Mixing
Proof that $2sqrt2 + sqrt7$ is an irrational number
$begingroup$
Prove that $2sqrt2 + sqrt7$ is an irrational number. I am trying to use contradiction to show that this is irrational. Also I am using the fact that $ 2sqrt2 + sqrt7 = frac{1}{2sqrt2 - sqrt7}$.
Assume $2sqrt2 + sqrt7 $ is rational. Then
$2sqrt2 + sqrt7 = frac{m}{n}$
and
$2sqrt2 - sqrt7 = frac{n}{m}$
if I add them together I get
$ 4sqrt2 = frac{m^2+n^2}{mn} $
then we can divide the 4 on both sides and label the new numerator and denominator as $ x , y$ then to prove $sqrt2$ is irrational is trivial. I am not sure if this is the correct way of proving this. Like why does it work to add it's reciprocal? Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?
irrational-numbers rationality-testing
edited Jan 27 at 5:32
Martin Sleziak
44.9k10122276
asked Oct 17 '15 at 21:47
ultrainstinctultrainstinct
2,1611725
$endgroup$
add a comment |
$begingroup$
Prove that $2sqrt2 + sqrt7$ is an irrational number. I am trying to use contradiction to show that this is irrational. Also I am using the fact that $ 2sqrt2 + sqrt7 = frac{1}{2sqrt2 - sqrt7}$.
Assume $2sqrt2 + sqrt7 $ is rational. Then
$2sqrt2 + sqrt7 = frac{m}{n}$
and
$2sqrt2 - sqrt7 = frac{n}{m}$
if I add them together I get
$ 4sqrt2 = frac{m^2+n^2}{mn} $
then we can divide the 4 on both sides and label the new numerator and denominator as $ x , y$ then to prove $sqrt2$ is irrational is trivial. I am not sure if this is the correct way of proving this. Like why does it work to add it's reciprocal? Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?
irrational-numbers rationality-testing
edited Jan 27 at 5:32
Martin Sleziak
44.9k10122276
asked Oct 17 '15 at 21:47
ultrainstinctultrainstinct
2,1611725
$endgroup$
add a comment |
$begingroup$
Prove that $2sqrt2 + sqrt7$ is an irrational number. I am trying to use contradiction to show that this is irrational. Also I am using the fact that $ 2sqrt2 + sqrt7 = frac{1}{2sqrt2 - sqrt7}$.
Assume $2sqrt2 + sqrt7 $ is rational. Then
$2sqrt2 + sqrt7 = frac{m}{n}$
and
$2sqrt2 - sqrt7 = frac{n}{m}$
if I add them together I get
$ 4sqrt2 = frac{m^2+n^2}{mn} $
then we can divide the 4 on both sides and label the new numerator and denominator as $ x , y$ then to prove $sqrt2$ is irrational is trivial. I am not sure if this is the correct way of proving this. Like why does it work to add it's reciprocal? Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?
irrational-numbers rationality-testing
edited Jan 27 at 5:32
Martin Sleziak
44.9k10122276
asked Oct 17 '15 at 21:47
ultrainstinctultrainstinct
2,1611725
$endgroup$
Prove that $2sqrt2 + sqrt7$ is an irrational number. I am trying to use contradiction to show that this is irrational. Also I am using the fact that $ 2sqrt2 + sqrt7 = frac{1}{2sqrt2 - sqrt7}$.
