A-Level Mathematical argument and notation Practice Questions

Free A-Level Mathematical argument and notation practice questions with full step-by-step worked solutions. Covers implication, notation, converse, counterexample. Practise exam-style problems and check your method.

implicationnotationconversecounterexampleequivalenceparity
A-Level70 questionsStep-by-step solutions
Question 1
1 markeasy
Which single symbol correctly fills the box in the statement x=3    x2=9x=3 \;\square\; x^2=9, given that the relationship works in one direction only?
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Worked solution

  1. Read the statement

    x=3    x2=9x=3 \;\square\; x^2=9

    We are told the link only works one way: knowing x=3x=3 lets us conclude x2=9x^2=9, but not the other way round.

  2. Test the forward direction

    x=3x2=9x=3 \Rightarrow x^2=9

    If x=3x=3 then squaring gives 99, so the arrow from left to right is true.

  3. Test the reverse direction

    x2=9x=3 is false (x could be 3)x^2=9 \Rightarrow x=3 \text{ is false } (x \text{ could be } -3)

    The reverse fails because xx could be 3-3. So this is NOT a two-way \Leftrightarrow.

  4. Choose the symbol

    \Rightarrow

    A one-way link uses the implication arrow \Rightarrow (‘implies’).

Answer
\Rightarrow
Question 2
2 markseasy
In the line ‘3x=123x=12, hence x=4x=4’, is the word ‘hence’ used correctly?
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Worked solution

  1. Read the argument

    3x=12    x=43x=12 \;\Rightarrow\; x=4

    ‘Hence’ signals that the second line follows logically from the first.

  2. Check the deduction

    3x3=123x=4\frac{3x}{3}=\frac{12}{3} \Rightarrow x=4

    Dividing both sides by 33 genuinely gives x=4x=4, so the conclusion does follow.

  3. Conclude

    Yes, used correctly\text{Yes, used correctly}

    ‘Hence’ is correct because x=4x=4 is a valid consequence of 3x=123x=12.

Answer
Yes, used correctly
Question 3
3 marksintermediate
For real numbers a,ba,b, which connective correctly completes ab=0    (a=0 or b=0)ab=0 \;\square\; (a=0 \text{ or } b=0)?
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Worked solution

  1. Forward direction

    ab=0a=0 or b=0ab=0 \Rightarrow a=0 \text{ or } b=0

    If a product of two real numbers is zero, at least one factor must be zero (the zero-product property).

  2. Reverse direction

    (a=0 or b=0)ab=0(a=0 \text{ or } b=0) \Rightarrow ab=0

    If either factor is zero then the product is zero, so the reverse holds too.

  3. Both directions hold

    \Leftrightarrow

    Since each side implies the other, the link is a full equivalence.

  4. Why it matters

    basis of solving by factorising\text{basis of solving by factorising}

    This equivalence is exactly what lets us solve (x2)(x3)=0(x-2)(x-3)=0 by setting each bracket to zero.

  5. Note the real-number condition

    needs no zero divisors\text{needs no zero divisors}

    This relies on the reals having no ‘zero divisors’; it would fail in some other number systems.

  6. Conclude

    ab=0(a=0 or b=0)ab=0 \Leftrightarrow (a=0 \text{ or } b=0)

    The correct connective is \Leftrightarrow.

Answer
\Leftrightarrow
Question 4
4 markshard
Is the implication ‘for all real x,yx,y: x>yx2>y2x>y \Rightarrow x^2>y^2’ true? Justify your answer.
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Worked solution

  1. Read the claim

    x>yx2>y2x>y \Rightarrow x^2>y^2

    It claims that a bigger number always has a bigger square.

  2. Try positive numbers

    x=3,y=2:9>4  x=3,y=2: 9>4 \;\checkmark

    For positive values it seems to work.

  3. Test with negatives

    x=1,y=2x=1, y=-2

    The danger with squaring is negative numbers, so test x=1x=1, y=2y=-2.

  4. Check the hypothesis

    1>2  1 > -2 \;\checkmark

    Here x>yx>y is true.

  5. Check the conclusion

    x2=1,  y2=4x^2=1, \; y^2=4

    But x2=1x^2=1 and y2=4y^2=4.

  6. Compare the squares

    141 \not> 4

    The conclusion x2>y2x^2>y^2 is FALSE here.

  7. Identify the counterexample

    (x,y)=(1,2)(x,y)=(1,-2)

    This single case breaks the implication.

  8. Explain the reason

    squaring ignores sign\text{squaring ignores sign}

    Squaring depends on magnitude, not order, so ordering is not preserved when negatives are involved.

  9. State the corrected version

    x>yx2>y2|x|>|y| \Rightarrow x^2>y^2

    The correct statement compares magnitudes, not the numbers themselves.

  10. Conclude

    False\text{False}

    The implication is false; a counterexample such as x=1,y=2x=1,y=-2 disproves it.

Answer
False
Question 5
5 markschallenging
‘All primes are odd. 22 is prime. Therefore 22 is odd.’ The conclusion is false, yet the argument is logically valid in form. Explain the situation precisely.
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Worked solution

  1. Symbolise the argument

    primeodd;  2 prime;  2 odd\text{prime} \Rightarrow \text{odd}; \; 2 \text{ prime}; \; \therefore 2 \text{ odd}

    This is a valid deductive form (modus ponens).

  2. Check the logical form

    PQ,  PQP\Rightarrow Q,\; P \vdash Q

    If the premises were true, the conclusion would follow — so the FORM is valid.

  3. Test the first premise

    all primes are odd?\text{all primes are odd?}

    Examine whether ‘all primes are odd’ is actually true.

  4. Find a counterexample

    2 is prime and even2 \text{ is prime and even}

    22 is a prime number that is even, so the first premise is FALSE.

  5. Distinguish validity from truth

    validtrue premises\text{valid} \ne \text{true premises}

    Validity is about the structure; it does not guarantee the premises are true.

  6. Define soundness

    sound=valid+true premises\text{sound} = \text{valid} + \text{true premises}

    An argument is sound only if it is valid AND all premises are true.

  7. Classify this argument

    valid but not sound\text{valid but not sound}

    Here the form is valid, but a false premise makes it unsound, so the conclusion can be false.

  8. Explain the false conclusion

    garbage in, garbage out\text{garbage in, garbage out}

    A valid argument from a false premise can produce a false conclusion.

  9. State precisely

    valid, unsound; premise ‘all primes odd’ is false\text{valid, unsound; premise ‘all primes odd’ is false}

    The argument is valid in form but unsound because the premise is false.

  10. Define a valid argument

    true premisestrue conclusion\text{true premises} \Rightarrow \text{true conclusion}

    Validity means: IF the premises were true, the conclusion would have to be true.

  11. Define a sound argument

    valid+true premises\text{valid} + \text{true premises}

    Soundness additionally requires the premises actually to be true.

  12. Identify the modus ponens form

    PQ,  PQP\Rightarrow Q,\; P \vdash Q

    The argument follows the valid modus ponens pattern.

  13. Test the questionable premise

    are all primes odd?\text{are all primes odd?}

    We must check whether ‘all primes are odd’ is true.

  14. Produce the counterexample

    2 is prime and even2 \text{ is prime and even}

    22 is prime yet even, so the premise is false.

  15. Draw the distinction

    valid but unsound\text{valid but unsound}

    Hence the argument is valid in form but unsound, so a false conclusion is possible.

Answer
Valid in form but unsound: the premise ‘all primes are odd’ is false

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