Mixing
Let $X ={n^3 + 3n^2 +3n | n>0}$ and $Y = { n^3 -1 | n>0}.$ Prove that $X =Y$.
$begingroup$
Let $X={n^3+3n^2+3nmid n>0},$ and $Y={n^3-1mid n>0 }$. Prove that $X=Y$.
I tried proving it by adding some values to $X$ and $Y$, but I could not make it equal.
elementary-number-theory derivatives
edited Jan 21 at 20:14
jordan_glen
1
asked Jan 21 at 18:21
user634874user634874
13
$endgroup$
|
show 1 more comment
$begingroup$
Let $X={n^3+3n^2+3nmid n>0},$ and $Y={n^3-1mid n>0 }$. Prove that $X=Y$.
I tried proving it by adding some values to $X$ and $Y$, but I could not make it equal.
elementary-number-theory derivatives
edited Jan 21 at 20:14
jordan_glen
1
asked Jan 21 at 18:21
user634874user634874
13
$endgroup$
2
$begingroup$
Try $$n=1$$, then your Statement is false.
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 18:23
$begingroup$
Observe that $n^3+3n^3+3n=n^3+3n^3+3n+1-1=(n+1)^3-1;$ . Check you did copy correctly the statemnet of hte problem.
$endgroup$
– DonAntonio
Jan 21 at 18:24
$begingroup$
Please use MathJax for formatting
$endgroup$
– J. W. Tanner
Jan 21 at 18:27
1
$begingroup$
X = { n^3 + 3n^2 + 3n | n is greater than or equal to zero. Y = { n^3 - 1| n>0} prove that X = Y
$endgroup$
– user634874
Jan 21 at 19:10
$begingroup$
in X {n is greater than or equal to zero}
$endgroup$
– user634874
Jan 21 at 19:10
|
show 1 more comment
$begingroup$
Let $X={n^3+3n^2+3nmid n>0},$ and $Y={n^3-1mid n>0 }$. Prove that $X=Y$.
I tried proving it by adding some values to $X$ and $Y$, but I could not make it equal.
elementary-number-theory derivatives
edited Jan 21 at 20:14
jordan_glen
1
asked Jan 21 at 18:21
user634874user634874
13
$endgroup$
Let $X={n^3+3n^2+3nmid n>0},$ and $Y={n^3-1mid n>0 }$. Prove that $X=Y$.
I tried proving it by adding some values to $X$ and $Y$, but I could not make it equal.
elementary-number-theory derivatives
elementary-number-theory derivatives
edited Jan 21 at 20:14
jordan_glen
1
asked Jan 21 at 18:21
user634874user634874
13
edited Jan 21 at 20:14
jordan_glen
1
asked Jan 21 at 18:21
user634874user634874
13
edited Jan 21 at 20:14
jordan_glen
1
edited Jan 21 at 20:14
jordan_glen
1
edited Jan 21 at 20:14
jordan_glen
1
1
asked Jan 21 at 18:21
user634874user634874
13
asked Jan 21 at 18:21
user634874user634874
13
asked Jan 21 at 18:21
user634874user634874
13
13
2
$begingroup$
Try $$n=1$$, then your Statement is false.
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 18:23
$begingroup$
Observe that $n^3+3n^3+3n=n^3+3n^3+3n+1-1=(n+1)^3-1;$ . Check you did copy correctly the statemnet of hte problem.
$endgroup$
– DonAntonio
Jan 21 at 18:24
$begingroup$
Please use MathJax for formatting
$endgroup$
– J. W. Tanner
Jan 21 at 18:27
1
$begingroup$
X = { n^3 + 3n^2 + 3n | n is greater than or equal to zero. Y = { n^3 - 1| n>0} prove that X = Y
$endgroup$
– user634874
Jan 21 at 19:10
$begingroup$
in X {n is greater than or equal to zero}
$endgroup$
– user634874
Jan 21 at 19:10
|
show 1 more comment
2
$begingroup$
Try $$n=1$$, then your Statement is false.
