Question

Prove using the notion without loss of generality that\({\bf{5x + 5y}}\)is an odd integer when x and y are integers of opposite parity.

Short answer

\(5x + 5y\)is an odd integer.

Step-by-step solution

Without loss of generality

Assume that x is odd and y is even without loss of generality.

Without loss of generality is satisfied here as when \(x\)and \(y\) are interchanged in the given expression, the value of original expression would remain the same.

Find the two addends of the given expression

Let \(x = 2k + 1\) , where \(k\) is some integer.

And, \(y = 2i\) , where \(i\) is some integer.

Now, \(5x + 5y\) can be written as:

\(\begin{aligned}5x + 5y &= 5\left( {2k + 1} \right) + 5\left( {2i} \right)\\ &= 10k + 5 + 10i\\ &= 10k + 10i + 4 + 1\\ &= 2\left( {5k + 5i + 2} \right) + 1\end{aligned}\)

The above expansion has two addends:

The first one is even since it is multiplied by 2 and the second one is odd (i.e.,1).

So, the overall sum is also odd.

Thus, \(5x + 5y\)must be an odd integer.