Mixing
How could possibly that two functions are equal to each other?
$begingroup$
We're given such a function
$$f(x) = f(x+5)$$
$$f(x) = 3x+2$$
And asked to compute
$$f(12)+f(18)-f(9)$$
I thought I could directly evaluate the required values using the second equation. In such case, I suspect why we're given the first equation if we're not supposed to use it. Below is the relationship I noticed
$$Q: f(12) = f(17) equiv 0 implies f(12) neq f(17)$$
However, two functions cannot be equal to each other without knowing more about it. Hence, that is false. Could you please assist me with understanding the fact?
Regards
functions
asked Jan 4 at 18:39
EnzoEnzo
1617
$endgroup$
add a comment |
$begingroup$
We're given such a function
$$f(x) = f(x+5)$$
$$f(x) = 3x+2$$
And asked to compute
$$f(12)+f(18)-f(9)$$
I thought I could directly evaluate the required values using the second equation. In such case, I suspect why we're given the first equation if we're not supposed to use it. Below is the relationship I noticed
$$Q: f(12) = f(17) equiv 0 implies f(12) neq f(17)$$
However, two functions cannot be equal to each other without knowing more about it. Hence, that is false. Could you please assist me with understanding the fact?
Regards
functions
asked Jan 4 at 18:39
EnzoEnzo
1617
$endgroup$
9
$begingroup$
The two functions you gave are not equal. The first is periodic with period $5$, while the second one is monotonic increasing.
$endgroup$
– Tyler6
Jan 4 at 18:42
2
$begingroup$
Using your first function definition in your second one gives that $fleft(xright) = 3x + 2 = fleft(x + 5right) = 3left(x + 5right) + 2 = 3x + 17$, but this gives the ridiculous conclusion that $15 = 0$! Thus, as pointed out by Tyler6, they can't be equal.
$endgroup$
– John Omielan
Jan 4 at 18:44
$begingroup$
@Tyler6 But my teacher calculated something using modular arithmetics.
$endgroup$
– Enzo
Jan 4 at 18:47
add a comment |
$begingroup$
We're given such a function
$$f(x) = f(x+5)$$
$$f(x) = 3x+2$$
And asked to compute
$$f(12)+f(18)-f(9)$$
I thought I could directly evaluate the required values using the second equation. In such case, I suspect why we're given the first equation if we're not supposed to use it. Below is the relationship I noticed
$$Q: f(12) = f(17) equiv 0 implies f(12) neq f(17)$$
However, two functions cannot be equal to each other without knowing more about it. Hence, that is false. Could you please assist me with understanding the fact?
Regards
functions
asked Jan 4 at 18:39
EnzoEnzo
1617
$endgroup$
We're given such a function
$$f(x) = f(x+5)$$
$$f(x) = 3x+2$$
And asked to compute
$$f(12)+f(18)-f(9)$$
I thought I could directly evaluate the required values using the second equation. In such case, I suspect why we're given the first equation if we're not supposed to use it. Below is the relationship I noticed
$$Q: f(12) = f(17) equiv 0 implies f(12) neq f(17)$$
However, two functions cannot be equal to each other without knowing more about it. Hence, that is false. Could you please assist me with understanding the fact?
Regards
functions
functions
asked Jan 4 at 18:39
EnzoEnzo
1617
asked Jan 4 at 18:39
EnzoEnzo
1617
edited Jan 4 at 18:48
Enzo
asked Jan 4 at 18:39
EnzoEnzo
1617
asked Jan 4 at 18:39
EnzoEnzo
1617
asked Jan 4 at 18:39
EnzoEnzo
1617
1617
9
$begingroup$
The two functions you gave are not equal. The first is periodic with period $5$, while the second one is monotonic increasing.
$endgroup$
– Tyler6
Jan 4 at 18:42
2
$begingroup$
Using your first function definition in your second one gives that $fleft(xright) = 3x + 2 = fleft(x + 5right) = 3left(x + 5right) + 2 = 3x + 17$, but this gives the ridiculous conclusion that $15 = 0$! Thus, as pointed out by Tyler6, they can't be equal.
$endgroup$
– John Omielan
Jan 4 at 18:44
$begingroup$
@Tyler6 But my teacher calculated something using modular arithmetics.
$endgroup$
– Enzo
Jan 4 at 18:47
add a comment |
9
$begingroup$
The two functions you gave are not equal. The first is periodic with period $5$, while the second one is monotonic increasing.
