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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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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**Pi Man****Guest**

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.

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,332

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)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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,

so:

n^2=n.n => n|n^2.

and n^2|m^2, so we get that :

n|m^2.

But:

2. If a|(bc) then (a|b)||(a|c),

.

then we get that:

n^2|m^2 => n|m^2 =>n|m.m=>(n|m)or(n|m)=>n|m.

But, by definition, n !| m, so :

n^2 can't divide m^2

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

(a|b) means: there exists an integer k, for which ak=b:

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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

but , so for i=1 to infty

So for int i:

Now let see int k=n^2/m^2:

By (1) delta_i-gamma_i is always >=0, so is always a square.

But the product of squares is square, so

But then , int j and:

We know that:

But now a,b>=0, so

.

So

,

where j,m,n are int, but this implies that:

.

*Last edited by krassi_holmz (2006-04-23 20:53:19)*

IPBLE: Increasing Performance By Lowering Expectations.

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