[math] \text {Proof that } a^2=b^3 \text { has solutions}, a, b = I > 1 [/math]

 

[math] a^2=b^3 (1) [/math]

 

[math] \sqrt{a^2}=\sqrt{b^3} [/math]

 

[math] a=b^\frac{3}{2} [/math]

 

[math] a=\sqrt{b}^3 (2) [/math]

 

[math] b=4: (2) a=\sqrt{4}^3 [/math]

[math] =2^3 [/math]

[math] =8 [/math]

 

[math] (1) a^2=b^3 [/math]

[math] 8^2=4^3 [/math]

[math] 64=64 [/math]

[math] LS=RS [/math]

 

[math] \text{Therefore proven} [/math]

 

[math] b=9: (2) a=\sqrt{9}^3 [/math]

[math] =3^3 [/math]

[math] =27 [/math]

 

[math] (1) 27^2=9^3 [/math]

[math] 729=729 [/math]

[math] LS=RS [/math]

 

[math] \text{Therefore proven} [/math]

I would have considered 27^2 = 9^3 a proof for the claim, already. Or [math](x^3)^2 = (x^2)^3[/math] (in conjunction with the claim that there exists an x for which x^2 and x^3 exist). What's that fuzz about?

[math] \text {Proof that } a^2=b^3 \text { has solutions}, a, b = I > 1 [/math]

 

[math] a^2=b^3 (1) [/math]

 

[math] \sqrt{a^2}=\sqrt{b^3} [/math]

 

[math] a=b^\frac{3}{2} [/math]

 

[math] a=\sqrt{b}^3 (2) [/math]

 

[math] b=4: (2) a=\sqrt{4}^3 [/math]

[math] =2^3 [/math]

[math] =8 [/math]

 

[math] (1) a^2=b^3 [/math]

[math] 8^2=4^3 [/math]

[math] 64=64 [/math]

[math] LS=RS [/math]

 

[math] \text{Therefore proven} [/math]

 

[math] b=9: (2) a=\sqrt{9}^3 [/math]

[math] =3^3 [/math]

[math] =27 [/math]

 

[math] (1) 27^2=9^3 [/math]

[math] 729=729 [/math]

[math] LS=RS [/math]

 

[math] \text{Therefore proven} [/math]

 

In a broader sense, I think you've basically proven that the solution set includes any a and b where

[math] a=\sqrt{b}^3[/math]

 

So a = 64 and b = 16 is also a solution, as is a = 125, b = 25, etc.

or, if you like, if (x^3)^2=(x^2)^3 (as already shown by timo)

then x^6=X^6.

 

Therefore any whole number (X) to the power of 6 has a whole number square root (X^3) and a whole number cube root (X^2).

Edited June 6, 201213 yr by Joatmon