Compare Surds by Squaring Both Sides — Monotonic for Non-Negatives

Here is one of the most frequent mini-problems inside JEE and Board algebra: the question secretly wants you to compare two non-negative numbers, at least one of which is a surd, and you have to say which is larger. Typical instances:

Which is larger, \sqrt{10} or 3.2?
Is \sqrt{5} + \sqrt{3} larger or smaller than \sqrt{15}?
For the inequality \sqrt{2x + 1} < 5, what happens when you "square both sides"?

The fast reflex is the same every time: square both sides, because squaring preserves order when both sides are non-negative. The radical disappears and the comparison becomes a computation on rational numbers you can do in one line.

The trigger

You are asked to compare two expressions A and B with A, B \geq 0, and at least one of them contains a square root. That is enough — square both sides, drop the radical, compare the results.

The guarantee you are using is a property of the squaring function on the non-negatives: squaring is monotonically increasing for x \geq 0. That is, if 0 \leq A \leq B, then A^2 \leq B^2, and vice versa. In symbols:

A \leq B \iff A^2 \leq B^2 \quad \text{(both } A, B \geq 0\text{)}.

So squaring preserves the direction of the inequality. The radical on one side turns into a rational number. The comparison becomes trivial.

The worked reflex: √10 vs 3.2

Spot the trigger: compare \sqrt{10} and 3.2. Both are non-negative. Square both sides.

(\sqrt{10})^2 = 10 \quad \text{and} \quad (3.2)^2 = 10.24.

Why: 3.2^2 = 3.2 \times 3.2 = 10.24. Compute it as (3 + 0.2)^2 = 9 + 2 \cdot 3 \cdot 0.2 + 0.04 = 9 + 1.2 + 0.04 = 10.24.

Since 10 < 10.24, by monotonicity \sqrt{10} < 3.2. Done in one line.

Compare this to the alternative — computing \sqrt{10} \approx 3.162\ldots by hand or from memory, then checking if 3.162 < 3.2. The squaring move skips the estimation entirely. The radical is gone; the arithmetic is clean.

Example 2: √5 + √3 vs √15

This is a trap. The tempting wrong move is "split": \sqrt{5 + 3} = \sqrt{8} versus \sqrt{15}, and conclude the right side is bigger. That comparison is correct, but only because the tempting splitting step is wrong — \sqrt{5} + \sqrt{3} \neq \sqrt{8}.

The right move: compare A = \sqrt{5} + \sqrt{3} and B = \sqrt{15} directly by squaring. Both are non-negative.

A^2 = (\sqrt{5} + \sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}.

B^2 = (\sqrt{15})^2 = 15.

Now compare 8 + 2\sqrt{15} with 15, i.e., compare 2\sqrt{15} with 7. Square again:

(2\sqrt{15})^2 = 4 \cdot 15 = 60, \qquad 7^2 = 49.

Since 60 > 49, 2\sqrt{15} > 7, so A^2 > B^2, so A > B. That is, \sqrt{5} + \sqrt{3} > \sqrt{15}.

Why the double-square worked: after the first square the comparison was still "rational vs surd" (specifically 7 vs 2\sqrt{15}). Both of those are non-negative, so squaring one more time finishes the job. The monotonicity guarantee is the same; you just apply it twice.

The non-negativity guardrail

The squaring trick only preserves the inequality when both sides are non-negative. If one of them is negative, squaring can flip the sign. Example: -3 < 2, but (-3)^2 = 9 > 4 = 2^2. The direction reversed, because squaring makes the -3 further from zero than 2.

For surds, this is rarely a problem because square roots are always non-negative (principal root convention), and the other side of the comparison is typically a positive rational or another surd. But stay alert whenever a comparison involves an expression like a - \sqrt{b}, which could be negative — check sign before squaring.

The inequality-solving use

The same reflex is the workhorse for solving radical inequalities. Example: solve \sqrt{2x + 1} < 5.

Step 1: both sides are non-negative (the square root is non-negative; 5 > 0). So squaring preserves the inequality.

(\sqrt{2x+1})^2 < 5^2 \iff 2x + 1 < 25 \iff x < 12.

Step 2: combine with the domain requirement that the radicand is non-negative: 2x + 1 \geq 0, i.e., x \geq -1/2. Final answer: -1/2 \leq x < 12.

The squaring step is justified only because both sides were non-negative. For an inequality like \sqrt{2x+1} < -5, squaring would give 2x + 1 < 25, the same condition — but the original inequality has no solutions (a square root cannot be less than a negative number). The monotonicity argument breaks when the comparison direction includes a negative value. Check sides before squaring.

When the comparison is with an integer

Special case that comes up constantly: comparing \sqrt{n} with an integer k. Square both sides and compare n with k^2. So:

\sqrt{50} vs 7? Compare 50 with 49. 50 > 49, so \sqrt{50} > 7.
\sqrt{23} vs 5? Compare 23 with 25. 23 < 25, so \sqrt{23} < 5.
\sqrt{99} vs 10? Compare 99 with 100. 99 < 100, so \sqrt{99} < 10.

This is the exact move powering the "sandwich \sqrt{n} between two integers" trick from an earlier satellite — find the integer squares on either side of n, and by monotonicity \sqrt{n} sits between their square roots.

Why it beats estimation

Three reasons the squaring reflex is better than estimating the decimal value of each surd:

No memory load. You do not need to remember that \sqrt{10} \approx 3.162. You only need to square 3.2, which is one line of arithmetic.
No rounding errors. Decimal estimates can get close calls wrong (\sqrt{10} vs 3.163 is delicate, but squaring gives an exact answer).
It scales. When the comparison is \sqrt{5} + \sqrt{3} vs \sqrt{15}, estimating each surd and adding is an exercise in accumulating rounding errors. Squaring gives an exact comparison in two moves.

The compressed reflex

See a comparison involving a square root → check both sides are non-negative → square both sides → compare the rational results. The radical is gone, the inequality direction is preserved, and the problem is now ordinary arithmetic. One idea, many uses.