Real Number Proof Workshop

From Inequalities to Proofs

The previous tutorial developed useful inequalities for estimating products and sums. The next step is to use such facts inside complete proofs: identify exactly what is assumed, choose a route from those assumptions to the conclusion, and justify each comparison. A proof is not a list of plausible observations. Every step must follow from a hypothesis, a definition, or a result established earlier in this course.

In this workshop, we will practice three habits. First, read an implication in the correct direction: prove the conclusion from the stated hypothesis, not from a related statement that merely seems similar. Second, when signs matter, use trichotomy to separate the possible cases. Third, check the hypotheses before dividing or multiplying an inequality. These habits will be useful throughout real analysis.

Plan a Direct Proof Before Calculating

A direct proof starts with the assumptions in the statement and uses them to reach the conclusion. For a claim of the form “if $P$, then $Q$,” the proof assumes $P$ and establishes $Q$. It does not need to establish that $Q$ implies $P$. That reverse claim is called the converse, and it may be false even when the original implication is true.

For example, if the hypothesis is $x>2$, then $x-2>0$. Also $x+2>0$, since $x>2$ implies $x+2>4>0$. Both factors are positive, so their product is positive. This approach uses the factorization of a difference of squares and makes the sign of each factor explicit.

Worked Example: Proving a Difference-of-Squares Bound

Prove that if $x>2$, then $x^2>4$. Assume $x>2$. Subtracting $2$ gives $x-2>0$, and adding $2$ gives $x+2>4>0$. The product of two positive real numbers is positive, so

$$ 0<(x-2)(x+2)=x^2-4. $$

Adding $4$ to both sides gives $4<x^2$, or $x^2>4$, as required. The factorization identifies the signs that make the conclusion follow; simply observing that $x$ is “large” would not be a proof.

The converse would say that $x^2>4$ implies $x>2$. It is false: $x=-3$ satisfies $x^2=9>4$, but $-3$ is not greater than $2$. A single verified counterexample is enough to disprove a universal claim. It is not enough to test a few examples in order to prove a universal claim: examples can reveal a pattern, but a proof must cover every value allowed by the hypotheses.

Proof habit. For an implication, write down the hypothesis you are allowed to assume and the conclusion you must establish. Do not silently replace the statement with its converse. To disprove a universal claim, one counterexample satisfying its hypothesis and violating its conclusion is sufficient.

Use Trichotomy to Organize Sign Arguments

Trichotomy says that for any real number $x$, exactly one of $x<0$, $x=0$, or $x>0$ holds. When a proof depends on signs, these alternatives provide a complete case division. The important point is to account for every case and explain why the assumptions rule out the cases that cannot occur.

Theorem (Positive Sum and Product Determine Both Signs). If $x,y\in\mathbb{R}$, $xy>0$, and $x+y>0$, then $x>0$ and $y>0$.

Proof. Since $xy>0$, neither $x$ nor $y$ is zero: if either were zero, their product would be zero. By trichotomy, each must therefore be positive or negative. They cannot have opposite signs, because the product of a positive number and a negative number is negative. Thus either both are positive or both are negative. If both were negative, their sum would be negative, contradicting $x+y>0$. The only remaining possibility is that both are positive. $\square$

The two assumptions do different jobs. A positive product alone tells us that the nonzero factors have the same sign, but does not tell us which sign: for instance, $(-2)(-3)>0$. A positive sum then rules out the possibility that both are negative. Keeping track of what each hypothesis contributes makes the proof easier to understand and helps prevent a missing case.

Worked Example: Deducing Signs from Two Conditions

Suppose $a,b\in\mathbb{R}$, $ab=12$, and $a+b=7$. Since $12>0$, we have $ab>0$; since $7>0$, we have $a+b>0$. The theorem therefore gives $a>0$ and $b>0$. In fact, the two conditions also determine the numbers: the identity

$$ (a-b)^2=(a+b)^2-4ab $$

follows by expanding both sides, because $(a+b)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=(a-b)^2$. Substituting the given values yields $(a-b)^2=49-48=1$. Thus $a-b=1$ or $a-b=-1$. Combined with $a+b=7$, these alternatives give $(a,b)=(4,3)$ or $(3,4)$. In either case, direct substitution gives product $12$ and sum $7$.

This example illustrates a useful order of work: first establish the signs that follow from the hypotheses, then use algebra to obtain more detailed information. The sign conclusion is not needed to solve the equations here, but it is a direct application of the theorem and reinforces the distinction between a product’s sign and the signs of its factors.

Check Denominators Before Comparing Fractions

A common source of errors in real-number proofs is multiplying an inequality by an expression whose sign has not been established. Multiplication by a positive number preserves the direction of an inequality; multiplication by a negative number reverses it. A safe proof either establishes the sign first or avoids the step that would require an unknown sign.

Theorem (Bounds for a Positive Fraction). If $x>0$, then $$ 0<\frac{x}{1+x}<1. $$

Proof. Since $x>0$, we have $1+x>1>0$, so the denominator $1+x$ is positive. The numerator $x$ is positive, and therefore $x/(1+x)>0$. Also,

1-\frac{x}{1+x} =\frac{1+x}{1+x}-\frac{x}{1+x} =\frac{1}{1+x}>0.

