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    Hey how would I go about this question from Solomon K paper?
    (b) Use proof by contradiction to prove that log2 3 is irrational. (6)
    Never seen it before any explanations would be greatly appreciated!
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    (Original post by Icyytea)
    Hey how would I go about this question from Solomon K paper?
    (b) Use proof by contradiction to prove that log2 3 is irrational. (6)
    Never seen it before any explanations would be greatly appreciated!
    start by assuming it is true, so log(base2)3 = p/q where p and q are rational numbers
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    You know that logab = c means ac = b so go from there
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    I get 2^(p/q) = 3 now i'm lost, i've never done this before
    is it even c3?
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    (Original post by Icyytea)
    I get 2^(p/q) = 3 now i'm lost, i've never done this before
    is it even c3?
    Try rearranging. Write 2^(p/q) as (2^p)^(1/q)=3 and then rearrange further.
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    (Original post by ViralJZ)
    Try rearranging. Write 2^(p/q) as (2^p)^(1/q)=3 and then rearrange further.
    ah i see and then since q/p are rational they must both be integers but for any integers the 2^p = 3^q equation isn't true?
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    (Original post by Icyytea)
    ah i see and then since q/p are rational they must both be integers but for any integers the 2^p = 3^q equation isn't true?
    Indeed. 2^p where p is an integer is always an even number multiplied by an even number. 3^q where q is an integer is always an odd number multiplied by an odd number.
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    Awesome guys thanks
 
 
 
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