Question

Prove the following using the specified technique:A) Letxandybe two real numbers such thatx yis rational. Prove by contrapositive that ifxis irrational, thenx−yis irrational.B) Prove by contradiction that for any positive two real numbers,xandy, ifx·y≤50, theneitherx <8 ory <8.

Answer 1

Answer:

Following are the solution to this question:

Explanation:

In point A:

The predicate of the whole argument is if x+y is rational and x-y is reasonable, then x is reasonable as well" To see the sum of the two logical estimates is often sound. Let us be two real $\frac{p}{9}$ .rs amount then seeing how.  

$\frac{p}{q}+\frac{r}{s} = \frac{(ps+rq)}{qs}$ is logical, too. x Thus if$x + y$ is logical and $x- y$ is true, then $x + y +x- y = 2x$ should be sound.  

Let's assume $2x=\frac{p}{q}.$ Therefore the rational $x=\frac{p}{2q}$proves the necessary statement in the query.  

In point A:

Let $x \cdot y \leq 50 \ and \ x \geq 8, \ but \ \geq 8$, both to demonstrate it through contradiction. So,$x \cdot y \geq 8\cdot 8 = 64$. This implies that both $x \cdot y \ and \ \leq 50 \ and x \cdot y \ and \geq 64$ are valid in the very same period. The argument is valid.