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1. prove that if

2^x = 5 , then x is irrational .

Need a little help

thnx
2. Assume that x is rational, so can be written in the form a/b where a/b are coprime integers. Follow that to lead to a contradiction, showing that your assumption was false.
3. (Original post by dbmag9)
Assume that x is rational, so can be written in the form a/b where a/b are coprime integers. Follow that to lead to a contradiction, showing that your assumption was false.

x= a/b

x^2 = a^2/b^2 = 5

where to go from that
4. (Original post by A.braham)
x= a/b

x^2 = a^2/b^2 = 5

where to go from that
Is your question x^2 or 2^x?

For x^2, what you've written is a good start. Then rearrange and think about the factors of the numbers on each side. Notice that the starting assumption involved the numbers being coprime (sharing no factors, so a/b is in its simplest form).
5. sorry it is 2^x

i'm really stuck and don't know how to start solvin it

help appreciated
6. (Original post by A.braham)
sorry it is 2^x

i'm really stuck and don't know how to start solvin it

help appreciated
What are you expecting links to?

Suppose that x is rational, say x = a/b, where a and b are positive integers, and that 2^x = 5. We want to use this to get a contradiction. Obviously the two above facts combine to give us 2^(a/b) = 5. Raise both sides to the power of b, notice that the numbers on both sides are positive whole numbers, and look at prime factors.

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