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I don't even understand what this proof is asking, let alone how to do it.
here it is:
Show that if $x>1$ is a real number and if $a<b$ are rational numbers, then $0\le x^a \le x^b$.
any hints or help would be awesome!

Swapnil Tripathi
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Zoë Soriano
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  • You can use \lefor $\le$ – Swapnil Tripathi Dec 03 '14 at 03:12
  • This is simply saying: prove that if a,b are both real numbers and a is strictly smaller than b,and if x is strictly larger than 1 then $0$ is smaller than or equal to $x^a$ is smaller than or equal to $x^b$ So think about what happens when a,b are both integers and think about what happens if a,b are both smaller than one etc. – 123 Dec 03 '14 at 03:17

1 Answers1

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Hint

  1. What does $a<b$ mean in terms of integers?
  2. Recall the definition of $x^{p/q}$ where $p,q$ are integers.
user1537366
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