$endgroup$
– Tyler6
Jan 4 at 18:42
2
$begingroup$
Using your first function definition in your second one gives that $fleft(xright) = 3x + 2 = fleft(x + 5right) = 3left(x + 5right) + 2 = 3x + 17$, but this gives the ridiculous conclusion that $15 = 0$! Thus, as pointed out by Tyler6, they can't be equal.
$endgroup$
– John Omielan
Jan 4 at 18:44
$begingroup$
@Tyler6 But my teacher calculated something using modular arithmetics.
$endgroup$
– Enzo
Jan 4 at 18:47
9
9
$begingroup$
The two functions you gave are not equal. The first is periodic with period $5$, while the second one is monotonic increasing.
$endgroup$
– Tyler6
Jan 4 at 18:42
$begingroup$
The two functions you gave are not equal. The first is periodic with period $5$, while the second one is monotonic increasing.
$endgroup$
– Tyler6
Jan 4 at 18:42
2
2
$begingroup$
Using your first function definition in your second one gives that $fleft(xright) = 3x + 2 = fleft(x + 5right) = 3left(x + 5right) + 2 = 3x + 17$, but this gives the ridiculous conclusion that $15 = 0$! Thus, as pointed out by Tyler6, they can't be equal.
$endgroup$
– John Omielan
Jan 4 at 18:44
$begingroup$
Using your first function definition in your second one gives that $fleft(xright) = 3x + 2 = fleft(x + 5right) = 3left(x + 5right) + 2 = 3x + 17$, but this gives the ridiculous conclusion that $15 = 0$! Thus, as pointed out by Tyler6, they can't be equal.
$endgroup$
– John Omielan
Jan 4 at 18:44
$begingroup$
@Tyler6 But my teacher calculated something using modular arithmetics.
$endgroup$
– Enzo
Jan 4 at 18:47
$begingroup$
@Tyler6 But my teacher calculated something using modular arithmetics.
$endgroup$
– Enzo
Jan 4 at 18:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
My thought was your function is 3x+2 with a period of 5. So that you can take any interval of length 5 and translate that graph.This gives us a piecewise continuous function. Note the graph will look different depending on which length 5 segment we start with. As an example the graph of f(x) would look like 3x+2 on [0,5) and we would have copies of this open line segment from [5n,5(n+1)).
So we could subtract enough multiples from our x values until they lie in [0,5). So f(12) = f(2) = 8, f(18)=f(3)=11, and f(9)=f(4)=14.
Thus 8+11-14=5.
answered Jan 4 at 19:07
Joel PereiraJoel Pereira
73519
$endgroup$
add a comment |
$begingroup$
It seems you are working in a modulo class from your line portion of
$$fleft(12right) = fleft(17right) equiv 0$$
You also stated in a later comment that this is the case. As such, it seems from the first equation of $fleft(xright) = fleft(x + 5right)$ that it is modulo $5$, i.e., a system where $2$ numbers can be "equal" (often referred to as being "equivalent" to help avoid confusion, plus this is why the "$equiv$" symbol is used instead of "=") to each other if they both have the same remainder when divided by $5$. Thus, $fleft(xright) = 3x + 2$ does work as $fleft(x + 5right) = 3x + 17 equiv 3x + 2 pmod 5$. You can just plug the values into to the second equation to determine the result. Thus, we have that
$$fleft(12right) + fleft(18right) - fleft(9right) = 38 + 56 - 29 = 65 equiv 0 pmod 5$$
You could also have simplified the values somewhat as well. In particular, in modulo arithmetic, we could also have used the first equation more directly. You could note that
$$fleft(12right) + fleft(18right) - fleft(9right) equiv fleft(2right) + fleft(3right) - fleft(4right) = 8 + 11 - 14 = 5 equiv 0 pmod 5$$
As you can see, in modulo arithmetic, due to the periodic nature of it, the values repeat as only the remainder when divided by the modulus value is used, so you can often use a simpler set, which are the same modulo the value, to make a calculation. The particular example equation is relatively simple, but this concept can sometimes help a lot to simplify calculations in more complicated situations.
I believe your original statement of $fleft(xright) = fleft(x + 5right)$ would be better expressed as $fleft(xright) equiv fleft(x + 5right)$ as the values are not "equal" in the normal sense, but they are "equivalent" when you consider them modulo $5$, i.e., both sides always give the same remainder when divided by $5$. Also, I suspect this would make it less confusing to anybody looking at it.
