A University-Level Understanding of Number with an eye towards elementary education

\(60=10_{time}\) because 60 seconds = 1 minute (one “ten” in mixed radix); \(3600=100_{time}\) since 3600 seconds = 60 minutes = 1 hour.

\(1234\) (1 day, 2 h, 3 min, 4 s) in seconds: \(1\cdot 86400+2\cdot 3600+3\cdot 60+4=93784\text{.}\)

No, there is no zero, and place does not mean anything (as \(1^a=1\) for any natural number \(a\text{.}\)

Interpretation (consistent mixed-radix): radices increase by 3 each level starting from 5, so weights are \(w_1=1\text{,}\) \(w_2=5\text{,}\) \(w_3=5\cdot 8=40\text{,}\) \(w_4=5\cdot 8\cdot 11=440\text{,}\) \(w_5=5\cdot 8\cdot 11\cdot 14=6160\text{.}\) Allowed digits: 1st: \(0..4\text{,}\) 2nd: \(0..7\text{,}\) 3rd: \(0..10\text{,}\) 4th: \(0..13\text{,}\) etc.

Note. If one instead reads “4th place: up to ten 3rd places” literally (max digit 10), many numbers would become unrepresentable without the next place; the mixed-radix interpretation above ensures full coverage.

No. Tally marks are not positional: the value does not depend on place, only on count. Thus they are not a place-value system (though they form a unary counting system).

Example system: “Dozens-and-sixes” mixed radix. Digits allowed: 0–Z with \(Z=35\text{.}\) Places (right→left): ones (\(1\)), sixes (\(6\) ones), dozens (\(12\) sixes = \(72\) ones), grosses (\(12\) dozens), etc. Examples:

\(51\) means \(5\cdot 6+1=31\) ones. \(2:30\) (i.e., \(230\)) means \(2\cdot 72+3\cdot 6+0=162\text{.}\) \(1:0:4\) means \(1\cdot 12\cdot 72+0\cdot 72+4\cdot 6= 864+24=888\text{.}\) \(10\) equals one six: \(6\text{.}\) \(1:00\) equals one dozen of sixes: \(72\text{.}\)

Addition example: \(51+25 = (5\cdot 6+1)+(2\cdot 6+5)=7\cdot 6+6= (1\ \text{carry})\ 16 \rightarrow 114\) (i.e., \(1\cdot 6^2+1\cdot 6+4= 36+6+4=46\) in base-ten).

With 10 fingers and each finger as one “digit”: base two counts to \(2^{10}-1=1023\text{;}\) base three to \(3^{10}-1=59048\text{;}\) base four to \(4^{10}-1=1{,}048{,}575\) (assuming you can reliably distinguish 2 or 3 or 4 states per finger).

\(77,\;500,\;126,\;64,\;0\)

\(37+25+63+75=200\text{,}\) \(250\cdot 16=4000\text{,}\) \((300-1)\cdot 48=14352\text{.}\)

Associative; Zero property; Distributive; Identity; Commutative.

Therefore, \((-1)(-a)=a\text{.}\)

Fill-ins: \(-1+ \big((-1)(-1)\big)=0\text{,}\) hence \(((-1)(-1))=1\text{.}\)

All equalities hold: \(-28=-28\text{,}\) \(-30=-30\text{,}\) \(30=30\text{.}\)

Place matters. In \(108_{10}\text{,}\) the \(8\) contributes \(8\cdot 10^0=8\text{.}\) In \(108_{9}\text{,}\) it contributes \(8\cdot 9^0=8\) as well, but the \(1\) is \(1\cdot 9^2=81\) so the whole value differs. In \(108_{2}\) the digit \(8\) is invalid (base two allows only 0 and 1). Thus the same glyph “8” can mean different things—or be illegal—depending on base.

Blocks: in base five, a “long” is 5 ones. \(302_{5}\) means 3 longs of longs (i.e., \(3\cdot 25\)), 0 longs, 2 ones → \(77_{10}\text{.}\) \(32_{5}\) means 3 longs and 2 ones → \(17_{10}\text{.}\) Regrouping pictures cannot turn three 25-blocks into three 5-blocks.

Dividing by 5 with remainders peels off the least-significant base-five digit: the remainder is how many ones remain after forming groups of 5; the quotient counts how many groups. Repeating with the quotient mirrors regrouping ones→longs→squares: each step counts how many blocks of the next place you can make, producing the same digits in reverse order.

Agree: \(1000-398=(1000+2)-(398+2)=1002-400\text{.}\) On a number line, shifting both endpoints right by the same amount preserves the distance (difference). This works for adding the same number to both terms (or subtracting the same number), but not for arbitrary unequal changes.

Both use distributivity: A expands \(27(10+6)=27\cdot 10+27\cdot 6\text{.}\) B uses \((30-3)16=30\cdot 16-3\cdot 16\text{.}\) Commutativity/associativity justify rearrangements; each yields \(432\text{.}\)

If \(b\) is even, \((2a)\cdot \dfrac{b}{2}=a\cdot (2\cdot \tfrac{b}{2})=ab\) by associativity. If \(b\) is odd, halving leaves a fraction (e.g., \(24\cdot 75=(48)\cdot 37.5\)), so in integer-only contexts you must adjust differently (e.g., double one factor and “almost” halve the other using distributivity).

An identity leaves elements unchanged (\(0\) for addition; \(1\) for multiplication). An inverse “undoes” an element: additive inverse of \(-5\) is \(5\text{;}\) of \(0\) is \(0\text{;}\) of \(1\) is \(-1\text{.}\) Multiplicative inverses within \(\mathbb{Z}\) exist only for \(\pm 1\) (e.g., \(-5\) has none in \(\mathbb{Z}\text{,}\) while in \(\mathbb{Q}\) its inverse is \(-1/5\)).

Partition a rectangle of height \(a\) into widths \(b\) and \(c\text{.}\) The total area \(a(b+c)\) equals the sum of the two sub-rectangles \(ab+ac\text{.}\) With two cuts, one vertical (at \(b\)/\(c\)) and one horizontal (at \(a\)=\(a_1+a_2\)), the big rectangle splits into four: \((a+b)(c+d)=ac+ad+bc+bd\text{.}\)

\(0\times n=0\) because repeated addition of zero gives zero (and \(n\cdot 0=0\) by distributivity). Division \(n\div 0\) would require a number \(x\) with \(0\cdot x=n\text{,}\) impossible unless \(n=0\text{;}\) hence undefined.

“Subtract a negative” means “add the opposite”: \(-12-(-9)=-12+9=-3\text{.}\) On a number line, start at \(-12\) and move 9 to the right to reach \(-3\text{.}\)

Distribute \(-1\) over \(1+(-1)=0\text{:}\) \(-1\cdot 1+(-1)\cdot(-1)=0\text{.}\) Since \(-1\cdot 1=-1\text{,}\) we have \(-1+((-1)(-1))=0\text{,}\) so \(((-1)(-1))=1\text{.}\) Each step uses identity, distributive property, and additive inverse.

Ordering by absolute value ignores sign. On the number line, larger negatives are less. Correct order: \(-12<-3<2<10\) (since \(-12\) is farthest left).