a)prove if a l b and k is any integer then a l ( b +ak) b)prove square root of 3

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a|b=>b=ak where k is an integer now b+ak=b+b as b+b =2b a|b so a also divides 2b that is b+ak so a|( b +ak) This is the same as proving the square root of 3, which I will write sqrt(3) is irrational. When I prove that, I am proving there is not rational number whose square is 3. So here goes: I put some extra help comments in (extra help) like this. Just ignore them if you don't need them. Let sqrt(3) = a/b where a and b are integers and they are in lowest terms, meaning they have no common factors, WE know if they had common factors we can always find another expression that has no common factors by canceling out the common ones. So a/b does exist. squaring both sides of sqrt(3) = a/b gives us 3 = a2/b2 so a2 = 3b2 Now a2 must have its factors as even powers of primes.(because of the exponent 2 on the left and both sides must have the same total power since they are equal). So it must have 32 as one of its factors. (can't be just 3 since that is 31 which is an odd exponent and when added to the 2 in b2 gives us an odd exponents also) This means a must have 3 as a factor. ( if 32 is a factor then 3 must be also) Now 3b2 has a factor 32 so b2 has a factor 3. But if b2 has a factor 3, then since powers must be even, it has 32 as a factor. This means that b must have 3 as a factor also. So we have shown that both a and b have 3 as a factor. But we stated that a/b has no common factors and we just found one, namely 3. So we have a contradiction, and it follows that we cannot express sqrt(3) as a rational fraction.

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