Finally, note that the proper relation in your question is not that they are equivalent to $0$, but instead should be something like
$$fleft(12right) equiv fleft(17right) equiv 3$$
because $fleft(12right) = 38$ and $fleft(17right) = 53$, and while they are not "equal" in the regular sense, they are "equivalent" to each other, and with $3$, modulo $5$ because they all have the same remainder of $3$ when divided by $5$.
answered Jan 4 at 18:50
John OmielanJohn Omielan
1,61629
$endgroup$
add a comment |
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2 Answers
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$begingroup$
My thought was your function is 3x+2 with a period of 5. So that you can take any interval of length 5 and translate that graph.This gives us a piecewise continuous function. Note the graph will look different depending on which length 5 segment we start with. As an example the graph of f(x) would look like 3x+2 on [0,5) and we would have copies of this open line segment from [5n,5(n+1)).
So we could subtract enough multiples from our x values until they lie in [0,5). So f(12) = f(2) = 8, f(18)=f(3)=11, and f(9)=f(4)=14.
Thus 8+11-14=5.
answered Jan 4 at 19:07
Joel PereiraJoel Pereira
73519
$endgroup$
add a comment |
$begingroup$
My thought was your function is 3x+2 with a period of 5. So that you can take any interval of length 5 and translate that graph.This gives us a piecewise continuous function. Note the graph will look different depending on which length 5 segment we start with. As an example the graph of f(x) would look like 3x+2 on [0,5) and we would have copies of this open line segment from [5n,5(n+1)).
So we could subtract enough multiples from our x values until they lie in [0,5). So f(12) = f(2) = 8, f(18)=f(3)=11, and f(9)=f(4)=14.
Thus 8+11-14=5.
answered Jan 4 at 19:07
Joel PereiraJoel Pereira
73519
$endgroup$
add a comment |
$begingroup$
My thought was your function is 3x+2 with a period of 5. So that you can take any interval of length 5 and translate that graph.This gives us a piecewise continuous function. Note the graph will look different depending on which length 5 segment we start with. As an example the graph of f(x) would look like 3x+2 on [0,5) and we would have copies of this open line segment from [5n,5(n+1)).
So we could subtract enough multiples from our x values until they lie in [0,5). So f(12) = f(2) = 8, f(18)=f(3)=11, and f(9)=f(4)=14.
Thus 8+11-14=5.
answered Jan 4 at 19:07
Joel PereiraJoel Pereira
73519
$endgroup$
My thought was your function is 3x+2 with a period of 5. So that you can take any interval of length 5 and translate that graph.This gives us a piecewise continuous function. Note the graph will look different depending on which length 5 segment we start with. As an example the graph of f(x) would look like 3x+2 on [0,5) and we would have copies of this open line segment from [5n,5(n+1)).
So we could subtract enough multiples from our x values until they lie in [0,5). So f(12) = f(2) = 8, f(18)=f(3)=11, and f(9)=f(4)=14.
Thus 8+11-14=5.
answered Jan 4 at 19:07
Joel PereiraJoel Pereira
73519
answered Jan 4 at 19:07
Joel PereiraJoel Pereira
73519
answered Jan 4 at 19:07
Joel PereiraJoel Pereira
73519
answered Jan 4 at 19:07
Joel PereiraJoel Pereira
73519
73519
add a comment |
add a comment |
$begingroup$
It seems you are working in a modulo class from your line portion of
$$fleft(12right) = fleft(17right) equiv 0$$
You also stated in a later comment that this is the case. As such, it seems from the first equation of $fleft(xright) = fleft(x + 5right)$ that it is modulo $5$, i.e., a system where $2$ numbers can be "equal" (often referred to as being "equivalent" to help avoid confusion, plus this is why the "$equiv$" symbol is used instead of "=") to each other if they both have the same remainder when divided by $5$. Thus, $fleft(xright) = 3x + 2$ does work as $fleft(x + 5right) = 3x + 17 equiv 3x + 2 pmod 5$. You can just plug the values into to the second equation to determine the result. Thus, we have that
$$fleft(12right) + fleft(18right) - fleft(9right) = 38 + 56 - 29 = 65 equiv 0 pmod 5$$
You could also have simplified the values somewhat as well. In particular, in modulo arithmetic, we could also have used the first equation more directly. You could note that
$$fleft(12right) + fleft(18right) - fleft(9right) equiv fleft(2right) + fleft(3right) - fleft(4right) = 8 + 11 - 14 = 5 equiv 0 pmod 5$$
As you can see, in modulo arithmetic, due to the periodic nature of it, the values repeat as only the remainder when divided by the modulus value is used, so you can often use a simpler set, which are the same modulo the value, to make a calculation. The particular example equation is relatively simple, but this concept can sometimes help a lot to simplify calculations in more complicated situations.
