5.2: Proof by Contradiction

The integer 100 cannot be written as the sum of three integers, an odd number of which are odd.

Assume, to the contrary, that 100 can be written as the sum of three integers, a, b, and c, an odd number of which are odd. We consider two cases. Case 1: exactly one of a, b, and c is odd. 100=a+b+c=(2x+1)+2y+2z=2(x+y+z)+1 Since x, y, z ∈ Z, the integer 100 is odd, producing a contradiction. Case 2: All of a, b, and c are odd. 100=a+b+c=2(x+y+z+1)+1 since x+y+z+1 ∈ Z, the integer 100 is odd, again a contradiction.

The sum of a rational number and an irrational number is irrational.

Assume, to the contrary, that there exist a rational number x and an irrational number y whose sum is a rational number z. Thus, x+y=z, where x = a/b and z=c/d for some integers a,b,c,d ∈ Z and b,d ≠ 0. This implies that y = c/d-a/b=(bc-ad)/bd Since bc-ad and bd are integers and bd≠0, it follows that y is rational, which is a contradiction.

If a is an even integer and b is an odd integer, then 4 does not divide (a²+2b²).

Assume, to the contrary, that there exist an even integer a and an odd integer b such that 4|(a²+2b²). Thus a=2x, b=2y+1, and a²+2b²=4z for some integers x, y, and z. Hence (2x)²+2(2y+1)²=4z. Simplifying, we obtain 4x²+8y²+8y+2=4z or, equivalently 2=4z-4x²-8y²-8y=4(z-x²-2y²-2y) Since z-x²-2y²-2y is an integer, 4|2, which is impossible.

For every integer m such that 2|m and 4 doesn't divide m, there exist no integers x and y for which x²+3y²=m.

Assume, to the contrary, that there exist an integer m such that 2|m and 4 doesn't divide m and integers x and y for which x²+3y²=m. Since 2|m, it follows that m is even. By theorem 3.16 x² and 3y² are of the same parity. We consider two cases. Case 1: x² and 3y² are even. 4|m which produces a contradiction. Case 2: x² and 3y² are odd. 4|m which produces a contradiction

No odd integer can be expressed as the sum of three even integers.

Assume, to the contrary, that there exists an odd integer n which can be expressed as the sum of three even integers x, y, and z. Then x = 2a, y = 2b, and z = 2c with a, b, c ∈ Z. Therefore: n=x+y+z=2a+2b+2c=2(a+b+c) Since a+b+c is an integer, n is even. This is a contradiction.

There is no smallest positive real number. (proof)

Assume, to the contrary, that there is a smallest positive real number, say r. Since 0<r/2<r, it follows that r/2 is a positive real number that is smaller than r. This, however, is a contradiction.

the real number √2 is irrational (proof)

Assume, to the contrary, that √2 is rational. Then √2=a/b, where a,b∈Z and b≠0. We may further assume that a/b has been expressed in (or reduced to) lowest terms. Then 2=a²/b²; so a²=2b². Since b² is an integer, a² is even. By Theorem 3.12, a is even. So a=2c, where c∈Z. Thus (2c)²=2b², and so 4c²=2b². Therefore b²=2c². Because c² is an integer, b² is even, which implies by theorem 3.12 that b is even. Since a and b are even, each has 2 as a divisor, which is a contradiction since a/b has been reduced to lowest terms.

There is no smallest positive real number. (strategy)

Proof Strategy: In a proof by contradiction, we begin by assuming that the statement is false and attempt to show that this leads us to a contradiction. Hence we begin by assuming that there is a smallest positive real number. It is useful to represent this number by a symbol, say r. Our goal is to produce a contradiction. How do we go about doing this? Of course, if we could think of a positive real number that is less than r, then this would give us a contradiction.

The real number √2 is irrational. (strategy)

Proof Strategy: In the proof of this result, we will use Theorem 3.12 which states that an integer x is even if and only if x² is even. Also, in the proof, it will be useful to express a rational number m/n, where m,n ∈ Z and n≠0, in lowest terms, which means m and n contain no common divisor greater than 1.