Q. \( \text{Positive} + \text{positive} = \text{positive} \)

Q: Why is a positive plus a positive positive?

If a>0 and b>0, add the inequalities a>0 and b>0. By order axioms of the reals, a+b>0. So the sum of two strictly positive numbers is strictly positive.

Q: Can the sum of two positives be zero?

No. If a > 0 and b > 0, then a + b > 0. Zero is not strictly positive, so a sum of strictly positive terms cannot equal 0.

Q: Is a+b larger than each addend?

Yes. From b>0 we get a+b>a; and from a>0 we get a+b>b. So a+b exceeds both a and b.

Q: What about nonnegative numbers (zero allowed)?

For a\ge 0 and b\ge 0, the sum satisfies a+b\ge 0. It can be zero if both are zero; otherwise it is positive.

Q: What about infinite sums (series) of positive terms?

For a series with positive terms a_n > 0, partial sums S_N = \sum_{n=1}^N a_n form an increasing sequence. The series either converges to a nonnegative limit or diverges to +\infty.

Answer

Let \(a>0\) and \(b>0\). Then \(a+b>0\), since the sum of two positive numbers is positive. Final result: \(a+b>0\).

Detailed Explanation

Detailed, step-by-step explanation

State the hypothesis. Let \(x\) and \(y\) be real numbers with

\(x > 0\) and \(y > 0\).
Use the order-preserving property of addition. The real-number order has the property that if \(a < b\), then for any real \(c\) we have \(a + c < b + c\). Apply this property to the inequality \(0 < y\) with \(c = x\). Adding \(x\) to both sides gives

\(0 + x < y + x\).

Simplify both sides to obtain

\(x < x + y\).
Combine the inequalities by transitivity. From the hypothesis we have \(x > 0\). From the previous step we obtained \(x < x + y\), which is equivalent to \(x + y > x\). Now chain the inequalities:

\(x + y > x\) and \(x > 0\) together imply \(x + y > 0\).

The transitivity of the strict inequality gives the final relation.
Conclude the result. Therefore, the sum of two positive real numbers is positive, that is,

\[x + y > 0.\]

In words: positive plus positive equals positive.

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FAQs

Why is a positive plus a positive positive?

If \(a>0\) and \(b>0\), add the inequalities \(a>0\) and \(b>0\). By order axioms of the reals, \(a+b>0\). So the sum of two strictly positive numbers is strictly positive.

Can the sum of two positives be zero?

No. If \(a > 0\) and \(b > 0\), then \(a + b > 0\). Zero is not strictly positive, so a sum of strictly positive terms cannot equal 0.

Is \(a+b\) larger than each addend?

Yes. From \(b>0\) we get \(a+b>a\); and from \(a>0\) we get \(a+b>b\). So \(a+b\) exceeds both \(a\) and \(b\).

Does this hold in all number systems?

It holds in ordered rings/fields like natural numbers, integers, rationals, and reals. It fails where "positive" is undefined (complex numbers) or in structures with different orders (some matrices, rings without order).

What about nonnegative numbers (zero allowed)?

For \(a\ge 0\) and \(b\ge 0\), the sum satisfies \(a+b\ge 0\). It can be zero if both are zero; otherwise it is positive.

How to prove it using inequality rules?

Are there counterexamples with vectors or matrices?

Positivity for vectors/matrices is a partial order; two "positive" matrices (e.g., positive semidefinite) sum to a positive semidefinite matrix, but definitions vary. In non-ordered algebraic systems the statement may be meaningless or require a specific order.

What about infinite sums (series) of positive terms?

For a series with positive terms \(a_n > 0\), partial sums \(S_N = \sum_{n=1}^N a_n\) form an increasing sequence. The series either converges to a nonnegative limit or diverges to \(+\infty\).

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