Teaching Kids Programming: Videos on Data Structures and Algorithms
How to Prove √2 Is Irrational — Two Methods (Proof by Contradiction & Geometric Infinite Descent)
Introduction
The statement “√2 is irrational” means there are no integers 
and 
with 
such that 
.
Method 1 — Proof by Contradiction
Assume, for contradiction, that 
is rational. Then there exist integers and 
with and such that
Squaring both sides gives:
From 
we see that 
is even, so must be even. Write 
for some integer 
.
Substitute back:
Thus 
is even, so is even.
Therefore both and are even, contradicting the assumption that . This contradiction shows the assumption was false; hence is irrational.
Method 2 — Geometric Infinite Descent (Median / Perpendicular Construction) geometric construction and argument
isosceles right triangle to prove sqr root of 2 is irrational
Consider an isosceles right triangle 
with right angle at 
, legs 
and hypotenuse 
, so 
.
Let 
be the midpoint of 
, and draw the line through perpendicular to the hypotenuse 
, meeting at 
.
Then 
is an isosceles right triangle similar to . Denote the small triangle’s hypotenuse and leg by 
and 
.
, confirming 
.
The crucial point is similarity (a fixed scale), not literal subtraction of whole sides. The small triangle is a constant fraction of the original.
From the 
, takes the middle point 
, and draw a perpendicular line to the hypotenuse 
. So the is similar to 
. We use the repeative subtraction to compute the 
. 
because 
is equal to 
, Thus 
.
Then 
. Thus: 
which is 
where 
is 
and 
is 
. And we can continue this process infinity as the 
is similar to 
.
Hence every time you repeat the same construction (take the midpoint of a triangle leg and draw the perpendicular to the hypotenuse), you obtain a new isosceles right triangle similar to the original with all side lengths multiplied by a factor 
. Thus the hypotenuse and legs shrink geometrically by a fixed factor at each step.
This produces the infinite-descent contradiction in the integer setting: if you started with integer side lengths satisfying 
, the construction (or the equivalent integer-preserving midpoint-of-hypotenuse → midpoint-of-
construction that yields a strictly smaller positive integer solution of the same Diophantine equation. Repeating indefinitely would produce an infinite strictly decreasing sequence of positive integers, which is impossible. Therefore no such integer solution exists and is irrational.
Conclusion
Both methods prove that cannot be written as a ratio of integers.
- Method 1 (proof by contradiction) uses parity to show any assumed fraction in lowest terms leads to both numerator and denominator being even.
- Method 2 (geometric infinite descent) uses a similarity construction inside an isosceles right triangle to produce a smaller integer solution, contradicting minimality.
Either argument gives a clear, rigorous proof that is irrational.
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