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by phoenixreader
1149 days ago
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I was interested in the proof of this ("if x is a rational number, and x raised to the power of an integer is an integer, then x is an integer"), so I asked the new Bing. It actually gave a correct proof (despite no relevant links found in Google)! ------BELOW IS BING'S OUTPUT This statement is true. If x is a rational number and x raised to the power of an integer is an integer, then x is an integer. This statement is called the rational root theorem. The proof of this theorem is based on the fact that if x is not an integer, then it can be expressed as a fraction p/q where p and q are integers and q ≠ 0. Then, x raised to the power of an integer can be expressed as (p/q)^n which can be simplified to p^n/q^n. Since p^n and q^n are integers, p^n/q^n is a rational number. If x raised to the power of an integer is an integer, then q^n must divide p^n. Since p and q are relatively prime, this implies that q = 1. Therefore, x is an integer. |
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That is just wrong. It is the completely standard proof and can be found everywhere (the proof is over 2000 years old). Add to that, that it forgets to choose p and q as relatively prime, but then uses that at the end.... Not impressive.