Rational + irrational = always irrational?

Question

I had a little back and forth with my logic professor earlier today about proving a number is irrational. I proposed that 1 + an irrational number is always irrational, thus if I could prove that 1 + irrational number is irrational, then it stood to reason that was also proving that the number in question was irrational.

Eg. $\sqrt2 + 1$ can be expressed as a continuous fraction, and through looking at the fraction, it can be assumed $\sqrt2 + 1$ is irrational. I suggested that because of this, $\sqrt2$ is also irrational.

My professor said this is not always true, but I can't think of an example that suggests this.

If $x+1$ is irrational, is $x$ always irrational?

Actually, a better question is: if $x$ is irrational, is $x+n$ irrational, provided $n$ is a rational number?

Answer 1

Suppose $x$ is irrational and $x+\dfrac pq=\dfrac mn$ then, $x=\dfrac mn-\dfrac pq=\dfrac{mq-np}{nq}$ so, $x$ would then be rational. :)

Answer 2

Look at the contrapositive: If $x$ is rational, then $x+n$ is rational. Clearly this is a true statement.

Answer 3

A proof in the style of "mathematics made difficult": Note that a number $r$ is rational if and only if $\mathbb Q(r) = \mathbb Q$. Now it is easy to see that $\mathbb Q(\gamma) = \mathbb Q(r+\gamma)$ for all rational $r$ and arbitrary $\gamma$. So if $r+\gamma$ is irrational, then $\gamma$ is also.

Answer 4

In fact, for any rational number $ r $ it is true that the irrationality of $ x+r $ implies the irrationality of $ x $. This is due to the fact that the rationals are closed under addition. Assume that $ x+r $ is irrational and (for contradiction) that $ x $ is rational, by the fact that the rationals are closed under addition ($\mathbb {Q}$ is a field) you get that $ x+r $ is rational. Contradiction.

Answer 5

Note that the sum of two rationals is always rational, and that if $n$ is rational then $-n$ is rational. Now suppose that $x$ is any number and $n$ is rational.

Suppose $x+n$ is rational, then $(x+n)+(-n)$ is rational. Therefore $x+(n+(-n))$ is rational. Therefore $x+0$ is rational, and finally $x$ is rational.

Answer 6

Yes, it is true that $x+1$ being irrational implies $x$ is irrational. Given that $x+1$ is irrational, assume $x=\frac ab$ with $a,b$ integers. Then $x+1=\frac {a+b}b$ would be rational as well.

Answer 7

The rational numbers are closed under addition and subtraction. Let $w$ be any irrational number and $r$ be a rational number. Since $$ (r + w) - r = w$$ and $w$ is irrational, one of the subtrahends here is irrational. Since $r$ is rational, the irrational quantity must by $r + w$.

Answer 8

This follows essentially from the fact that as additive abelian groups $\mathbb Q$ is a normal subgroup of $\mathbb R$, which is true since both groups are abelian and the latter contains the former. Hence we have a quotient homomorphism $\varphi: \mathbb R \rightarrow \mathbb R / \mathbb Q$. $r \in \mathbb R$ is rational iff $\varphi (r) = 0_{\mathbb R / \mathbb Q}$, where $0_{\mathbb R / \mathbb Q}$ is the identity element of $\mathbb R / \mathbb Q$.

Consider then for $x \in \mathbb R - \mathbb Q$, $r \in \mathbb Q$, that \begin{align*} \varphi (x+r)&= \varphi(x) + \varphi(r) \\ &= \varphi(x) + 0_{\mathbb R / \mathbb Q} \\ &= \varphi(x) \\ & \ne 0_{\mathbb R / \mathbb Q} \text{ by assumption that }x \not\in \mathbb Q, \end{align*}

so $x+r \not \in \mathbb Q$.

Of course, the above is just reinterpreting the above elementary proofs in a more general context, but this lets us apply the same line of reasoning to a wide variety of things including modular arithmetic, rotations/reflections of geometric objects, Rubik's cube moves, matrix multiplication, permutations of sets of numbers, exotic differentiable structures on spheres, ...

Answer 9

Let x be rational and y be irrational. So let us assume that x + y is rational.
(x+y) - x will also be a rational since rational - rational is always rational.
therefore y is also rational. but that is a contradiction. Hence proved!

Answer 10

Lemma:

If $a$ is rational, and $b$ is rational, then $a+b$ is rational.

Proof:

Let $a=p/q$, and $b=r/s$, where $p,q,r,s$ are integers. Then, $a+b=(ps+qr)/(qs)\blacksquare$

Now, let $a$ be rational, so that $-a$ is rational. If $a+x$ is rational, then $x=(a+x)+(-a)$ is rational, by the lemma. The desired result is the contrapositive: if $x$ is irrational, then $a+x$ cannot be rational, for then $x$ would be rational, which is a contradiction; hence, if $x$ is irrational, then $a+x$ is irrational.

This is a proof by contrapositive: when you prove an implication $P\implies Q$ by first proving that $\neg Q\implies \neg P$. Here, we assume that $a$ is rational. Then, we prove that $$ \text{$a+x$ is rational} \implies \text{$x$ is rational} $$ Then, to prove the implication $$ \text{$x$ is irrational} \implies \text{$a+x$ is irrational} $$ we assume, for the sake of contradiction, that if $x$ is rational, then $a+x$ can be rational. This would imply that $x$ is irrational, which is a contradiction. Hence, $a+x$ is irrational.

Proofs by contrapositive are similar in nature to proofs by contradiction, except that the proof of $\neg Q\implies\neg P$ is a direct proof. This means that any intermediate results deduced using this direct proof can be used in other proofs. This is why, generally speaking, mathematicians favour contraposition over contradiction. See here for further discussion.

Answer 11

Let $j$ be an irrational number, and $x$ a rational number, such that $x+1=j$.

There are coprime integers $z_1$ and $z_2$ such that $x=\frac{z_1}{z_2}$

Substitute| $$\frac{z_1}{z_2}+1=\frac{z_1}{z_2}+\frac{z_2}{z_2}=\frac{z_1+z_2}{z_2}=j$$

The sum of two integers is always an integer, so $j$ is shown to be the quotient of two integers, contradicting the assumption that $j$ is irrational.

So no such irrational number exists of the form $x+1$, with $x$ rational.