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 18:23
$begingroup$
Observe that $n^3+3n^3+3n=n^3+3n^3+3n+1-1=(n+1)^3-1;$ . Check you did copy correctly the statemnet of hte problem.
$endgroup$
– DonAntonio
Jan 21 at 18:24
$begingroup$
Please use MathJax for formatting
$endgroup$
– J. W. Tanner
Jan 21 at 18:27
1
$begingroup$
X = { n^3 + 3n^2 + 3n | n is greater than or equal to zero. Y = { n^3 - 1| n>0} prove that X = Y
$endgroup$
– user634874
Jan 21 at 19:10
$begingroup$
in X {n is greater than or equal to zero}
$endgroup$
– user634874
Jan 21 at 19:10
2
2
$begingroup$
Try $$n=1$$, then your Statement is false.
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 18:23
$begingroup$
Try $$n=1$$, then your Statement is false.
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 18:23
$begingroup$
Observe that $n^3+3n^3+3n=n^3+3n^3+3n+1-1=(n+1)^3-1;$ . Check you did copy correctly the statemnet of hte problem.
$endgroup$
– DonAntonio
Jan 21 at 18:24
$begingroup$
Observe that $n^3+3n^3+3n=n^3+3n^3+3n+1-1=(n+1)^3-1;$ . Check you did copy correctly the statemnet of hte problem.
$endgroup$
– DonAntonio
Jan 21 at 18:24
$begingroup$
Please use MathJax for formatting
$endgroup$
– J. W. Tanner
Jan 21 at 18:27
$begingroup$
Please use MathJax for formatting
$endgroup$
– J. W. Tanner
Jan 21 at 18:27
1
1
$begingroup$
X = { n^3 + 3n^2 + 3n | n is greater than or equal to zero. Y = { n^3 - 1| n>0} prove that X = Y
$endgroup$
– user634874
Jan 21 at 19:10
$begingroup$
X = { n^3 + 3n^2 + 3n | n is greater than or equal to zero. Y = { n^3 - 1| n>0} prove that X = Y
$endgroup$
– user634874
Jan 21 at 19:10
$begingroup$
in X {n is greater than or equal to zero}
$endgroup$
– user634874
Jan 21 at 19:10
$begingroup$
in X {n is greater than or equal to zero}
$endgroup$
– user634874
Jan 21 at 19:10
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
It is not true. Presumably $n$ ranges over the natural numbers. $Y$ includes $0$ from $n=1$ while $X$ does not. Otherwise the sets are equal because the expression for $X$ is $(n+1)^3-1$, so every element that is in $X$ will be in $Y$. It is made harder to see by the fact that $n$ is used to define both $X$ and $Y$, but it is only a dummy variable. Make the definition $Y={m^3-1|mgt 0}$ and you can say the element of $X$ that is caused by $n$ is also an element of $Y$ that is caused by $m=n+1$.
The statement would be true if the definition of $X$ were changed to include $n ge 0$
answered Jan 21 at 18:55
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
X={ n^3+3n^2+3n | n≥0} and Y={n^3−1|n>0}. Prove that X=Y.
$endgroup$
– user634874
Jan 21 at 19:14
$begingroup$
That is exactly what I did. Any element of $X$ is also in $Y$ and vice versa. It is easier to say if you use $m$ for the definition of $Y$.
$endgroup$
– Ross Millikan
Jan 21 at 19:16
$begingroup$
So final answer proves X =Y? Will that be satisfied?
$endgroup$
– user634874
Jan 21 at 21:06
$begingroup$
That is right. Have you just calculated the first few elements of each set?
