1.4 Disproof by counter example — Practise with help

1. $15 = 3 \times 5$, so $15$ is not prime.
2. $5^2 + 5 + 11 = 25 + 5 + 11 = 41$.
3. "If a number is even, then it ends in $0$." This converse is false: $4$ is even and does not end in $0$.

Look at the smallest prime. It has no factor other than $1$ and itself.

Write down the first few primes and check each one against the word "odd".

The number $2$ is prime, because its only factors are $1$ and $2$. The assumption holds. But $2$ is even, so it is not odd, and the conclusion fails. The claim is false.

The number $2$ is the only even prime. It is the counter example, and after it every prime is odd.

The constant is $11$. Look for a value of $n$ that makes the constant combine with the other terms into a square.

Try $n = 11$, and work out $11^2 - 11 + 11$.

Take $n = 11$. Then

But

so the value is not prime. The assumption holds, since $11$ is a whole number, and the conclusion fails. The claim is false.

Squaring hides the sign. Try making $x$ negative and larger in size than $y$ but less than $y$ as a number.

Try $x = -3$ and $y = 2$. Work out $x^2$ and $y^2$, then compare $x$ and $y$.

Take $x = -3$ and $y = 2$. Then

so $x^2 > y^2$, and the assumption holds. But $x = -3 < 2 = y$, so the conclusion fails. The claim is false.

Look again at the smallest prime, the one that is even.

Add $2$ and $3$, then decide whether the total is even.

Take the primes $2$ and $3$. Both are prime, so the assumption holds. Their sum is

which is odd, so the conclusion fails. The claim is false. It fails because one of the two primes can be $2$, the only even prime.

Test a number between $0$ and $1$. Multiplying by another factor smaller than $1$ makes the result smaller.

Try $x = \frac{1}{2}$. Work out $x^2$ and $x^3$, then compare them.

Take $x = \frac{1}{2}$. Then

Now

so $x^3 > x^2$ is false. The assumption holds, since $\frac{1}{2}$ is a real number, and the conclusion fails. The claim is false.

A right angle needs Pythagoras: the square of the longest side must equal the sum of the squares of the other two.

Take the sides $2$, $3$ and $4$, and compare $4^2$ with $2^2 + 3^2$.

Take the triangle with sides $2$, $3$ and $4$. The longest side is $4$, so a right angle would need

But

and $16 \neq 13$. The sides are whole numbers, so the assumption holds, and there is no right angle, so the conclusion fails. The claim is false.