Assume $2sqrt2 + sqrt7 $ is rational. Then
$2sqrt2 + sqrt7 = frac{m}{n}$
and
$2sqrt2 - sqrt7 = frac{n}{m}$
if I add them together I get
$ 4sqrt2 = frac{m^2+n^2}{mn} $
then we can divide the 4 on both sides and label the new numerator and denominator as $ x , y$ then to prove $sqrt2$ is irrational is trivial. I am not sure if this is the correct way of proving this. Like why does it work to add it's reciprocal? Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?
irrational-numbers rationality-testing
irrational-numbers rationality-testing
edited Jan 27 at 5:32
Martin Sleziak
44.9k10122276
asked Oct 17 '15 at 21:47
ultrainstinctultrainstinct
2,1611725
edited Jan 27 at 5:32
Martin Sleziak
44.9k10122276
asked Oct 17 '15 at 21:47
ultrainstinctultrainstinct
2,1611725
edited Jan 27 at 5:32
Martin Sleziak
44.9k10122276
edited Jan 27 at 5:32
Martin Sleziak
44.9k10122276
edited Jan 27 at 5:32
Martin Sleziak
44.9k10122276
44.9k10122276
asked Oct 17 '15 at 21:47
ultrainstinctultrainstinct
2,1611725
asked Oct 17 '15 at 21:47
ultrainstinctultrainstinct
2,1611725
asked Oct 17 '15 at 21:47
ultrainstinctultrainstinct
2,1611725
2,1611725
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The proof works because if your number is a rational $frac mn$, then its reciprocal, $frac nm$ is also rational: it is the quotient of two integers. (The only way that it wouldn't be rational is in the case $m=0$, but that would imply that $2sqrt2 = -sqrt7$, which is clearly false.)
Similarly, when you add $frac mn$ to $frac nm$, you get another rational number, $frac{m^2+n^2}{mn}$.
answered Oct 17 '15 at 22:23
ThéophileThéophile
20.3k13047
$endgroup$
$begingroup$
In short, the answer to the last two questions of your post ("Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?") is yes.
$endgroup$
– Théophile
Oct 17 '15 at 22:26
add a comment |
$begingroup$
It is pure luck that $2sqrt2 + sqrt7 = frac{1}{2sqrt2 - sqrt7}$, down to the fact that $(2sqrt 2)^2-(sqrt 7)^2 = 1$.
The following approach is more genral:
Suppose $2sqrt2 + sqrt7$ is rational. Then its square, $15+4sqrt {14}$, is still rational. So $sqrt{14}$ must be rational. Which it isn't.
edited Oct 17 '15 at 23:11
Akiva Weinberger
13.9k12268
answered Oct 17 '15 at 22:23
TonyKTonyK
43.7k358136
$endgroup$
add a comment |
$begingroup$
Here's another proof more general that doesn't rely on friendly symmetry. $sqrt 2$ and $sqrt 7$ are easy to show to be irrational. So for $2sqrt 2 + sqrt 7 = r$ then $2sqrt 2 - r = -sqrt 2$ so $8 + r^2 + 2sqrt 2 = 7$ therefore $sqrt 2$ is rational.
It's the same idea, really. you just don't need to rely on the not so likely fact that their conjugates are reciprical.
answered Oct 18 '15 at 4:44
fleabloodfleablood
73.1k22789
$endgroup$
$begingroup$
I have noticed the following mistakes in your answer. $2sqrt(2) - r$ is $-sqrt(7)$ and not $-sqrt(2)$. Also $8+r^2 + 2sqrt(2)$ is $23+2sqrt(2)+4sqrt(14)$ and not $7$.
$endgroup$
– Bysshed
Apr 2 '16 at 22:58
add a comment |
$begingroup$
From
$$2sqrt2 + sqrt7 = frac{n}{m}$$ and $$2sqrt2 - sqrt7 = frac{m}{n},$$ by addition, you get
$$4sqrt{2}=frac{m}{n}+frac{n}{m}.$$ That is, $sqrt{2}$ should be rational.
${}{}{}{}$
edited Apr 2 '16 at 23:31
Bysshed
1,06311023
answered Oct 17 '15 at 22:26
mflmfl
26.9k12142
$endgroup$
$begingroup$
I think that, on the second equailty, it should go $m/n$ instead of $n/m$. Same in the last part.
$endgroup$
– Miguelgondu
Oct 17 '15 at 23:39
$begingroup$
@Miguelgondu Typo corrected. Thank you for noticing it.
