**Im sorry this is the second time this happened... the \"2r\" is 2raised to the

Question

**Im sorry this is the second time this happened... the "2r" is 2raised to the r power, and the 2(p/q) is 2 raised to the power of(p/q). and further 2p and 3q are each raised to the power of p andq respectively. Prove that there is no rational number r suchthat 2r=3 So far i have: Assume there exists a rational number p/q (p,q are integers)with no common factors, so that
2(p/q)=3 => 2p=3q and here is where i am stuck.... Please help me with the nextsteps... 2r=3 So far i have: Assume there exists a rational number p/q (p,q are integers)with no common factors, so that
2(p/q)=3 => 2p=3q and here is where i am stuck.... Please help me with the nextsteps...

Explanation / Answer

Here are the three cases Case 1: x=y it can be easy seen that if x=y, then the equation does not hold.(It holds of course for the case x=y=0, but that would giveus a fraction of 0/0, which is not allowed) Case 2: x>y In this case, put both sides of the equation to the power of 1/y,giving us 2x-y=3, Now since x and y are integers, thepower that 2 is put to is also an integer, and 21=2,22=4, etc.., so the equation does not hold Case 3: y>x In this case, do the same as above, but in reverse, giving us2=3y-x, and since 31=3, 32=9, thenit is easily seen that 3 to any integer can not be 2 Since all possible combinations have been exercised, it has beenproven that 2r != 3

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