Hard A-Level Mathematical argument and notation Questions

Symbolise the argument

This is a valid deductive form (modus ponens).

Check the logical form

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

Test the first premise

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

Find a counterexample

$2$ is a prime number that is even, so the first premise is FALSE.

Distinguish validity from truth

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

Define soundness

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

Classify this argument

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

Explain the false conclusion

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

State precisely

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

Define a valid argument

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

Define a sound argument

Soundness additionally requires the premises actually to be true.

Identify the modus ponens form

The argument follows the valid modus ponens pattern.

Test the questionable premise

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

Produce the counterexample

$2$ is prime yet even, so the premise is false.

Draw the distinction

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

Follow the first step

Adding $4$ to both sides is valid, so $x^2=4$ is correct.

Examine the final step

Square-rooting must allow both signs, so $x=2$ alone is incomplete.

Recover the missing root

The correct deduction is $x=2$ or $x=-2$.

Verify x=-2

Substituting $x=-2$ satisfies the original equation.

Name the flaw

The word ‘therefore’ was applied to only one square root, dropping $x=-2$.

Explain the logical point

The correct relationship is an equivalence to $x=\pm2$, not just $x=2$.

Write the solution set

Using set notation records the complete solution.

Interpret

The solution set has two elements, not one.

State the correct conclusion

That is the fully correct statement.

Restate the equation

We start again from the correct first line.

Factorise as a difference of squares

An alternative valid route factorises directly.

Apply the zero-product property

Setting each factor to zero exposes both roots.

Solve each factor

This recovers the two solutions cleanly.

Explain the original slip

Taking only the positive square root is what lost the root $-2$.

Write the full solution set

The complete, correct answer in set notation.

Recall the definition of modulus

The modulus equals the number itself when non-negative, and its negative otherwise.

Reverse direction: assume x \ge 0

By definition, for non-negative $x$ the modulus is $x$ itself. Reverse holds.

Forward direction: assume |x| = x

Now suppose the modulus equals $x$ and deduce the sign of $x$.

Use non-negativity of modulus

The modulus is always $\ge 0$.

Combine

Since $x$ equals a non-negative quantity, $x \ge 0$. Forward holds.

Check the boundary x=0

At $x=0$ both sides are consistent.

Check a negative x

For negative $x$, $|x| \ne x$, and indeed $x\ge0$ is false — consistent.

Both directions established

Each implication holds, so the biconditional is true.

Conclude

The statement $|x|=x \Leftrightarrow x\ge0$ is true for all real $x$.

Recall modulus is non-negative

The modulus never returns a negative value.

Split by definition

The modulus is defined piecewise, so we reason in two cases.

Case non-negative

For $x\ge0$ the modulus equals $x$ itself.

Case negative

For $x<0$ the modulus is $-x$, which differs from $x$.

Combine into the biconditional

Only the non-negative case gives equality, matching both directions.

State the conclusion

The biconditional is therefore true for all real $x$.

Solve the target inequality

First find exactly when $x^2<4$: it is the open interval $(-2,2)$.

Recall 'necessary'

A necessary condition must be implied BY $x^2<4$.

Recall 'not sufficient'

But it must not, on its own, force $x^2<4$.

Test candidate 'x<2'

Every $x$ with $x^2<4$ certainly has $x<2$, so ‘$x<2$’ is necessary.

Check it is not sufficient

But $x<2$ allows $x=-5$, where $x^2<4$ fails; so ‘$x<2$’ is not sufficient.

Reject 'x<3' as too weak? test necessity

‘$x<3$’ is also necessary, but let us compare with the sufficiency requirement.

Reject '-2<x<2' (that is sufficient)

The exact interval is necessary AND sufficient, so it is the wrong answer here.

Reject '0<x<2' (not necessary)

‘$0<x<2$’ is not necessary because negative $x$ can satisfy $x^2<4$.

Reject '|x|<3'? it is necessary not sufficient too

Both ‘$x<2$’ and ‘$|x|<3$’ qualify; pick the offered one.

Identify the correct option

‘$|x|<3$’ is implied by $x^2<4$ (necessary) but $x=2.5$ shows it is not sufficient.

Verify necessity of |x|<3

If $|x|<2$ then certainly $|x|<3$.

Verify not sufficient

So $|x|<3$ does not force $x^2<4$.

Recall necessary vs sufficient

A necessary condition is implied by the target; a sufficient one implies it.

Restate the target set

The target holds exactly on the open interval $(-2,2)$.

Confirm the choice

Since $(-2,2)$ sits strictly inside $\{|x|<3\}$, that condition is necessary but not sufficient.

Assess a direct approach

Starting directly from ‘$ab$ odd’ is awkward, so consider the contrapositive.

Write the contrapositive

Negate both sides and reverse.

Simplify the negations

‘Not both odd’ means at least one is even; ‘not odd’ means even.

Assume at least one is even (WLOG a)

Without loss of generality suppose $a$ is even, so $a=2k$.

Multiply

Then $ab$ is $2$ times an integer.

Conclude the product is even

So $ab$ is even, proving the contrapositive.

Note the symmetric case

The case ‘$b$ even’ is identical by symmetry.

Invoke equivalence

Since a statement and its contrapositive are equivalent, the original is proved.

Conclude

The result follows.

Reject checking examples

Testing $3\times5=15$ etc. does not prove a general statement.

Reject the converse

That is the converse (also true) but it is a different statement.

Why the contrapositive is easier

Starting from ‘$ab$ odd’ is awkward, so we flip to the contrapositive.

Recall the even representation

Any even integer can be written as $2k$ for some integer $k$.

Verify with an example

Testing an even $a$ confirms the product is even.

State the logical justification

Proving the contrapositive proves the original by logical equivalence.