The last inequality holds because the reciprocal of a positive real number is positive. Thus $1-x/(1+x)>0$, which is equivalent to $x/(1+x)<1$. Combining the two strict inequalities proves the result. $\square$

The proof checks the denominator before using the fraction. One could also compare $x$ with $1+x$, but the denominator’s positivity must still be part of the reasoning that justifies the comparison of the quotients. In either route, the hypothesis $x>0$ supplies exactly the sign information required.

Worked Example: Applying the Fraction Bound

Let $t=5$. Since $5>0$, the theorem gives $0<5/(1+5)<1$. Evaluating the fraction confirms the bound:

\frac{5}{1+5}=\frac56,\qquad 0<\frac56<1.

The last comparison follows because $5>0$ and $5<6$, so dividing each by the positive number $6$ gives $0<5/6<1$. The general theorem is more useful than this calculation: it proves the same bound for every positive real $t$, without requiring a separate computation for each value.

A Fraction That Preserves Order

The same denominator check supports a stronger comparison. On positive inputs, the expression $x/(1+x)$ preserves strict order: a larger positive input gives a larger value. Rather than relying on a graph or an intuition about the expression, we can prove the comparison by subtracting the two fractions and checking the signs of the resulting numerator and denominator.

Theorem (Order Preservation by a Positive Fraction). If $0<x<y$, then $$ \frac{x}{1+x}<\frac{y}{1+y}. $$

Proof. The assumptions $x>0$ and $y>0$ imply $1+x>0$ and $1+y>0$. Therefore $(1+x)(1+y)>0$. We compute the difference, using this common positive denominator:

\frac{y}{1+y}-\frac{x}{1+x} =\frac{y(1+x)-x(1+y)}{(1+x)(1+y)} =\frac{y+xy-x-xy}{(1+x)(1+y)} =\frac{y-x}{(1+x)(1+y)}.

Because $x<y$, we have $y-x>0$. The denominator is also positive, so the final quotient is positive. Hence $y/(1+y)-x/(1+x)>0$, which is equivalent to $x/(1+x)<y/(1+y)$. $\square$

Notice how the proof avoids cross-multiplying without explanation. Cross-multiplication is valid here because both denominators are positive; the proof states that fact and then shows that the difference has a positive numerator and denominator. If either input were allowed to make a denominator zero or negative, this argument would need to change.

Worked Example: Comparing Two Values Exactly

Compare $2/3$ and $4/5$ using the order-preservation theorem. The inputs $2$ and $4$ satisfy $0<2<4$, and the corresponding fractions are

\frac{2}{1+2}=\frac23,\qquad \frac{4}{1+4}=\frac45.

The theorem gives $2/3<4/5$. We can verify the comparison directly by bringing the fractions to a common denominator:

\frac45-\frac23=\frac{12-10}{15}=\frac{2}{15}>0.

Thus the larger input does indeed produce the larger fraction in this instance. The theorem proves this pattern for every pair of positive inputs in the stated order, rather than only for these two values.

Turn Proof Habits into a Routine

Before writing a proof, identify the exact hypotheses and conclusion. Then choose a route suited to the structure of the claim: factor an algebraic expression when its factors have useful signs, split into sign cases when trichotomy applies, or compare two quantities by examining their difference. In each route, state why the signs needed for multiplication, division, or comparison hold.

A frequent pitfall is to treat a true conclusion as if it automatically proved the hypothesis. For example, the implication $x>2\Rightarrow x^2>4$ is true, but its converse is false. Another is to divide by a quantity without checking whether it is zero; a third is to multiply an inequality by an expression without knowing its sign. These are not merely stylistic concerns: each can invalidate an otherwise plausible argument.

The results in this workshop show how short arguments can be made complete. Trichotomy handles every possible sign, while a carefully chosen difference exposes the order between two fractions. In later proofs, the same discipline applies to more involved expressions: make the assumptions visible, use each one for a specific purpose, and verify that every transformation preserves the claim being proved.

Key takeaway. A sound real-number proof follows from its stated hypotheses, covers every relevant case, and checks signs before multiplying or dividing inequalities. To disprove a universal claim, give a counterexample that satisfies its hypothesis but fails its conclusion.

Check Your Understanding

Use the proof strategies and results from this workshop to answer the questions.

Give a counterexample to the converse of “if $x>2$, then $x^2>4$.” Verify both parts of the counterexample.
If $xy>0$ and $x+y<0$, what can you conclude about the signs of $x$ and $y$? Justify your answer using trichotomy.
Use the positive-fraction theorem to state a strict upper and lower bound for $7/(1+7)$.
In the proof that $x/(1+x)<y/(1+y)$ when $0<x<y$, why is $(1+x)(1+y)$ positive?
For $0<u<v$, write the difference $v/(1+v)-u/(1+u)$ as a single fraction and explain why it is positive.

What You'll Learn