Is this a typo in Section 1.8.1 Mathematics for Computer Science?How can I learn about proofs for computer science?How can I start to learn proof theory?Proving the existence of a proof without actually giving a proofProof by well ordering: Every positive integer greater than one can be factored as a product of primes.Gentzen and computer scienceElegant demonstration with minimum category theory knowledgeDifferent definitions of Natural Numbers and Proof by ContradictionProve $3^2n-5$ is a multiple of $4$Defining new symbols in a proof, when is this justified?Prove convergence rate to zero increases as n increases
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Is this a typo in Section 1.8.1 Mathematics for Computer Science?
How can I learn about proofs for computer science?How can I start to learn proof theory?Proving the existence of a proof without actually giving a proofProof by well ordering: Every positive integer greater than one can be factored as a product of primes.Gentzen and computer scienceElegant demonstration with minimum category theory knowledgeDifferent definitions of Natural Numbers and Proof by ContradictionProve $3^2n-5$ is a multiple of $4$Defining new symbols in a proof, when is this justified?Prove convergence rate to zero increases as n increases
$begingroup$
Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?
This is from MIT's Mathematics for Computer Science
proof-explanation proof-theory
asked Apr 26 at 1:14
doctopusdoctopus
1414
$endgroup$
add a comment |
$begingroup$
Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?
This is from MIT's Mathematics for Computer Science
proof-explanation proof-theory
asked Apr 26 at 1:14
doctopusdoctopus
1414
$endgroup$
$begingroup$
Generally: If $m,n$ are positive integers and $m^1/n$ is not an integer then $m^1/n$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
Apr 26 at 1:48
1
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
Apr 26 at 7:31
add a comment |
$begingroup$
Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?
This is from MIT's Mathematics for Computer Science
proof-explanation proof-theory
asked Apr 26 at 1:14
doctopusdoctopus
1414
$endgroup$
Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?
This is from MIT's Mathematics for Computer Science
proof-explanation proof-theory
proof-explanation proof-theory
asked Apr 26 at 1:14
doctopusdoctopus
1414
asked Apr 26 at 1:14
doctopusdoctopus
1414
asked Apr 26 at 1:14
doctopusdoctopus
1414
asked Apr 26 at 1:14
doctopusdoctopus
1414
asked Apr 26 at 1:14
doctopusdoctopus
1414
1414
$begingroup$
Generally: If $m,n$ are positive integers and $m^1/n$ is not an integer then $m^1/n$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
Apr 26 at 1:48
1
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
Apr 26 at 7:31
add a comment |
$begingroup$
Generally: If $m,n$ are positive integers and $m^1/n$ is not an integer then $m^1/n$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
Apr 26 at 1:48
1
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
Apr 26 at 7:31
$begingroup$
Generally: If $m,n$ are positive integers and $m^1/n$ is not an integer then $m^1/n$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
Apr 26 at 1:48
$begingroup$
Generally: If $m,n$ are positive integers and $m^1/n$ is not an integer then $m^1/n$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
Apr 26 at 1:48
1
1
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
Apr 26 at 7:31
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
Apr 26 at 7:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
answered Apr 26 at 1:21
John DoeJohn Doe
12.6k11441
$endgroup$
$begingroup$
This makes sense thanks
$endgroup$
– doctopus
2 days ago
add a comment |
$begingroup$
No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
answered Apr 26 at 1:19
RandallRandall
10.9k11431
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
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$begingroup$
No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
answered Apr 26 at 1:21
John DoeJohn Doe
12.6k11441
$endgroup$
$begingroup$
This makes sense thanks
$endgroup$
– doctopus
2 days ago
add a comment |
$begingroup$
No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
answered Apr 26 at 1:21
John DoeJohn Doe
12.6k11441
$endgroup$
$begingroup$
This makes sense thanks
$endgroup$
– doctopus
2 days ago
add a comment |
$begingroup$
No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
answered Apr 26 at 1:21
John DoeJohn Doe
12.6k11441
$endgroup$
No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
answered Apr 26 at 1:21
John DoeJohn Doe
12.6k11441
answered Apr 26 at 1:21
John DoeJohn Doe
12.6k11441
answered Apr 26 at 1:21
John DoeJohn Doe
12.6k11441
answered Apr 26 at 1:21
John DoeJohn Doe
12.6k11441
12.6k11441
$begingroup$
This makes sense thanks
$endgroup$
– doctopus
2 days ago
add a comment |
$begingroup$
This makes sense thanks
$endgroup$
– doctopus
2 days ago
$begingroup$
This makes sense thanks
$endgroup$
– doctopus
2 days ago
$begingroup$
This makes sense thanks
$endgroup$
– doctopus
2 days ago
add a comment |
$begingroup$
No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
answered Apr 26 at 1:19
RandallRandall
10.9k11431
$endgroup$
add a comment |
$begingroup$
No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
answered Apr 26 at 1:19
RandallRandall
10.9k11431
$endgroup$
add a comment |
$begingroup$
No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
answered Apr 26 at 1:19
RandallRandall
10.9k11431
$endgroup$
No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
answered Apr 26 at 1:19
RandallRandall
10.9k11431
answered Apr 26 at 1:19
RandallRandall
10.9k11431
answered Apr 26 at 1:19
RandallRandall
10.9k11431
answered Apr 26 at 1:19
RandallRandall
10.9k11431
10.9k11431
add a comment |
add a comment |
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$begingroup$
Generally: If $m,n$ are positive integers and $m^1/n$ is not an integer then $m^1/n$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
Apr 26 at 1:48
1
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
Apr 26 at 7:31