$endgroup$
– Ross Millikan
Jan 21 at 21:52
add a comment |
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$begingroup$
It is not true. Presumably $n$ ranges over the natural numbers. $Y$ includes $0$ from $n=1$ while $X$ does not. Otherwise the sets are equal because the expression for $X$ is $(n+1)^3-1$, so every element that is in $X$ will be in $Y$. It is made harder to see by the fact that $n$ is used to define both $X$ and $Y$, but it is only a dummy variable. Make the definition $Y={m^3-1|mgt 0}$ and you can say the element of $X$ that is caused by $n$ is also an element of $Y$ that is caused by $m=n+1$.
The statement would be true if the definition of $X$ were changed to include $n ge 0$
answered Jan 21 at 18:55
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
X={ n^3+3n^2+3n | n≥0} and Y={n^3−1|n>0}. Prove that X=Y.
$endgroup$
– user634874
Jan 21 at 19:14
$begingroup$
That is exactly what I did. Any element of $X$ is also in $Y$ and vice versa. It is easier to say if you use $m$ for the definition of $Y$.
$endgroup$
– Ross Millikan
Jan 21 at 19:16
$begingroup$
So final answer proves X =Y? Will that be satisfied?
$endgroup$
– user634874
Jan 21 at 21:06
$begingroup$
That is right. Have you just calculated the first few elements of each set?
$endgroup$
– Ross Millikan
Jan 21 at 21:52
add a comment |
$begingroup$
It is not true. Presumably $n$ ranges over the natural numbers. $Y$ includes $0$ from $n=1$ while $X$ does not. Otherwise the sets are equal because the expression for $X$ is $(n+1)^3-1$, so every element that is in $X$ will be in $Y$. It is made harder to see by the fact that $n$ is used to define both $X$ and $Y$, but it is only a dummy variable. Make the definition $Y={m^3-1|mgt 0}$ and you can say the element of $X$ that is caused by $n$ is also an element of $Y$ that is caused by $m=n+1$.
The statement would be true if the definition of $X$ were changed to include $n ge 0$
answered Jan 21 at 18:55
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
X={ n^3+3n^2+3n | n≥0} and Y={n^3−1|n>0}. Prove that X=Y.
$endgroup$
– user634874
Jan 21 at 19:14
$begingroup$
That is exactly what I did. Any element of $X$ is also in $Y$ and vice versa. It is easier to say if you use $m$ for the definition of $Y$.
$endgroup$
– Ross Millikan
Jan 21 at 19:16
$begingroup$
So final answer proves X =Y? Will that be satisfied?
$endgroup$
– user634874
Jan 21 at 21:06
$begingroup$
That is right. Have you just calculated the first few elements of each set?
$endgroup$
– Ross Millikan
Jan 21 at 21:52
add a comment |
$begingroup$
It is not true. Presumably $n$ ranges over the natural numbers. $Y$ includes $0$ from $n=1$ while $X$ does not. Otherwise the sets are equal because the expression for $X$ is $(n+1)^3-1$, so every element that is in $X$ will be in $Y$. It is made harder to see by the fact that $n$ is used to define both $X$ and $Y$, but it is only a dummy variable. Make the definition $Y={m^3-1|mgt 0}$ and you can say the element of $X$ that is caused by $n$ is also an element of $Y$ that is caused by $m=n+1$.
The statement would be true if the definition of $X$ were changed to include $n ge 0$
answered Jan 21 at 18:55
Ross MillikanRoss Millikan
299k24200374
$endgroup$
It is not true. Presumably $n$ ranges over the natural numbers. $Y$ includes $0$ from $n=1$ while $X$ does not. Otherwise the sets are equal because the expression for $X$ is $(n+1)^3-1$, so every element that is in $X$ will be in $Y$. It is made harder to see by the fact that $n$ is used to define both $X$ and $Y$, but it is only a dummy variable. Make the definition $Y={m^3-1|mgt 0}$ and you can say the element of $X$ that is caused by $n$ is also an element of $Y$ that is caused by $m=n+1$.
