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Proof 0=4

  1. May 12, 2012 #1
    I found this in my math book. It gives this supposed proof that 0=4 and asks where the error is. Note that this isn't homework. I found it at the end of the unit among other bonus problems to supposedly test ingenuity. Here it is:

    [itex]sin^2 \theta + cos^2 \theta = 1 \Rightarrow cos \theta = \sqrt{1-sin^2 \theta}[/itex]​

    Now we add 1 to both sides and then square them.

    [itex]1 + cos \theta = 1 + \sqrt{1-sin^2 \theta}[/itex]

    [itex](1 + cos \theta)^2 = (1 + \sqrt{1-sin^2 \theta})^2[/itex]​

    Next we substitute theta with pi.
    [itex](1 + cos \pi)^2 = (1 + \sqrt{1-sin^2 \pi})^2[/itex]​

    Given that

    [itex]cos \pi=-1[/itex]

    [itex]sin \pi=0[/itex]​

    it follows:

    [itex](1-1)^2=(1+ \sqrt{1-0^2})^2[/itex]

    [itex]0=2^2 \Rightarrow 0=4[/itex]​

    So, what's the problem here?

    Thanks for taking time to read this and hopefully solve it.

    Have a nice day.
     
  2. jcsd
  3. May 12, 2012 #2
    I have a similar question in my book too, very fundamental error right in the first step.


    For all real numbers x,

    c372c85e83015d1494b8cdaac9125b12.png

    Therefore,

    [itex]\sqrt{cos^{2}(\theta)} = |cos(\theta)|[/itex]
     
    Last edited: May 12, 2012
  4. May 12, 2012 #3

    Mentallic

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    Homework Helper

    This rule when stated usually implies that [tex]\theta\in \left[-\frac{\pi}{2},\frac{\pi}{2}\right][/tex] but it works for whenever [itex]\cos\theta \geq 0[/itex]

    Remember that the general rule is [tex]\sin^2\theta+\cos^2\theta=1[/tex] and so if we re-arrange to solve for [itex]\cos\theta[/itex] we would get [tex]\cos^2\theta=1-\sin^2\theta[/tex] and at this point, if we're to take the square root of both sides, we need to keep in mind that there is also a negative square root value as well, mainly, [tex]\cos\theta=\pm \sqrt{1-\sin^2\theta}[/tex] where we take the positive value when [itex]\cos\theta >0[/itex] and the negative when [itex]\cos\theta <0[/itex].
     
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