MYP 3 · Maths

LAWS OF ALGEBRA

The zero index law

QUESTION 1 [4 marks] — Criterion A Medium
Evaluate each expression:
a. $5^0$
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$$1$$
b. $(-3)^0$
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$$1$$
c. $4x^0$ (where $x\ne0$)
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$$4 \times 1 = 4$$
d. $(4x)^0$ (where $x\ne0$)
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$$1$$
QUESTION 2 [3 marks] — Criterion B Medium
Investigate WHY any non-zero number raised to the power 0 equals 1, using the index law for division.
a. Simplify $\dfrac{x^3}{x^3}$ two ways: (i) using the division index law $\dfrac{x^m}{x^n}=x^{m-n}$, and (ii) by direct cancellation (since anything divided by itself is 1).
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(i) $x^{3-3}=x^0$. (ii) $\frac{x^3}{x^3}=1$ directly, since any nonzero number divided by itself is 1.
b. Since both methods calculate the SAME expression, what must $x^0$ equal?
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Since both results describe $\frac{x^3}{x^3}$, we must have $x^0=1$.
QUESTION 3 [4 marks] — Criterion C Medium
A student thinks $x^0=0$, reasoning that 'anything to the power of nothing should give nothing'.
a. Explain why this reasoning is incorrect, using the division-based proof.
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The 'power of nothing gives nothing' intuition is a misleading guess, not based on any actual rule. The real reason comes from the index laws: $\frac{x^n}{x^n}=x^{n-n}=x^0$, but also $\frac{x^n}{x^n}=1$ directly — so $x^0$ must equal 1, not 0.
b. State the one exception where the zero index law does NOT apply, and explain why.
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The law does not apply when the base is 0 (i.e. $0^0$ is undefined) — this is a special case because $\frac{0^n}{0^n}$ involves division by zero, which is never allowed.
QUESTION 4 [3 marks] — Criterion D Medium
A scientist writes a very small measurement as $3.2 \times 10^{-4}$ metres, and a colleague asks what $10^0$ would represent if the exponent were 0 instead.
a. Evaluate $3.2 \times 10^0$.
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$$3.2 \times 1 = 3.2$$
b. Explain what this tells you about the role of $10^0$ in scientific notation generally.
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$10^0=1$ means that when the exponent is 0, the number is simply left unchanged (multiplying by 1 does nothing) — it represents 'no scaling at all', which is why $10^0$ acts as the 'neutral' starting point in the powers-of-ten scale.
QUESTION 5 [5 marks] — Criterion A Medium
Evaluate each expression:
a. $8^0 + 3^0$
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$$1+1=2$$
b. $7x^0$ (where $x\ne0$)
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$$7\times1=7$$
c. $(7x)^0$ (where $x\ne0$)
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$$1$$
d. $5^0 \times 5^3$
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$$1\times125=125$$
QUESTION 6 [5 marks] — Criterion B Medium
Investigate why $(ab)^0=1$ but $ab^0$ (without brackets) can give a DIFFERENT result, using $a=3, b=5$.
a. Evaluate $(3\times5)^0$ and $3\times5^0$ separately.
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$(3\times5)^0=15^0=1$. $3\times5^0=3\times1=3$.
b. Explain why these give different results, referring to WHAT exactly the zero exponent applies to in each case.
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In $(3\times5)^0$, the exponent 0 applies to the ENTIRE product $(3\times5)=15$, giving $15^0=1$. In $3\times5^0$ (no brackets), the exponent 0 applies ONLY to the 5, giving $5^0=1$, which is then multiplied by the separate factor of 3, giving $3\times1=3$ — brackets determine exactly what the exponent 'reaches', and this changes the result significantly.
QUESTION 7 [5 marks] — Criterion C Hard
A student believes $0^0=1$ 'because everything to the power 0 is 1', while another believes $0^0=0$ 'because 0 to any power is 0'.
a. Explain why BOTH students' reasoning, while each individually plausible-sounding, leads to a genuine mathematical CONTRADICTION when applied to the special case $0^0$.
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The first student's rule ('anything to power 0 is 1') and the second student's rule ('0 to any power is 0') are both generally true patterns — but they directly CONTRADICT each other specifically at $0^0$, since it can't be BOTH 1 and 0 simultaneously. This is exactly why $0^0$ is typically treated as a special, undefined (or context-dependent) case in mathematics, rather than following either general pattern automatically.
b. State the correct convention taught at this level: is $x^0=1$ valid for ALL values of $x$, or does it require an exception?
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The rule $x^0=1$ specifically requires $x\ne0$ — it is valid for every nonzero value of $x$, but the case $x=0$ (i.e. $0^0$) is excluded as a special undefined case, precisely because of the contradiction shown in part (a).
QUESTION 8 [2 marks] — Criterion C Medium
A classmate simplifies $\dfrac{x^5}{x^5}$ using the division law and gets $x^0$, then writes: 'so the answer is $x^0$', without simplifying further.
a. Explain why '$x^0$' is not fully simplified, and complete the simplification to a single number.
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$x^0$ is not a FINAL simplified numerical answer — since $x\ne0$ is implied (as $\frac{x^5}{x^5}$ requires $x\ne0$ to be defined), $x^0=1$, and a fully simplified answer should state this final numerical value, not leave it in index form when it evaluates to a specific constant.
QUESTION 9 [5 marks] — Criterion D Medium
A chemistry experiment involves a substance where the concentration formula includes a term $C_0 \times k^0$ at time zero (before any reaction has started), where $k$ is the reaction rate constant.
a. Explain, using the zero index law, why the term $k^0$ always simplifies to 1 REGARDLESS of the specific numerical value of the reaction rate constant $k$ (as long as $k\ne0$).
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By the zero index law, ANY nonzero base raised to the power 0 equals 1 — this holds true no matter what specific value $k$ takes (whether $k=0.5$, $k=3.7$, or any other nonzero rate constant), so $k^0=1$ universally in this formula.
b. Given this, simplify the full expression $C_0\times k^0$, and explain what this simplified result tells a chemist about the concentration at time zero, REGARDLESS of the specific reaction rate.
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$C_0\times k^0=C_0\times1=C_0$. This tells the chemist that the concentration at time zero is simply $C_0$ (the initial concentration), completely UNAFFECTED by the reaction rate constant $k$ — which makes physical sense, since no reaction has had time to occur yet at $t=0$, so the rate at which the reaction proceeds shouldn't influence the STARTING concentration at all.