The statement would be true if the definition of $X$ were changed to include $n ge 0$
answered Jan 21 at 18:55
Ross MillikanRoss Millikan
299k24200374
answered Jan 21 at 18:55
Ross MillikanRoss Millikan
299k24200374
answered Jan 21 at 18:55
Ross MillikanRoss Millikan
299k24200374
answered Jan 21 at 18:55
Ross MillikanRoss Millikan
299k24200374
299k24200374
$begingroup$
X={ n^3+3n^2+3n | n≥0} and Y={n^3−1|n>0}. Prove that X=Y.
$endgroup$
– user634874
Jan 21 at 19:14
$begingroup$
That is exactly what I did. Any element of $X$ is also in $Y$ and vice versa. It is easier to say if you use $m$ for the definition of $Y$.
$endgroup$
– Ross Millikan
Jan 21 at 19:16
$begingroup$
So final answer proves X =Y? Will that be satisfied?
$endgroup$
– user634874
Jan 21 at 21:06
$begingroup$
That is right. Have you just calculated the first few elements of each set?
$endgroup$
– Ross Millikan
Jan 21 at 21:52
add a comment |
$begingroup$
X={ n^3+3n^2+3n | n≥0} and Y={n^3−1|n>0}. Prove that X=Y.
$endgroup$
– user634874
Jan 21 at 19:14
$begingroup$
That is exactly what I did. Any element of $X$ is also in $Y$ and vice versa. It is easier to say if you use $m$ for the definition of $Y$.
$endgroup$
– Ross Millikan
Jan 21 at 19:16
$begingroup$
So final answer proves X =Y? Will that be satisfied?
$endgroup$
– user634874
Jan 21 at 21:06
$begingroup$
That is right. Have you just calculated the first few elements of each set?
$endgroup$
– Ross Millikan
Jan 21 at 21:52
$begingroup$
X={ n^3+3n^2+3n | n≥0} and Y={n^3−1|n>0}. Prove that X=Y.
$endgroup$
– user634874
Jan 21 at 19:14
$begingroup$
X={ n^3+3n^2+3n | n≥0} and Y={n^3−1|n>0}. Prove that X=Y.
$endgroup$
– user634874
Jan 21 at 19:14
$begingroup$
That is exactly what I did. Any element of $X$ is also in $Y$ and vice versa. It is easier to say if you use $m$ for the definition of $Y$.
$endgroup$
– Ross Millikan
Jan 21 at 19:16
$begingroup$
That is exactly what I did. Any element of $X$ is also in $Y$ and vice versa. It is easier to say if you use $m$ for the definition of $Y$.
$endgroup$
– Ross Millikan
Jan 21 at 19:16
$begingroup$
So final answer proves X =Y? Will that be satisfied?
$endgroup$
– user634874
Jan 21 at 21:06
$begingroup$
So final answer proves X =Y? Will that be satisfied?
$endgroup$
– user634874
Jan 21 at 21:06
$begingroup$
That is right. Have you just calculated the first few elements of each set?
$endgroup$
– Ross Millikan
Jan 21 at 21:52
$begingroup$
That is right. Have you just calculated the first few elements of each set?
$endgroup$
– Ross Millikan
Jan 21 at 21:52
add a comment |
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2
$begingroup$
Try $$n=1$$, then your Statement is false.
$endgroup$
– Dr. Sonnhard Graubner
Jan 21 at 18:23
$begingroup$
Observe that $n^3+3n^3+3n=n^3+3n^3+3n+1-1=(n+1)^3-1;$ . Check you did copy correctly the statemnet of hte problem.
$endgroup$
– DonAntonio
Jan 21 at 18:24
$begingroup$
Please use MathJax for formatting
$endgroup$
– J. W. Tanner
Jan 21 at 18:27
1
$begingroup$
X = { n^3 + 3n^2 + 3n | n is greater than or equal to zero. Y = { n^3 - 1| n>0} prove that X = Y
$endgroup$
– user634874
Jan 21 at 19:10
$begingroup$
in X {n is greater than or equal to zero}
$endgroup$
– user634874
Jan 21 at 19:10