Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

Fermat's Last theorems! can you show that the negative real exponent solutions are in one-to-one correspondenc?

Show that the solutions of Fermat's last theorem for negative real exponent "-n" are in one-to-one correspondence with solutions for positive n.

Fermat's Last Theorem states that "(x^n)+(y^n)=(z^n)" has no integer solutions for n>2 and x,y,z≠0.

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  • 9 years ago

    Begin by putting it into an equation form:

    (x^-n)+(y^-n)-(z^-n)=0

    Multiply both sides by (x^n)(y^n)(z^n)

    (y^n)(z^n)+(x^n)(z^n)-(x^n)(y^n)=0

    Now use one of the exponent rules (a^n)(b^n)=(ab)^n to get:

    (yz)^n+(xz)^n-(xy)^n=0

    Substitute in yz=a,xz=b,xy=c:

    a^n+b^n-c^n=0

    a^n+b^n=c^n

    As a,b,c are integers as they are the product of two integers we have shown that they are inn one-to-one correspondance.

    Source(s): Maths A-Level
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