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It seems like such an easy proof, but I just can't seem to get it...
Prove that if m does not divide n, then m² does not divide n².
I think the way it should be done is contrapositive. That is, prove if m² divides n², then m divides n.
If we assume m² divides n², then the following are true:
m divides n²
m ≤ n
Oh, and by the way, you can't use anything like the square root of a non-perfect square is irrational.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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You can also assume that m is strictly less than n, if it helps.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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This is outside my area but maybe this is a starting point for you.
Since M does not divide into N, then we can say:
N / M = X + c where X is some integer and c is a positive number less than one, i.e. a fraction. Example 9 / 4 = 2 + .25
Multiply both sides by N / M (i.e., squaring both sides)...
(N / M) * (N / M) = (X + c) * (N / M)
N**2 / M**2 = (X+c)**2
= X**2 + 2Xc + c**2
So you need to prove the last line above is not an integer. X squared is an integer and c**2 is definitely not (a positive number less than 1 squared is an even smaller number). So for that entire sum to be an integer, 2Xc can't be integer but when added to c**2, the result is an integer. I don't have any idea how to prove that and I'm probably barking up the wrong tree, but hopefully it will give you some ideas.
you just need to factor m and n into products of prime numbers.
m= p[sub]1[/sub][sup]i1[/sup] p[sub]2[/sub][sup]i2[/sup]... p[sub]l[/sub][sup]il[/sup]
n= q[sub]1[/sub][sup]j1[/sup] q[sub]2[/sub][sup]j2[/sup] ... q[sub]k[/sub][sup]jk[/sup]
m²=...
n²= ...
compare the two sets above, latter will contain former.and then take square roots
Last edited by George,Y (2006-04-22 15:49:12)
X'(y-Xβ)=0
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Contraversial:
n !| m => n^2 !| m^2
let n !| m and n^2 | m^2. But:
1. if a|b and b|c then a|c,
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(a|b) means: there exists an integer k, for which ak=b:
IPBLE: Increasing Performance By Lowering Expectations.
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If a|(bc) then (a|b)||(a|c)
6 | (3*2) but 6 !| 3 and 6 !| 2
The above statement is false.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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thank you Ricky for correcting me.
Oh, and by the way, you can't use anything like the square root of a non-perfect square is irrational.
Why? If we could, the proof is easy:
m^2|n^2=>m|n.
1. m^2|n^2 => km^2=n^2, so k must be square,because:
√(km^2)=m√k=√(n^2)=n.
m√k=n, so k is square. But then √k is integer and exists int k'=√k that:
mk'=n, so m|n.
IPBLE: Increasing Performance By Lowering Expectations.
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Because I'm using this proof to prove that:
m/n = √x, where x is not a prefect square
m² = x * n², so n² | m²
Since x is not a perfect square, so √x is not an integer.
m = √x * n, so n !| m
Thus, n² !| m², but that's a contradiction, so the root of a non-perfect square is irrational.
Edit:
Oh, and c | ab => c | a or c | b is true for primes. Just thought I'd add that.
Last edited by Ricky (2006-04-23 13:05:13)
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Oh, and c | ab => c | a or c | b is true for primes. Just thought I'd add that.
6 | (3*2) but 6 !| 3 and 6 !| 2
The above statement is false.
You have to add and c<ab, because for all integers (not only primes) a,b we have:
ab|ab, but ab!|a and ab!|b.
Can't you use that if a square m^2 divides square n^2, then the quotient m^2/n^2 is also square?
ok. Let
Last edited by krassi_holmz (2006-04-23 20:53:19)
IPBLE: Increasing Performance By Lowering Expectations.
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