Is this a typo in Section 1.8.1 Mathematics for Computer Science?How can I learn about proofs for computer...
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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
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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 yesterday
doctopusdoctopus
1393
$endgroup$
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 yesterday
doctopusdoctopus
1393
$endgroup$
$begingroup$
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
yesterday
1
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
yesterday
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 yesterday
doctopusdoctopus
1393
$endgroup$
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 yesterday
doctopusdoctopus
1393
asked yesterday
doctopusdoctopus
1393
asked yesterday
doctopusdoctopus
1393
asked yesterday
doctopusdoctopus
1393
asked yesterday
doctopusdoctopus
1393
1393
$begingroup$
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
yesterday
1
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
yesterday
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
yesterday
1
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
1
1
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
yesterday
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
yesterday
add a comment |
2 Answers
2
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oldest
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$begingroup$
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 yesterday
RandallRandall
10.9k11431
$endgroup$
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 yesterday
John DoeJohn Doe
12.3k11341
$endgroup$
add a comment |
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$begingroup$
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 yesterday
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 yesterday
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 yesterday
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 yesterday
RandallRandall
10.9k11431
answered yesterday
RandallRandall
10.9k11431
answered yesterday
RandallRandall
10.9k11431
answered yesterday
RandallRandall
10.9k11431
10.9k11431
add a comment |
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 yesterday
John DoeJohn Doe
12.3k11341
$endgroup$
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 yesterday
John DoeJohn Doe
12.3k11341
$endgroup$
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 yesterday
John DoeJohn Doe
12.3k11341
$endgroup$
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 yesterday
John DoeJohn Doe
12.3k11341
answered yesterday
John DoeJohn Doe
12.3k11341
answered yesterday
John DoeJohn Doe
12.3k11341
answered yesterday
John DoeJohn Doe
12.3k11341
12.3k11341
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
yesterday
1
$begingroup$
What makes you think this is a mistake?
$endgroup$
– Servaes
yesterday