thus: hence therefore i.e.
back to your question: first let such that then thus by definition:
now fix and take of both sides over all rational numbers to get now take of both sides of this
inequality over all rational numbers to get
next let and choose such that and let such that
then: so: taking of both sides over all rational numbers gives
us: now let then using we'll get: which with complete the proof. Q.E.D.
Note: throughout the proof i've assumed that we know all basic properties for rational powers of a real number.