Prove that √5 + √3 is irrational
(No.4 of Problem Set 10 in Numbers: Rational and Irrational by Ivan Niven)
Solution:
The argument made here is a proof by contradiction or reductio ad absurdum; That is, assuming that the proposition is false and then derive a contradiction from this assumption.
Note: My solutions are also parallel to Prof. Niven’s
First, suppose that √5 + √3 is rational, say a: √5 + √3 = a
By squaring, so that radicals would reduce and make the proof shorter:
5 + 2√15 + 3 = a2 ….. √15 = (a2 -8)/2
—————————————————————————————-
Note: Rational numbers are closed under the four fundamental operations. But…..
…..proving that √15 is a rational should be done first to verify the assumption (√5 + √3 is rational)
—————————————————————————————–
Second, assume that √15 = x/y, where x and y are integers having no common factor other than 1; That is, in lowest term
Squaring, we would obtain: 15 = (x2)/(y2) ….. x2 = 15y2
15y2 is divisible by 15, so x2 is divisible by 15, and hence x is also divisible by 15. Let’s say x = 15w
Then, x2 = 15y2 ….. (15w)2 = 15y2 ….. 15w2 = y2
It implies that y2 is divisible by 15, so y is also divisible by 15.
—————————————————————————————-
Now, it tells that x and y have another common factor, 15; Which is contradict to the second assumption that x/y is a rational fraction in lowest term.. Hence, √15 can’t be represented as a ratio of two integers; An irrational number.
Therefore, it follows that √5 + √3 is also an irrational number.
——————————————————————————————
Prove that cube root of 3 is irrational
Solution: Let cube root of 3 = r/s, where r/s is a rational fraction in lowest term
By cubing, to eliminate radical sign: 3 = (r3)/(s3) ….. 3s3 = r3
It implies that r3 is divisible by 3, so r is also divisible by 3; Supposing r = 3c
Then, 3s3 = (3c)3 ….. s3 = 9c3
It tells that s3 is divisible by 9, so s is also divisible by 9. Since it is divisible by 9, s is divisible by 3.
Since r and s are both divisible by 3, hence it contradicts the assumption that r/s is a fraction in lowest term.
Therefore this contradiction lead us to conclude that cube root of 3 is irrational
Quote:
“Reductio ad absurdum, which Euclid loved so much, is one of a mathematician’s finest weapons. It is a far finer gambit than any chess gambit: a chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game.” – Godfrey Harold Hardy