sorry im new to talking about high level maths (my first year at uni), let me try to present the problem better.
I need to prove by contradiction that:
sqrt(2)+ cuberoot(3) is not rational.
I have made sqrt(2) + cuberoot(3) = a/b (a,b are from the set of intiger numbers)...
Did you notice, when they showed you the proof for the square root of two being irrational, that you started with a fraction
in lowest terms, and that this lowest-terms assumption was
crucial to the proof? You might want to consider using similar reasoning here.
Or, since you're working with something a bit more complex here, maybe we can dispose of the fractional notation. Let's just say the following:
. . . . .\(\displaystyle \sqrt{\strut 2\,}\,+\, \sqrt[3]{\strut 3\,}\, =\, a\, \mbox{ for }\, a\, \in\, \mathbb{Q}\)
...where \(\displaystyle \, \mathbb{Q}\, \) is the set of rational (that is, fractional and integral) numbers. Then we have:
. . . . .\(\displaystyle \sqrt[3]{\strut 3\,}\, =\, a\, -\, \sqrt{\strut 2\,}\)
. . . . .\(\displaystyle 3\, =\, \left(\, a\, -\, \sqrt{\strut 2\,}\, \right)^3\)
. . . . .\(\displaystyle 3\, =\, a^3\, -\, 3\sqrt{\strut 2\,}a\, +\, 6a^2\, -\, 2\sqrt{\strut 2\,}\)
. . . . .\(\displaystyle 3\, -\, a^3\, -\, 6a^2\, =\, (-3a\, -\, 2)\sqrt{\strut 2\,}\)
On the left-hand side, what can you say about the square of a rational number? the cube? What can you say about the product of two rational numbers? What can you say about the sum or difference of rational numbers? On the right-hand side, what can you say about the product of a ration number and the square root of two, the latter of which is known to be irrational?
What then can you conclude?
