Visual Lessons

Four triangles, one new square

Fold to the midpoints

The diagonal as a length

Tighten the bounds by squaring

1.$4^2$ = 1.96 < 2

1.$5^2$ = 2.25 > 2

1.4 < $\sqrt{2}$ < 1.5

Each interval is narrower. Neither finite endpoint becomes √2.

Follow the prime-factor contradiction

1. Suppose $\sqrt{2}$ = $\frac{m}{n}$ for positive integers m and n.

Step 1 of 5

Move the pieces; preserve the area

Two cuts separate W and X from the five-sided piece V. Total area: $3^2$ + $6^2$ = 45.

All four new sides are congruent hypotenuses. The two acute angles of a right triangle are complementary, so each new corner is 90°. This establishes a square before using its area formula.

From square areas to the theorem

The right angle is established by perpendicular legs. The side opposite it is c, the hypotenuse.

Scale and test a triple

$3^2$ + $4^2$ = $5^2$. gcd = 1: primitive.

Generate from an odd square

$m^2$ = 9; n = $\frac{\left(9+1\right)}{2}$ = 5.

$4^2$ + $3^2$ = $5^2$

Triple: (4, 3, 5). The consecutive entries 4 and 5 have gcd 1, so the triple is primitive.

This does not produce (8,15,17): neither leg is one less than 17.

Model the lotus, then solve

Assumptions: level water; initially perpendicular, straight stem; fixed root; unchanged stem length.

Build a square on grid points

Area = $1^2$ + $1^2$ = 2 square units. Perpendicular side steps: $\left(1,1\right)$ and (-1,1).

Area 3 is impossible: both integer step magnitudes would be at most 1, whose squared sum is at most 2.