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I need to prove that the sum, difference, and product of integers are also integers. In the question they said to use the previous exercise which had me proving that the difference between two natural numbers is also natural (of course given that the number being subtracted is less than the one we're subtracting from). And this is how integers are defined in the book:

We define the set of integers, denoted by $\mathbb{Z}$, to be the set of numbers consisting of the natural numbers, their negatives, and the number 0.

The book defines the set of real numbers as a field, then natural numbers as the intersection of all the inductive subsets of $\mathbb{R}$.

Now my first idea is to do it in cases, so prove that $a$ and $b$, which are integers, add up to integers when they are both positive, then one of them is zero, then one of them is negative and so on. I feel like this is more work than what's intended, also I don't have an idea of what to do when I get to the case where one of them is negative.

Frankly the definition is what's giving me trouble, and I don't know how deep I should go with the proof. Any ideas?

By the way, the book is Fitzpatrick's Advanced Calculus. And the exercise is 9 from section 1.1 . Also I would appreciate if someone can tag this with the proper tags, as I don't know which are appropriate.

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    $\begingroup$ I need to know how sum, product, etc, are defined. $\endgroup$ – steven gregory Jul 1 '16 at 19:35
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    $\begingroup$ Also I'm really interested in how you proved that the difference of two naturals is natural because $3-5$ is very much unnatural. $\endgroup$ – Squirtle Jul 1 '16 at 19:36
  • $\begingroup$ I would suggest spending more time thinking about how to prove each section, and then taking a look at this example: foolproof.pbworks.com/w/page/8978143/… $\endgroup$ – RonaldB Jul 1 '16 at 19:36
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    $\begingroup$ @Squirtle I edited the question. $\endgroup$ – Jean Jul 1 '16 at 19:38
  • $\begingroup$ @StevenGregory I'm sorry but what do you mean? The book starts with defining the real numbers as a field, and then natural numbers as the intersection of all the inductive subsets of the real numbers. $\endgroup$ – Jean Jul 1 '16 at 19:40
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Just do them by cases:

Case 1 $x$ natural; $y$ natural.

$x + y$ and $xy$ are natural and therefore integers.

If $x >y$ then $x -y$ is natural and there fore an integer. If $x = y$ then $x - y = 0$ and therefore an integer. If $x < y$ then $y-x$ is natural and $x - y = -(y-x)$, a negative of a natural and therefor an integer.

Case 2 $y$ is 0. (the same if $x$ is 0).

Then $x + 0 = 0 + x = x$ an integer.

$x - 0 = x$ and integer. $0 - x = -x$; if $x$ is natural that is an integer-- if $x = 0$ then $-x = -0 = $, an integer-- and if $x$ is negative then $x = -x'$, $x'$ natural, so $-x = -(-x') = x'$ natural and an integer.

$x0 = 0x = 0$ an integer.

Case 3: $x$ is natural and $y = -y'$ the negative of a natural. (Or vice versa)

$y + x = x + y = x - (-y')$ is the difference of two naturals and therefore an integer.

$xy = x(-y') = -(xy')$ is the negative of the product of two naturals so is an integer.

$x - y = x + y'$ the sum of two naturals and thus an integer.

$y - x = -(y' + x)$ the negative of the sum of two naturals, so the negative of a natural, so an integer.

Case 4: $x = -x'$ and $y = -y'$; $x'$ and $y'$ are natural.

$x + y = -(x'+y')$ then negative of a natural.

$xy = (-x')(-y') = x'y'$ a natural.

$x - y = y' - x'$ the difference of two naturals and thus an integer.

$y-x = x' - y'$ ditto.

Thus we are done.

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  • $\begingroup$ Thank you very much for taking the time! $\endgroup$ – Jean Jul 2 '16 at 1:00

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