Find the error with this proof and explain how it mat be corrected in order to clearly prove the equation.

Prove that if m is an odd integer, then m2 is odd.

Proposed proof: Assume m is an odd integer. By definition of odd integer, m=2k+1, for some integer k.
This means that (2k + 1)^2 = 4k^2 + 1, so m is odd.

Respuesta :

Answer:

Step-by-step explanation:

It is true that for any given odd integer, square of that integer will also be odd.

i.e if [tex]m[/tex] is and odd integer then [tex]m^{2}[/tex] is also odd.

In the given proof the expansion for [tex](2k + 1)^{2}[/tex] is incorrect.

By definition we know,

[tex](a+b)^{2} = a^{2} + b^{2} + 2ab[/tex]

∴ [tex](2k + 1)^{2} = (2k)^{2} + 1^{2} + 2(2k)(1)\\(2k + 1)^{2} = 4k^{2} + 1 + 4k[/tex]

Now, we know [tex]4k^{2}[/tex] and [tex]4k[/tex] will be even values

∴[tex]4k^{2} + 1 + 4k[/tex] will be odd

hence [tex](2k + 1)^{2}[/tex] will be odd, which means [tex]m^{2}[/tex] will be odd.