$endgroup$
– mfl
Oct 18 '15 at 4:01
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The proof works because if your number is a rational $frac mn$, then its reciprocal, $frac nm$ is also rational: it is the quotient of two integers. (The only way that it wouldn't be rational is in the case $m=0$, but that would imply that $2sqrt2 = -sqrt7$, which is clearly false.)
Similarly, when you add $frac mn$ to $frac nm$, you get another rational number, $frac{m^2+n^2}{mn}$.
answered Oct 17 '15 at 22:23
ThéophileThéophile
20.3k13047
$endgroup$
$begingroup$
In short, the answer to the last two questions of your post ("Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?") is yes.
$endgroup$
– Théophile
Oct 17 '15 at 22:26
add a comment |
$begingroup$
The proof works because if your number is a rational $frac mn$, then its reciprocal, $frac nm$ is also rational: it is the quotient of two integers. (The only way that it wouldn't be rational is in the case $m=0$, but that would imply that $2sqrt2 = -sqrt7$, which is clearly false.)
Similarly, when you add $frac mn$ to $frac nm$, you get another rational number, $frac{m^2+n^2}{mn}$.
answered Oct 17 '15 at 22:23
ThéophileThéophile
20.3k13047
$endgroup$
$begingroup$
In short, the answer to the last two questions of your post ("Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?") is yes.
$endgroup$
– Théophile
Oct 17 '15 at 22:26
add a comment |
$begingroup$
The proof works because if your number is a rational $frac mn$, then its reciprocal, $frac nm$ is also rational: it is the quotient of two integers. (The only way that it wouldn't be rational is in the case $m=0$, but that would imply that $2sqrt2 = -sqrt7$, which is clearly false.)
Similarly, when you add $frac mn$ to $frac nm$, you get another rational number, $frac{m^2+n^2}{mn}$.
answered Oct 17 '15 at 22:23
ThéophileThéophile
20.3k13047
$endgroup$
The proof works because if your number is a rational $frac mn$, then its reciprocal, $frac nm$ is also rational: it is the quotient of two integers. (The only way that it wouldn't be rational is in the case $m=0$, but that would imply that $2sqrt2 = -sqrt7$, which is clearly false.)
Similarly, when you add $frac mn$ to $frac nm$, you get another rational number, $frac{m^2+n^2}{mn}$.
answered Oct 17 '15 at 22:23
ThéophileThéophile
20.3k13047
answered Oct 17 '15 at 22:23
ThéophileThéophile
20.3k13047
answered Oct 17 '15 at 22:23
ThéophileThéophile
20.3k13047
answered Oct 17 '15 at 22:23
ThéophileThéophile
20.3k13047
20.3k13047
$begingroup$
In short, the answer to the last two questions of your post ("Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?") is yes.
$endgroup$
– Théophile
Oct 17 '15 at 22:26
add a comment |
$begingroup$
In short, the answer to the last two questions of your post ("Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?") is yes.
$endgroup$
– Théophile
Oct 17 '15 at 22:26
$begingroup$
In short, the answer to the last two questions of your post ("Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?") is yes.
$endgroup$
– Théophile
Oct 17 '15 at 22:26
$begingroup$
In short, the answer to the last two questions of your post ("Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?") is yes.
$endgroup$
– Théophile
Oct 17 '15 at 22:26
add a comment |
$begingroup$
It is pure luck that $2sqrt2 + sqrt7 = frac{1}{2sqrt2 - sqrt7}$, down to the fact that $(2sqrt 2)^2-(sqrt 7)^2 = 1$.
The following approach is more genral:
Suppose $2sqrt2 + sqrt7$ is rational. Then its square, $15+4sqrt {14}$, is still rational. So $sqrt{14}$ must be rational. Which it isn't.
edited Oct 17 '15 at 23:11
Akiva Weinberger
13.9k12268
answered Oct 17 '15 at 22:23
TonyKTonyK
43.7k358136
$endgroup$
add a comment |
$begingroup$
It is pure luck that $2sqrt2 + sqrt7 = frac{1}{2sqrt2 - sqrt7}$, down to the fact that $(2sqrt 2)^2-(sqrt 7)^2 = 1$.
