This is "problem" 6 in Sheldon Axler's Precalculus book, page $6$ chapter $0$. I do not quite understand how I could solve a question like this in a precalculus course. I'm usually just given a bunch of exercises questions. So my answer was the following:

Suppose $\sqrt 2$ is an irrational number and $0$ is a rational number. Because $$\sqrt 2 = \sqrt 2 + 0 $$ the sum of an irrational number and a rational number is an irrational number.

Is this correct?


closed as off-topic by Shailesh, Saad, Namaste, Isaac Browne, Aweygan May 23 '18 at 2:37

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  • 4
    $\begingroup$ An example is NOT a proof. $\endgroup$ – Arnaud Mortier May 23 '18 at 0:19
  • $\begingroup$ @ArnaudMortier so using variables is the correct way of writing a proof? $\endgroup$ – Alexander John May 23 '18 at 0:21
  • $\begingroup$ Well, no.... what if you the rational number is anything other than $0$. Or the irrational is anything other than $\sqrt 2$. $\endgroup$ – fleablood May 23 '18 at 0:24
  • $\begingroup$ "so using variables is the correct way of writing a proof?" Um... no. Writing an argument that convinces someone a statement is always true is the correct way of writing a proof. Looking for a cookbook list of directions on how to do that is the wrong way to do it. $\endgroup$ – fleablood May 23 '18 at 0:25
  • $\begingroup$ @AlexanderJohn It will not necessarily always work like this, but it is still worth giving a try. On the contrary, just looking at examples will never provide a complete proof unless there is a finite amount of cases to consider. Here, there are infinitely many numbers of both types, that's too many to check them all. $\endgroup$ – Arnaud Mortier May 23 '18 at 0:29

If you have two rational numbers, their difference (one minus the other) must also be rational. This can be proved easily by reduction to the definition of rationality, since a difference between two ratios of integers can always be written as a single ratio of integers. From this, it follows that it is impossible to get a rational number as the sum of a rational and an irrational; if this were possible, you could re-arrange this to get an irrational as the difference between two rationals.

Theorem: If $r$ is rational and $z+r$ is rational then $z$ must be rational.

Proof: Since $r$ is rational, this means there are integers $a$ and $b \neq 0$ such that $r = a/b$. Since $z+r$ is also rational, this means there are integers $c$ and $d \neq 0$ such that $z+r = c/d$. Hence, we have:

$$z = z+r - r = \frac{a}{b} - \frac{c}{d} = \frac{ad-bc}{bd}.$$

Since $a,b,c,d$ are all integers (and $b \neq 0$, $d \neq 0$), the numerator $ad-bc$ is an integer, and the denominator $bd \neq 0$ is a non-zero integer. Hence, $z$ can be written as a ratio of integers, and so $z$ is rational. $\blacksquare$

Corollary: If $z$ is irrational and $r$ is rational then $z+r$ is irrational.

Proof: Follows trivially from the above theorem using a proof-by-contradiction. $\blacksquare$


Suppose $z$ is irrational and $r$ is rational. Were $z + r$ rational, $z = z + r - r$ is also rational. That is a contradiction. $z + r$ is not rational.

  • $\begingroup$ Worth emphasising that this is because the set of rationals is closed under addition/subtraction. $\endgroup$ – Penguino May 23 '18 at 0:22
  • $\begingroup$ I'm not sure if this answers the question, since it appears the question is why the proof isn't good, not "what is the proof." $\endgroup$ – Jeremy Upsal May 23 '18 at 0:28

Let $a $ be an irrational and $r $ a rational $=\frac {p}{q}. $ Assume $a+r $ is a rational $=\frac {m}{n} $. Then

$$a+\frac {p}{q}=\frac {m}{n} $$ and

$$a=\frac {m}{n}-\frac {p}{q}=\frac {mq-pn}{qn}\in \mathbb Q $$

which is false.


The thing to realize is that the sum/difference/product/quotient of two rational numbers are always rational (if you aren't dividing by zero).

Because for integers $a,b,c,d; b\ne 0, d\ne 0$ then $\frac ab \pm \frac cd = \frac {ad \pm bc}{cd}$ and $ad \pm bc, cd$ are integers. Likewise $\frac ab\times \frac cd = \frac {ac}{bd}$ and $ac,bd$ are integers. And if $c\ne 0$ then $\frac {\frac ab}{\frac cd}=\frac {ad}{bc}$ and $ad,bc$ are integers.

So if $r$ and $z + r$ are rational then $(z + r) -r = z$ is rational. So $z$ irrational, $r$ rational means $z+r$ being rational is impossible.


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