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Yes. The proof is easy. Let x be the irrational number and assume there exists some rational number r = a/b where a and b are integers (that's what it means to be rational).

Now suppose x/r is a rational number. Then x/r = (b/a)x = c/d where c and d are some other integers.

Since (b/a)x=c/d, then

x = bd/ac

which means that x itself is rational, but we assumed it was irrational.

The contradiction proves that the assertion is wrong. An irrational divided by a rational must be irrational.

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Q: Is an irrational number divided by a rational number always irrational?
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