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C3 solomon paper, no clue what to do

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!

Reply 1

Katiee224

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

Reply 2

shawn_o1

You know that log_ab = c means a^c = b so go from there

Reply 3

Icyytea

I get 2^(p/q) = 3 now i'm lost, i've never done this before
is it even c3?

Reply 4

ViralJZ

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.

Reply 5

Icyytea

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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Reply 6

ViralJZ

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.