselves to take addition mod 2, that does not impose addition mod 2 on the subtraction operation of functions unless you are talking about something like ##(f  g) (mod 2)## which is not ##f  g##. 


#15
Jan1314, 08:18 PM

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Clearly the function f(x) = x^{2} + 1 is different from the function g(x) = 4, so the equation above could not possibly be interpreted as a statement that is identically true. However, as a conditional equation, it IS a true statement for the values of x that I listed. A final example comes from the technique of partial fractions, in which, for example, the goal is to write x/[(x + 2)(x + 3)] in the form A/(x + 2) + B/(x + 3). After multiplying both sides by (x + 2)(x + 3), obtaining x = A(x + 3) + B(x + 2). This equation is really an identity, meaning that it holds for all values of x, possibly except for x = 3 and x = 2. If this were written as x ##\equiv## A(x + 3) + B(x + 2), it might reinforce the idea that this is an identity rather than a conditional equation. 


#16
Jan1314, 08:52 PM

P: 1,287

I'm only saying that considering its truth value for particular values of x means that I am no longer evaluating whether or not two functions are equal by the definition of equal functions. I know, that you know, what it means for functions to be equal. I'm just pointing it out as a reason for my opinion. Equality of numbers means precisely one thing. Equality of functions means precisely one thing. Equality of vectors means precisely one thing. Equality of sets means precisely one thing. . . . That's why I remarked that in this case, the ambiguity of ##f(x)## leads to an ambiguity of '=.' There isn't any ambiguity in ##f: X \rightarrow Y = g: X \rightarrow Y##. I'm sorry if it sounds like I am being pedantic, I wasn't really expecting a followup debate or response about ##f(x)## being ambiguous, or having to defend why I said that. 


#17
Jan1414, 12:32 AM

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My example from post #9: Since the OP hasn't chimed in for a while, I'm going to close this thread. 


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