3.2 Direct proof

Many results can be established directly simply by starting from the given assumptions and proceeding through a succession of inferences to deduce the desired conclusion. Workshop exercises and exam questions which invite such a proof will often begin with words like ‘‘verify’’ or ‘‘check’’.

To establish an implication of the form ‘‘$p\Rightarrow q$’’ (which may alternatively be phrased ‘‘$p$ implies $q$’’, or ‘‘if $p$, then $q$’’) by direct proof, one should assume $p$ to be true and then deduce that $q$ is true.

Proof. Let $x=\sqrt{2+\sqrt{2}}$. Then we have $x^{2}=2+\sqrt{2}$, and so

$$x^{4}=(x^{2})^{2}=\bigl(2+\sqrt{2}\bigr)^{2}=2^{2}+\bigl(\sqrt{2}\bigr)^{2}+2\cdot 2\sqrt{2}=6+4\sqrt{2},$$

from which we conclude that

$$x^{4}-4x^{2}+2=\bigl(6+4\sqrt{2}\bigr)-4\bigl(2+\sqrt{2}\bigr)+2=6+4\sqrt{2}-8-4\sqrt{2}+2=0,$$

as required. $\Box$

Proof. Let $y=3\,e^{2x}-2\,e^{3x}$. Then we have

$$\frac{\mathrm{d}y}{\mathrm{d}x}=6\,e^{2x}-6\,e^{3x}\qquad\text{and}\qquad\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=12\,e^{2x}-18\,e^{3x},$$

so that

	$\displaystyle\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}-5\,\frac{\mathrm{d}y}{\mathrm{d}x}+6y$	$\displaystyle=(12\,e^{2x}-18\,e^{3x})-5(6\,e^{2x}-6\,e^{3x})+6(3\,e^{2x}-2\,e^{3x})$
		$\displaystyle=12\,e^{2x}-18\,e^{3x}-30\,e^{2x}+30\,e^{3x}+18\,e^{2x}-12\,e^{3x}=0,$

as required. $\Box$

Proof. Every even number is a multiple of $2$, and so we can write $m=2k$ for some $k\in\mathbb{Z}$. Consequently we have $mn=(2k)n=2(kn)$, which is even because $kn$ is an integer. $\Box$

In general, statements which begin with ‘‘for all’’ can be rephrased in the form ‘‘$p\Rightarrow q$’’ for suitable statements $p$ and $q$, because if $A$ is a set and $P(x)$ is any statement involving a variable $x$, then the statements ‘‘$(\forall x\in A)\bigl(P(x)\bigr)$’’ and ‘‘$(x\in A)\Rightarrow P(x)$’’ are logically equivalent.

Proof. Denote by $D$ the set of odd integers. We then observe that this statement can be rewritten as

$$(n\in D)\Rightarrow(n^{2}-2n\in D).$$

To prove this, let $n\in D$. Then we can write $n=2m+1$ for some $m\in\mathbb{Z}$, and consequently we have

	$\displaystyle n^{2}-2n$	$\displaystyle=(2m+1)^{2}-2(2m+1)$
		$\displaystyle=4m^{2}+4m+1-4m-2=4m^{2}-1=2(2m^{2})-1,$

so $n^{2}-2n$ is odd because $2(2m^{2})$ is even. $\Box$

Sometimes a proof may involve separate consideration of different cases, which between them cover all possibilities; this allows us to make extra assumptions within each of the cases. When there are only a few possibilities, such as in the context of truth tables or parity models, this is a reasonable way to proceed. When there are many cases to consider, proof by consideration of all cases can be tedious and unilluminating.

Proof. Let $n\in\mathbb{Z}$, and observe that $n^{2}+n=n(n+1)$. If $n$ is even, then so is the product $n(n+1)$ by Example 3.2.4. Otherwise $n$ is odd; then $n+1$ is even, and once again the product $n(n+1)$ is even by Example 3.2.4. Thus for every $n\in\mathbb{Z}$, $n^{2}+n$ is even. $\Box$

The statement $q$ in an implication of the form ‘‘$p\Rightarrow q$’’ may involve ‘‘$\exists$’’. In such cases the proof usually involves constructing the required element somehow.

Proof. We observe that this statement can be rewritten as

$$(n\in\mathbb{Z})\Rightarrow\bigl((\exists\,m\in\mathbb{Z})(n+m=0)\bigr).$$

To prove this, let $n\in\mathbb{Z}$, and define $m=-n$; then $m\in\mathbb{Z}$, and $n+m=n+(-n)=0$ as required. $\Box$

So far we have considered proofs of statements of the form ‘‘$p\Rightarrow q$’’. If we wish to prove a bi-implication, that is, a statement of the form ‘‘$p\Leftrightarrow q$’’, then we must prove both the forward implication ‘‘$p\Rightarrow q$’’ and its converse ‘‘$q\Rightarrow p$’’. These two parts of the proof may involve quite different arguments.

Proof. ‘‘$\Rightarrow$’’. Suppose that $a$ and $b$ are solutions to the equation $x^{2}+sx+t=0$. Since they are distinct, the standard formula for solving a quadratic equation implies that

$$a=\frac{-s+\sqrt{s^{2}-4t}}{2}\qquad\text{and}\qquad b=\frac{-s-\sqrt{s^{2}-4t}}{2}$$

(or the other way round, which would not affect our proof), and therefore we have

	$\displaystyle a+b$	$\displaystyle=\frac{-s+\sqrt{s^{2}-4t}}{2}+\frac{-s-\sqrt{s^{2}-4t}}{2}$
		$\displaystyle=\frac{-s+\sqrt{s^{2}-4t}-s-\sqrt{s^{2}-4t}}{2}=-s$

and

	$\displaystyle ab$	$\displaystyle=\biggl(\frac{-s+\sqrt{s^{2}-4t}}{2}\biggr)\biggl(\frac{-s-\sqrt{s^{2}-4t}}{2}\biggr)$
		$\displaystyle=\frac{(-s)^{2}-\bigl(\sqrt{s^{2}-4t}\bigr)^{2}}{4}=\frac{s^{2}-(s^{2}-4t)}{4}=\frac{4t}{4}=t,$

as required.

‘‘$\Leftarrow$’’. Conversely, suppose that $a+b=-s$ and $ab=t$. Then $b=-s-a$, and hence

$$t=ab=a(-s-a)=-as-a^{2}.$$

Rearranging this equation, we see that $a^{2}+sa+t=0$, so that $a$ is a solution to the equation $x^{2}+sx+t=0$, as desired. In a similar way we can verify that $b$ is a solution to this equation. $\Box$