Question

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

Answer 1

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 $m$ is and odd integer then $m^{2}$ is also odd.

In the given proof the expansion for $(2k + 1)^{2}$ is incorrect.

By definition we know,

$(a+b)^{2} = a^{2} + b^{2} + 2ab$

∴ $(2k + 1)^{2} = (2k)^{2} + 1^{2} + 2(2k)(1)\\(2k + 1)^{2} = 4k^{2} + 1 + 4k$

Now, we know $4k^{2}$ and $4k$ will be even values

∴$4k^{2} + 1 + 4k$ will be odd

hence $(2k + 1)^{2}$ will be odd, which means $m^{2}$ will be odd.