The following approach is more genral:
Suppose $2sqrt2 + sqrt7$ is rational. Then its square, $15+4sqrt {14}$, is still rational. So $sqrt{14}$ must be rational. Which it isn't.
edited Oct 17 '15 at 23:11
Akiva Weinberger
13.9k12268
answered Oct 17 '15 at 22:23
TonyKTonyK
43.7k358136
$endgroup$
add a comment |
$begingroup$
It is pure luck that $2sqrt2 + sqrt7 = frac{1}{2sqrt2 - sqrt7}$, down to the fact that $(2sqrt 2)^2-(sqrt 7)^2 = 1$.
The following approach is more genral:
Suppose $2sqrt2 + sqrt7$ is rational. Then its square, $15+4sqrt {14}$, is still rational. So $sqrt{14}$ must be rational. Which it isn't.
edited Oct 17 '15 at 23:11
Akiva Weinberger
13.9k12268
answered Oct 17 '15 at 22:23
TonyKTonyK
43.7k358136
$endgroup$
It is pure luck that $2sqrt2 + sqrt7 = frac{1}{2sqrt2 - sqrt7}$, down to the fact that $(2sqrt 2)^2-(sqrt 7)^2 = 1$.
The following approach is more genral:
Suppose $2sqrt2 + sqrt7$ is rational. Then its square, $15+4sqrt {14}$, is still rational. So $sqrt{14}$ must be rational. Which it isn't.
edited Oct 17 '15 at 23:11
Akiva Weinberger
13.9k12268
answered Oct 17 '15 at 22:23
TonyKTonyK
43.7k358136
edited Oct 17 '15 at 23:11
Akiva Weinberger
13.9k12268
edited Oct 17 '15 at 23:11
Akiva Weinberger
13.9k12268
edited Oct 17 '15 at 23:11
Akiva Weinberger
13.9k12268
13.9k12268
answered Oct 17 '15 at 22:23
TonyKTonyK
43.7k358136
answered Oct 17 '15 at 22:23
TonyKTonyK
43.7k358136
answered Oct 17 '15 at 22:23
TonyKTonyK
43.7k358136
43.7k358136
add a comment |
add a comment |
$begingroup$
Here's another proof more general that doesn't rely on friendly symmetry. $sqrt 2$ and $sqrt 7$ are easy to show to be irrational. So for $2sqrt 2 + sqrt 7 = r$ then $2sqrt 2 - r = -sqrt 2$ so $8 + r^2 + 2sqrt 2 = 7$ therefore $sqrt 2$ is rational.
It's the same idea, really. you just don't need to rely on the not so likely fact that their conjugates are reciprical.
answered Oct 18 '15 at 4:44
fleabloodfleablood
73.1k22789
$endgroup$
$begingroup$
I have noticed the following mistakes in your answer. $2sqrt(2) - r$ is $-sqrt(7)$ and not $-sqrt(2)$. Also $8+r^2 + 2sqrt(2)$ is $23+2sqrt(2)+4sqrt(14)$ and not $7$.
$endgroup$
– Bysshed
Apr 2 '16 at 22:58
add a comment |
$begingroup$
Here's another proof more general that doesn't rely on friendly symmetry. $sqrt 2$ and $sqrt 7$ are easy to show to be irrational. So for $2sqrt 2 + sqrt 7 = r$ then $2sqrt 2 - r = -sqrt 2$ so $8 + r^2 + 2sqrt 2 = 7$ therefore $sqrt 2$ is rational.
