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Hi all,
I am having some difficulty understanding the question below and did not quite get one part of the solution. It would be great if someone could guide me.
Qns:
Find the truth set of "
What i did and don't understand:
I know for a number m to be a prime number, m>1. I know that
Thanks
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Well, you basically solve it in your first step right there. Can a number be prime if it has a proper factor? If you plug n into 3n^2 - 4n + 1, what two factors must that number have (based solely upon the work you already did)?
"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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Hi Ricky,
Do pardon me if i didn't manage to catch what you were saying correctly. I know a number cannot be prime if it has a factor like numbers 4, 6 have factor like 2 and 3. And prime numbers can only be divided by themselves. Is that what you are trying to say? I know if i input values like n=3 or n=5 into the equation and it would not be prime answer. So i guess this is how it is proven?
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You factorised the expression yourself, to (3n-1)(n-1).
The only time that that will be a prime is when either 3n-1 or n-1 is equal to 1.
Otherwise it would have two proper factors and not be prime, by definition.
Why did the vector cross the road?
It wanted to be normal.
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The only time that that will be a prime is when either 3n-1 or n-1 is equal to 1.
Or −1. In some cases, it can be −1. For example, the problem might be when the following is a prime for n a natural number:
−n[sup]2[/sup]+10n−9 = (n−9)(1−n). n−9 = 1 and 1−n = 1 wont work; you must try −1 instead.
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Hi,
But how do you know that it will only be a prime when "3n-1 or n-1 is equal to 1"? Is something to do with the definition of prime, where m>1 and m=(r)(s), where r=1 or s=1?
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