Pythagorean theorem does require the Cauchy-Schwarz inequality?In Wikipedia I found a proof of Schwarz inequality making use of the Pythagorean Theorem. Anyway I'm wondering... isn't that a petitio principii? Isn't the Pythagorean Theorem in Hilbert spaces proved by the Cauchy-Schwarz inequality? Or effectively is the Pythagorean Theorem a stronger if not equivalent notion?

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Pythagorean theorem does require the Cauchy-Schwarz inequality?In Wikipedia I found a proof of Schwarz inequality making use of the Pythagorean Theorem. Anyway I&#039;m wondering... isn&#039;t that a petitio principii? Isn&#039;t the Pythagorean Theorem in Hilbert spaces proved by the Cauchy-Schwarz inequality? Or effectively is the Pythagorean Theorem a stronger if not equivalent notion?

Landyn Whitney · Accepted Answer

I think your confusion is due to notation used in Pythagorean theorem: we don&#039;t actually require any norm-property of       |        |    ⋅      |        |   induced by the inner product in proving it. It&#039;s only after we recognise       |        |    ⋅      |        |   as a norm that it has its usual realization.Consider an inner product space V over real field for simplicity.Let&#039;s denote            |              |        x        |            |        2    =      ⟨    x    ,    x    ⟩  The important thing to realize here is that although notation       |        |    ⋅      |        |   is used, at this stage we don&#039;t recognize       |        |    ⋅      |        |   as a &quot;norm&quot;. It&#039;s just a convenient notation to denote   ⟨  x  ,  x  ⟩.Suppose now that x,y are orthogonal. We prove the &quot;Pythagorean Theorem&quot;Claim : If x,y are orthogonal then      |        |    x  +  y      |              |        2    =      |        |    x      |              |        2    +      |        |    y      |              |        2  Proof :      ⟨    x    +    y    ,    x    +    y    ⟩    =            |              |        x        |            |        2    +            |              |        y        |            |        2    +  2      ⟨    x    ,    y    ⟩    =            |              |        x        |            |        2    +            |              |        y        |            |        2  by orthogonality, bilinearlity.So that didn&#039;t require CS inequality.So this implies CS inequality, and this CS inequality ensures that the function       |          |      ⋅      |        |   is indeed a norm. In short,                    I        P                                    ⟹                Pythagorean Theorem                ⟹                Cauchy-Schwarz                                                  ⟹                &quot;norm function&quot; is indeed a norm

Pythagorean theorem does require the Cauchy-Schwarz inequality? In Wikipedia I found a proof of Sch

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