It's the same idea, really. you just don't need to rely on the not so likely fact that their conjugates are reciprical.
answered Oct 18 '15 at 4:44
fleabloodfleablood
73.1k22789
$endgroup$
$begingroup$
I have noticed the following mistakes in your answer. $2sqrt(2) - r$ is $-sqrt(7)$ and not $-sqrt(2)$. Also $8+r^2 + 2sqrt(2)$ is $23+2sqrt(2)+4sqrt(14)$ and not $7$.
$endgroup$
– Bysshed
Apr 2 '16 at 22:58
add a comment |
$begingroup$
Here's another proof more general that doesn't rely on friendly symmetry. $sqrt 2$ and $sqrt 7$ are easy to show to be irrational. So for $2sqrt 2 + sqrt 7 = r$ then $2sqrt 2 - r = -sqrt 2$ so $8 + r^2 + 2sqrt 2 = 7$ therefore $sqrt 2$ is rational.
It's the same idea, really. you just don't need to rely on the not so likely fact that their conjugates are reciprical.
answered Oct 18 '15 at 4:44
fleabloodfleablood
73.1k22789
$endgroup$
Here's another proof more general that doesn't rely on friendly symmetry. $sqrt 2$ and $sqrt 7$ are easy to show to be irrational. So for $2sqrt 2 + sqrt 7 = r$ then $2sqrt 2 - r = -sqrt 2$ so $8 + r^2 + 2sqrt 2 = 7$ therefore $sqrt 2$ is rational.
It's the same idea, really. you just don't need to rely on the not so likely fact that their conjugates are reciprical.
answered Oct 18 '15 at 4:44
fleabloodfleablood
73.1k22789
answered Oct 18 '15 at 4:44
fleabloodfleablood
73.1k22789
answered Oct 18 '15 at 4:44
fleabloodfleablood
73.1k22789
answered Oct 18 '15 at 4:44
fleabloodfleablood
73.1k22789
73.1k22789
$begingroup$
I have noticed the following mistakes in your answer. $2sqrt(2) - r$ is $-sqrt(7)$ and not $-sqrt(2)$. Also $8+r^2 + 2sqrt(2)$ is $23+2sqrt(2)+4sqrt(14)$ and not $7$.
$endgroup$
– Bysshed
Apr 2 '16 at 22:58
add a comment |
$begingroup$
I have noticed the following mistakes in your answer. $2sqrt(2) - r$ is $-sqrt(7)$ and not $-sqrt(2)$. Also $8+r^2 + 2sqrt(2)$ is $23+2sqrt(2)+4sqrt(14)$ and not $7$.
$endgroup$
– Bysshed
Apr 2 '16 at 22:58
$begingroup$
I have noticed the following mistakes in your answer. $2sqrt(2) - r$ is $-sqrt(7)$ and not $-sqrt(2)$. Also $8+r^2 + 2sqrt(2)$ is $23+2sqrt(2)+4sqrt(14)$ and not $7$.
$endgroup$
– Bysshed
Apr 2 '16 at 22:58
$begingroup$
I have noticed the following mistakes in your answer. $2sqrt(2) - r$ is $-sqrt(7)$ and not $-sqrt(2)$. Also $8+r^2 + 2sqrt(2)$ is $23+2sqrt(2)+4sqrt(14)$ and not $7$.
$endgroup$
– Bysshed
Apr 2 '16 at 22:58
add a comment |
$begingroup$
From
$$2sqrt2 + sqrt7 = frac{n}{m}$$ and $$2sqrt2 - sqrt7 = frac{m}{n},$$ by addition, you get
$$4sqrt{2}=frac{m}{n}+frac{n}{m}.$$ That is, $sqrt{2}$ should be rational.
${}{}{}{}$
edited Apr 2 '16 at 23:31
Bysshed
1,06311023
answered Oct 17 '15 at 22:26
mflmfl
26.9k12142
$endgroup$
$begingroup$
I think that, on the second equailty, it should go $m/n$ instead of $n/m$. Same in the last part.
