Question

Prove Forall x elementof Z, x^2 - 5x - 1 is odd.

Answer 1

Answer with Step-by-step explanation:

Let Z be a set of integers

We have to prove that $x^2-5x-1$ is odd for all value of x $\in$Z.

There are two cases

1.When x is an odd integer number

2.when x is even integer number

1.When x is an odd integer  number

When x is positive odd integer

Then $x^2$  is positive odd number (because square of odd positive number is always positive an odd number )

5x is also odd number

-5x -1 = -even number

Suppose x=3

Then -15-1=-16

Odd number - even number=Odd number

Hence ,$x^2-5x-1$ is an odd number .

When x is negative odd number

Then  x square is positive odd number  and 5x is negative odd term term

Therefore,odd number + odd number -1=Even  number -1=Odd number

Hence $x^2-5x-1$ is an odd number.

2.When  x is an even number

When x is positive even number

x square is positive even number  and 5x is positive even number

Even number -Even number -1 =Even number -odd number =Odd number

Suppose x=4

$x^2-5x-1$=16-20-1=-5=Odd number

When x is negative even number

Then x square is positive even number  and 5x is negative even number

$x^2-5x-1$=Even number +Even number -1=Even number -1=Odd number

Hence, for all elements of Z$x^2-5x-1$is odd.