HA6 - Proof with algebra

HA6 - Proof with algebra

Learn to represent arbitrary integers and construct algebraic proofs about odd, even and divisible numbers. This develops the general proofs in HA6, including stating why a factor proves divisibility and using a counterexample to reject a false universal claim.

Representing integers

Is the product of any two odd integers odd, even when one is negative? Examples suggest an answer, but a proof must include every allowed pair. We need expressions that represent those pairs without choosing particular numbers.

An even integer can be written 2n2n, where nn is an integer. An odd integer can be written 2n+12n+1: one more than an even integer. Negative integers are included; for n=−2n=-2, these expressions give −4-4 and −3-3.

General integer representations

With integer parameters:

Integers describedGeneral representation
Any integernn
Any even integer2n2n
Any odd integer2n+12n+1
Three consecutive integersn, n+1, n+2n,\ n+1,\ n+2
Three consecutive even integers2n, 2n+2, 2n+42n,\ 2n+2,\ 2n+4
Three consecutive odd integers2n+1, 2n+3, 2n+52n+1,\ 2n+3,\ 2n+5

Consecutive integers differ by 11 and alternate between even and odd. Consecutive odd integers differ by 22, since an even integer lies between them. The letter must be declared an integer: 2n2n need not be even if nn is allowed to be, for example, 0.50.5.

Worked example 1: consecutive integers

Question. Write and simplify an expression for the sum of three consecutive integers, starting with nn.

Plan. Represent the next two integers by adding 11 and 22, then collect terms.

n+(n+1)+(n+2)=n+n+n+1+2use the three consecutive values=3n+3collect terms\begin{aligned} n+(n+1)+(n+2)&=n+n+n+1+2 &&\text{use the three consecutive values}\\ &=3n+3 &&\text{collect terms} \end{aligned}

Answer. 3n+33n+3, where nn is any integer.

Check. At n=−2n=-2, the integers are −2,−1,0-2,-1,0, with sum −3-3. The expression gives 3(−2)+3=−33(-2)+3=-3.

Now require all three integers to be odd. The step becomes 22, so n,n+1,n+2n,n+1,n+2 would no longer represent the required list.

Worked example 2: consecutive odd integers

Question. Write and simplify the sum of three consecutive odd integers, starting with 2n+12n+1.

Plan. Add 22 for each next odd integer, keeping the same integer parameter.

(2n+1)+(2n+3)+(2n+5)=6n+9collect the three terms and constants\begin{aligned} (2n+1)+(2n+3)+(2n+5)&=6n+9 &&\text{collect the three terms and constants} \end{aligned}

Answer. 6n+96n+9, where nn is an integer.

Check. With n=−2n=-2, the odd integers are −3,−1,1-3,-1,1, whose sum is −3-3. The expression gives −12+9=−3-12+9=-3.

Two arbitrary odd integers need not be consecutive or equal. Represent them by 2m+12m+1 and 2n+12n+1, with independent integers m,nm,n. Using the same expression twice would cover only equal odd integers.

Watch out

2n+1,2n+2,2n+32n+1,2n+2,2n+3 are consecutive integers, but the middle one is even. Three consecutive odd integers require gaps of 22: 2n+1,2n+3,2n+52n+1,2n+3,2n+5.

Start at a different odd expression and keep the step correct.

Divisibility and parity

A proof connects a general representation to the property being claimed. Expanding alone may be unfinished: your conclusion must explain what the final form guarantees.

Divisibility and parity

An integer is divisible by a positive integer dd when it equals dd times an integer. Its parity means whether it is even or odd. A form 2K2K is even and 2K+12K+1 is odd when KK is an integer. Zero and negative multiples count: 0=5×00=5\times0 and −15=5×(−3)-15=5\times(-3).

For example, a result 5(2n+1)5(2n+1) is a multiple of 55 because 2n+12n+1 is an integer whenever nn is an integer. The factor 55 supplies the divisor; the integer bracket supplies the quotient.

Worked example 3: a sum of odd integers

Question. Prove that the sum of any two odd integers is even.

Plan. Use independent integer parameters for the two odd integers and show their sum is twice an integer.

Let the odd integers be 2m+12m+1 and 2n+12n+1, where m,nm,n are integers.

(2m+1)+(2n+1)=2m+2n+2collect terms=2(m+n+1)factor out 2\begin{aligned} (2m+1)+(2n+1)&=2m+2n+2 &&\text{collect terms}\\ &=2(m+n+1) &&\text{factor out }2 \end{aligned}

Answer. Since m+n+1m+n+1 is an integer, the sum is twice an integer and is therefore even.

Check. The odd integers −3-3 and 77 have sum 4=2×24=2\times2. This illustrates the result; the algebra covers every pair.

Keep the same two arbitrary odd integers but change the operation to multiplication. The target form is now one more than twice an integer.

Worked example 4: a product of odd integers

Question. Prove that the product of any two odd integers is odd.

Plan. Expand the product, then isolate an even expression plus 11.

For integers m,nm,n:

(2m+1)(2n+1)=4mn+2m+2n+1four products=2(2mn+m+n)+1factor 2 from the first three terms\begin{aligned} (2m+1)(2n+1)&=4mn+2m+2n+1 &&\text{four products}\\ &=2(2mn+m+n)+1 &&\text{factor }2\text{ from the first three terms} \end{aligned}

Answer. 2mn+m+n2mn+m+n is an integer, so the product is of the form 2K+12K+1 and is odd. This answers the opening question for negative as well as positive odd integers.

Check. (−3)×7=−21=2(−11)+1(-3)\times7=-21=2(-11)+1, an odd integer.

For divisibility by a number other than 22, use the same final step: expose that number as a factor and justify the integer quotient.

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