$endgroup$
– Miguelgondu
Oct 17 '15 at 23:39
$begingroup$
@Miguelgondu Typo corrected. Thank you for noticing it.
$endgroup$
– mfl
Oct 18 '15 at 4:01
add a comment |
$begingroup$
From
$$2sqrt2 + sqrt7 = frac{n}{m}$$ and $$2sqrt2 - sqrt7 = frac{m}{n},$$ by addition, you get
$$4sqrt{2}=frac{m}{n}+frac{n}{m}.$$ That is, $sqrt{2}$ should be rational.
${}{}{}{}$
edited Apr 2 '16 at 23:31
Bysshed
1,06311023
answered Oct 17 '15 at 22:26
mflmfl
26.9k12142
$endgroup$
$begingroup$
I think that, on the second equailty, it should go $m/n$ instead of $n/m$. Same in the last part.
$endgroup$
– Miguelgondu
Oct 17 '15 at 23:39
$begingroup$
@Miguelgondu Typo corrected. Thank you for noticing it.
$endgroup$
– mfl
Oct 18 '15 at 4:01
add a comment |
$begingroup$
From
$$2sqrt2 + sqrt7 = frac{n}{m}$$ and $$2sqrt2 - sqrt7 = frac{m}{n},$$ by addition, you get
$$4sqrt{2}=frac{m}{n}+frac{n}{m}.$$ That is, $sqrt{2}$ should be rational.
${}{}{}{}$
edited Apr 2 '16 at 23:31
Bysshed
1,06311023
answered Oct 17 '15 at 22:26
mflmfl
26.9k12142
$endgroup$
From
$$2sqrt2 + sqrt7 = frac{n}{m}$$ and $$2sqrt2 - sqrt7 = frac{m}{n},$$ by addition, you get
$$4sqrt{2}=frac{m}{n}+frac{n}{m}.$$ That is, $sqrt{2}$ should be rational.
${}{}{}{}$
edited Apr 2 '16 at 23:31
Bysshed
1,06311023
answered Oct 17 '15 at 22:26
mflmfl
26.9k12142
edited Apr 2 '16 at 23:31
Bysshed
1,06311023
edited Apr 2 '16 at 23:31
Bysshed
1,06311023
edited Apr 2 '16 at 23:31
Bysshed
1,06311023
1,06311023
answered Oct 17 '15 at 22:26
mflmfl
26.9k12142
answered Oct 17 '15 at 22:26
mflmfl
26.9k12142
answered Oct 17 '15 at 22:26
mflmfl
26.9k12142
26.9k12142
$begingroup$
I think that, on the second equailty, it should go $m/n$ instead of $n/m$. Same in the last part.
$endgroup$
– Miguelgondu
Oct 17 '15 at 23:39
$begingroup$
@Miguelgondu Typo corrected. Thank you for noticing it.
$endgroup$
– mfl
Oct 18 '15 at 4:01
add a comment |
$begingroup$
I think that, on the second equailty, it should go $m/n$ instead of $n/m$. Same in the last part.
$endgroup$
– Miguelgondu
Oct 17 '15 at 23:39
$begingroup$
@Miguelgondu Typo corrected. Thank you for noticing it.
$endgroup$
– mfl
Oct 18 '15 at 4:01
$begingroup$
I think that, on the second equailty, it should go $m/n$ instead of $n/m$. Same in the last part.
$endgroup$
– Miguelgondu
Oct 17 '15 at 23:39
$begingroup$
I think that, on the second equailty, it should go $m/n$ instead of $n/m$. Same in the last part.
$endgroup$
– Miguelgondu
Oct 17 '15 at 23:39
$begingroup$
@Miguelgondu Typo corrected. Thank you for noticing it.
$endgroup$
– mfl
Oct 18 '15 at 4:01
$begingroup$
@Miguelgondu Typo corrected. Thank you for noticing it.
$endgroup$
– mfl
Oct 18 '15 at 4:01
add a comment |
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StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
