Just thought of one of my favourite proofs:
Theorem: An irrational number raised to the power of an irrational number can be rational.
Proof: \(\sqrt{2}\) is irrational. Consider \(\sqrt{2}^{\sqrt{2}}\) (don't you love how mathematicians use the word "consider"?). Either this number is rational, or irrational. If it is rational, we're done! We have found an irrational number raised to an irrational number which is rational. If it is irrational, then: \[\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2}\times \sqrt{2}} = 2\] is rational. Either way, we have found an irrational number raised to the power of an irrational number which is rational.
The thing I like about this proof is that it makes no pronouncements on \(\sqrt{2}^{\sqrt{2}}\) - whether it is rational or irrational does not matter.
Dave Neary /
npub1yv…za0un
2023-05-18 01:44:24
Author Public Key
npub1yvl5qa9235lqcf3t95wurmsc9nvu7rr9n44rxx4fkcxe04raqy9qtza0unPublished at
2023-05-18 01:44:24Event JSON
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"content": "Just thought of one of my favourite proofs:\nTheorem: An irrational number raised to the power of an irrational number can be rational.\nProof: \\(\\sqrt{2}\\) is irrational. Consider \\(\\sqrt{2}^{\\sqrt{2}}\\) (don't you love how mathematicians use the word \"consider\"?). Either this number is rational, or irrational. If it is rational, we're done! We have found an irrational number raised to an irrational number which is rational. If it is irrational, then: \\[\\left(\\sqrt{2}^{\\sqrt{2}}\\right)^{\\sqrt{2}} = \\sqrt{2}^{\\sqrt{2}\\times \\sqrt{2}} = 2\\] is rational. Either way, we have found an irrational number raised to the power of an irrational number which is rational.\n\nThe thing I like about this proof is that it makes no pronouncements on \\(\\sqrt{2}^{\\sqrt{2}}\\) - whether it is rational or irrational does not matter.",
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