I believe your original statement of $fleft(xright) = fleft(x + 5right)$ would be better expressed as $fleft(xright) equiv fleft(x + 5right)$ as the values are not "equal" in the normal sense, but they are "equivalent" when you consider them modulo $5$, i.e., both sides always give the same remainder when divided by $5$. Also, I suspect this would make it less confusing to anybody looking at it.
Finally, note that the proper relation in your question is not that they are equivalent to $0$, but instead should be something like
$$fleft(12right) equiv fleft(17right) equiv 3$$
because $fleft(12right) = 38$ and $fleft(17right) = 53$, and while they are not "equal" in the regular sense, they are "equivalent" to each other, and with $3$, modulo $5$ because they all have the same remainder of $3$ when divided by $5$.
answered Jan 4 at 18:50
John OmielanJohn Omielan
1,61629
$endgroup$
add a comment |
$begingroup$
It seems you are working in a modulo class from your line portion of
$$fleft(12right) = fleft(17right) equiv 0$$
You also stated in a later comment that this is the case. As such, it seems from the first equation of $fleft(xright) = fleft(x + 5right)$ that it is modulo $5$, i.e., a system where $2$ numbers can be "equal" (often referred to as being "equivalent" to help avoid confusion, plus this is why the "$equiv$" symbol is used instead of "=") to each other if they both have the same remainder when divided by $5$. Thus, $fleft(xright) = 3x + 2$ does work as $fleft(x + 5right) = 3x + 17 equiv 3x + 2 pmod 5$. You can just plug the values into to the second equation to determine the result. Thus, we have that
$$fleft(12right) + fleft(18right) - fleft(9right) = 38 + 56 - 29 = 65 equiv 0 pmod 5$$
You could also have simplified the values somewhat as well. In particular, in modulo arithmetic, we could also have used the first equation more directly. You could note that
$$fleft(12right) + fleft(18right) - fleft(9right) equiv fleft(2right) + fleft(3right) - fleft(4right) = 8 + 11 - 14 = 5 equiv 0 pmod 5$$
As you can see, in modulo arithmetic, due to the periodic nature of it, the values repeat as only the remainder when divided by the modulus value is used, so you can often use a simpler set, which are the same modulo the value, to make a calculation. The particular example equation is relatively simple, but this concept can sometimes help a lot to simplify calculations in more complicated situations.
I believe your original statement of $fleft(xright) = fleft(x + 5right)$ would be better expressed as $fleft(xright) equiv fleft(x + 5right)$ as the values are not "equal" in the normal sense, but they are "equivalent" when you consider them modulo $5$, i.e., both sides always give the same remainder when divided by $5$. Also, I suspect this would make it less confusing to anybody looking at it.
Finally, note that the proper relation in your question is not that they are equivalent to $0$, but instead should be something like
$$fleft(12right) equiv fleft(17right) equiv 3$$
because $fleft(12right) = 38$ and $fleft(17right) = 53$, and while they are not "equal" in the regular sense, they are "equivalent" to each other, and with $3$, modulo $5$ because they all have the same remainder of $3$ when divided by $5$.
answered Jan 4 at 18:50
John OmielanJohn Omielan
1,61629
$endgroup$
add a comment |
$begingroup$
It seems you are working in a modulo class from your line portion of
$$fleft(12right) = fleft(17right) equiv 0$$
You also stated in a later comment that this is the case. As such, it seems from the first equation of $fleft(xright) = fleft(x + 5right)$ that it is modulo $5$, i.e., a system where $2$ numbers can be "equal" (often referred to as being "equivalent" to help avoid confusion, plus this is why the "$equiv$" symbol is used instead of "=") to each other if they both have the same remainder when divided by $5$. Thus, $fleft(xright) = 3x + 2$ does work as $fleft(x + 5right) = 3x + 17 equiv 3x + 2 pmod 5$. You can just plug the values into to the second equation to determine the result. Thus, we have that
$$fleft(12right) + fleft(18right) - fleft(9right) = 38 + 56 - 29 = 65 equiv 0 pmod 5$$
You could also have simplified the values somewhat as well. In particular, in modulo arithmetic, we could also have used the first equation more directly. You could note that
$$fleft(12right) + fleft(18right) - fleft(9right) equiv fleft(2right) + fleft(3right) - fleft(4right) = 8 + 11 - 14 = 5 equiv 0 pmod 5$$
As you can see, in modulo arithmetic, due to the periodic nature of it, the values repeat as only the remainder when divided by the modulus value is used, so you can often use a simpler set, which are the same modulo the value, to make a calculation. The particular example equation is relatively simple, but this concept can sometimes help a lot to simplify calculations in more complicated situations.
