Question

Your class is in groups, working on an induction worksheet. One of the questions asks to prove that if n is even, then 8n is also even. Your group proves it, but you're then asked if the converse is also true. Your classmate suggests that it is. What would you say to explain how this suggestion is flawed, and how would you help correct them

Answer 1

The answer is "this statement is not true".

Explanation:

It is true because it can be achieved by using the initiation method. (Or at even a look, as n is still n=2m for the whole m, and the same number is 8n=8* (2*m)=2*(8*m). So it's confirmed !! )

Now the opposite declaration says, "If 8n would be an even amount then n is an even number", that's why the statement is not true.

An easy way of demonstrating why an argument also isn't valid is to show an example it against. Take into consideration the 24. 24=8*3. i.e. the shape is 8n, where it is n=3. Of course, 24 is also an equivalent number, but n=3 is an odd number here. So, we at least this one event, wherein the opposite statement could be sustained. (q.e.d).

Note:- "8n would be an even number per each integer n." is true since 8n=2*(4n), even for all integer n.