Question bank — Integers — operations, number line, absolute value

True or false, and why? ::: False — $3-5=-2$, which is not natural. Integers exist precisely to close this gap so every subtraction has an answer.

True or false, and why? ::: False — zero is neither positive nor negative. It is the origin of the number line and belongs to $\mathbb{Z}$, but it has no direction, so it is excluded from both ${\mathbb{Z}}^{+}$ and the negatives.

$-3-5$. True or false, and why? ::: False — that rule is only for multiplication/division. Here $-3-5=-3+(-5)=-8$: both are movements left (figure s01), which pile up, not cancel.

$|a|$ can ever be negative. True or false, and why? ::: False — $|a|$ is a distance from zero (figure s03), and distance is never negative. By definition $|a|\ge 0$ for every integer $a$.

$|a|=|b|$ then $a=b$. True or false, and why? ::: False — equal distances allow opposite directions, e.g. $|-5|=|5|$ but $-5\ne 5$. Absolute value forgets sign, so it cannot recover it.

$a-b=b-a$. True or false, and why? ::: False — subtraction is not commutative. $7-10=-3$ but $10-7=3$; only addition and multiplication are commutative.

$-a$ is always a negative number. True or false, and why? ::: False — $-a$ means "the additive inverse of $a$", not "a negative number". If $a=-4$, then $-a=4$, which is positive. The minus sign flips whatever it meets.

$a<b$ means $a$ is the smaller magnitude. True or false, and why? ::: False — $a<b$ means $a$ is to the left of $b$ (figure s01), regardless of magnitude. For example $-9<-2$ even though $9>2$; the more negative number is smaller.

$|a+b|=|a|+|b|$ for all integers. True or false, and why? ::: False in general — this is only an inequality ($\le$). Equality needs $a,b$ to share a sign (or one to be zero). E.g. $|3+(-3)|=0$ but $|3|+|-3|=6$; see 2.1.05-Inequalities for the triangle inequality.

$-3+(-5)=2$ "because two negatives make a positive". Find and fix the error. ::: They applied a multiplication rule to addition. Two leftward moves accumulate (figure s01): $-3+(-5)=-8$. Sign in addition comes from direction of movement, not a blanket rule.

$(-2)-(-6)=-8$. Find and fix the error. ::: They kept the second minus as "more negative". Subtracting a negative adds its opposite: $(-2)-(-6)=(-2)+6=4$ — a move right, not left.

$|3-8|=|3|-|8|=3-8=-5$. Find and fix the error. ::: Absolute value does not distribute over subtraction. Evaluate inside first: $3-8=-5$, then $|-5|=5$. A magnitude can never come out negative.

$8÷0=0$ "since anything with zero is zero". Find and fix the error. ::: Division asks "what $x$ makes $0\times x=8$?" No such $x$ exists, so $8÷0$ is undefined — it is not zero, and confusing it with $0÷8=0$ (which is zero) is the trap.

$-4^2=16$. Find and fix the error. ::: Without brackets the exponent binds tighter than the sign, so $-4^2=-(4^2)=-16$. Only $(-4{)}^2=16$. The bracket decides whether the minus is squared.

$(-3)\times (-5)=15$, then $(-3)+(-5)=15$ too. Find and fix the error. ::: The sign-cancelling behaviour is exclusive to multiplication (figure s02). In addition both moves go left: $(-3)+(-5)=-8$.

$(-)\times (-)=(+)$ forced, not just a convention? Explain. ::: Because $(-3)\times ((-5)+5)=(-3)\times 0=0$. Since $(-3)\times 5=-15$, the other term $(-3)\times (-5)$ must equal $+15$ to keep the sum at $0$. The distributive law leaves no other choice (see the diagonal of figure s02).

define $a-b$ as $a+(-b)$ instead of treating subtraction separately? Explain. ::: So one addition rule handles everything. "Take away $b$" is exactly "move left by $b$" (figure s01), i.e. add $-b$; this also explains why $-(-6)=+6$ flips subtracting-a-negative into adding-a-positive.

$a$ if $a\ge 0$, else $-a$)? Explain. ::: Because distance must come out non-negative (figure s03). A positive input is already its own distance; a negative input needs its sign flipped, and $-a$ (the inverse) does exactly that, e.g. $-(-5)=5$.

$\le$ and not $=$? Explain. ::: When $a,b$ point in opposite directions, part of one move cancels the other, so the final distance $|a+b|$ is less than the total distance walked $|a|+|b|$. Equality only when nothing cancels.

Explain. ::: They have no way to go below a starting point — no negatives and no zero-as-origin. Integers add a left-hand side to the number line so "5 below zero" or "owe $5" have honest addresses.

$a+(-b)$ when $a=b$ exactly (equal magnitudes, opposite signs)? Answer. ::: Exactly $0$. The rightward and leftward moves are the same length (figure s01), so they cancel and land you back on the origin — the tie case of the opposite-signs rule.

$0$ positive, negative, or does it have a sign at all in $|0|$? Answer. ::: $0$ has no sign; it is neither positive nor negative, and $|0|=0$ because it sits at the origin, zero steps away.

Answer. ::: There is no "larger", so the rule's tie-branch applies: the result is $0$. A tie lands you exactly on zero, which carries no sign.

$0÷5$ defined? Is $5÷0$ defined? Answer. ::: $0÷5=0$ is fine (the $x$ with $5x=0$ is $x=0$). But $5÷0$ is undefined — no $x$ satisfies $0\cdot x=5$. Only the divisor being zero breaks things.

$0÷0$ defined? (The subtler zero case.) Answer. ::: No — it is indeterminate. Here $0\times x=0$ holds for every $x$, so there is no single answer; too many candidates is just as fatal as none, so $0÷0$ stays undefined.

$|a|=a$ hold, and for which does $|a|=-a$? Answer. ::: $|a|=a$ whenever $a\ge 0$ (including $a=0$, the shared boundary); $|a|=-a$ whenever $a<0$. Zero satisfies both since $0=-0$.

$|-a|=|a|$ ever fail? Answer. ::: Never — $a$ and $-a$ are mirror images across the origin (figure s03), always the same distance from it. This holds even at $a=0$, where both sides are $0$.