I believe your original statement of $fleft(xright) = fleft(x + 5right)$ would be better expressed as $fleft(xright) equiv fleft(x + 5right)$ as the values are not "equal" in the normal sense, but they are "equivalent" when you consider them modulo $5$, i.e., both sides always give the same remainder when divided by $5$. Also, I suspect this would make it less confusing to anybody looking at it.
Finally, note that the proper relation in your question is not that they are equivalent to $0$, but instead should be something like
$$fleft(12right) equiv fleft(17right) equiv 3$$
because $fleft(12right) = 38$ and $fleft(17right) = 53$, and while they are not "equal" in the regular sense, they are "equivalent" to each other, and with $3$, modulo $5$ because they all have the same remainder of $3$ when divided by $5$.
answered Jan 4 at 18:50
John OmielanJohn Omielan
1,61629
$endgroup$
It seems you are working in a modulo class from your line portion of
$$fleft(12right) = fleft(17right) equiv 0$$
You also stated in a later comment that this is the case. As such, it seems from the first equation of $fleft(xright) = fleft(x + 5right)$ that it is modulo $5$, i.e., a system where $2$ numbers can be "equal" (often referred to as being "equivalent" to help avoid confusion, plus this is why the "$equiv$" symbol is used instead of "=") to each other if they both have the same remainder when divided by $5$. Thus, $fleft(xright) = 3x + 2$ does work as $fleft(x + 5right) = 3x + 17 equiv 3x + 2 pmod 5$. You can just plug the values into to the second equation to determine the result. Thus, we have that
$$fleft(12right) + fleft(18right) - fleft(9right) = 38 + 56 - 29 = 65 equiv 0 pmod 5$$
You could also have simplified the values somewhat as well. In particular, in modulo arithmetic, we could also have used the first equation more directly. You could note that
$$fleft(12right) + fleft(18right) - fleft(9right) equiv fleft(2right) + fleft(3right) - fleft(4right) = 8 + 11 - 14 = 5 equiv 0 pmod 5$$
As you can see, in modulo arithmetic, due to the periodic nature of it, the values repeat as only the remainder when divided by the modulus value is used, so you can often use a simpler set, which are the same modulo the value, to make a calculation. The particular example equation is relatively simple, but this concept can sometimes help a lot to simplify calculations in more complicated situations.
I believe your original statement of $fleft(xright) = fleft(x + 5right)$ would be better expressed as $fleft(xright) equiv fleft(x + 5right)$ as the values are not "equal" in the normal sense, but they are "equivalent" when you consider them modulo $5$, i.e., both sides always give the same remainder when divided by $5$. Also, I suspect this would make it less confusing to anybody looking at it.
Finally, note that the proper relation in your question is not that they are equivalent to $0$, but instead should be something like
$$fleft(12right) equiv fleft(17right) equiv 3$$
because $fleft(12right) = 38$ and $fleft(17right) = 53$, and while they are not "equal" in the regular sense, they are "equivalent" to each other, and with $3$, modulo $5$ because they all have the same remainder of $3$ when divided by $5$.
answered Jan 4 at 18:50
John OmielanJohn Omielan
1,61629
edited Jan 4 at 20:14
answered Jan 4 at 18:50
John OmielanJohn Omielan
1,61629
answered Jan 4 at 18:50
John OmielanJohn Omielan
1,61629
answered Jan 4 at 18:50
John OmielanJohn Omielan
1,61629
1,61629
add a comment |
add a comment |
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$begingroup$
The two functions you gave are not equal. The first is periodic with period $5$, while the second one is monotonic increasing.
$endgroup$
– Tyler6
Jan 4 at 18:42
2
$begingroup$
Using your first function definition in your second one gives that $fleft(xright) = 3x + 2 = fleft(x + 5right) = 3left(x + 5right) + 2 = 3x + 17$, but this gives the ridiculous conclusion that $15 = 0$! Thus, as pointed out by Tyler6, they can't be equal.
$endgroup$
– John Omielan
Jan 4 at 18:44
$begingroup$
@Tyler6 But my teacher calculated something using modular arithmetics.
$endgroup$
– Enzo
Jan 4 at 18:47