Theorem: allow a, b, and c it is in integers with a e 0 and also b e 0. If a|b and also b|c, then a|c.

In order to prove this statement, we very first need to recognize what the math notation colorreda|b implies.

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I have a separate lesson discussing the meaning of a|b.

To review, the mathematics notation a|b is check out as “a divides b “. The assumption is that both a and b space integers but a doesn’t equal zero, a e 0. In addition, the vertical bar in a|b is called pipe.


As it stands, the notation a|b is not useful to us due to the fact that in its present form, there’s no way that we can algebraically manipulate it. Us must convert it in an equation form.

Here’s the thing, a|b deserve to be written in the equation together b = ar where r is one integer.


For example, in 2|10, we understand that 2 same divides 10. That means there is one integer once multiplied to 2 offers a product the 10.

What could that number be? that is colorred5 since 2 imes 5 = 10.

Thus, we say 2|10 implies 10 = 2left( 5 ight)



Note: The objective of brainstorming in writing proof is for united state to understand what the organize is trying come convey; and also gather enough information to connect the dots, which will certainly be used to leg the hypothesis and the conclusion.

Since we room using the method of direct proof, we want to present that we deserve to manipulate the hypothesis to arrive at the conclusion.

Hypothesis: a divides b and also b divides c

Conclusion: a divides c


Now, let’s express each notation right into an equation. Us hope the by doing therefore will expose an opportunity so we deserve to proceed with our heat of reasoning.

a|bb = to be ← Equation #1m is one integer
b|cc = bn ← Equation #2n is an integer

What should we perform next? Well, we can substitute the expression for b that Equation #1 into the b the Equation #2.


After substitution, we obtain the one below.

c = left( am ight)n

Apply the Associative building of Multiplication. Notification that the group symbol (parenthesis) moves from am to mn.

The Associative property of Multiplication assures that when multiplying numbers, the product is constantly the same no matter just how we team the numbers. Thus, left( am ight)n = aleft( mn ight).

This property allows us to rewrite the equation there is no breaking any type of math laws since the 2 equations may look different however they are essentially the very same or equivalent.

I expect you deserve to see currently why we have to perform such slight adjustment utilizing the Associative Property.

c = left( am ight)n → c = aleft( mn ight)

After we substitute the expression the largeb from Equation #1 right into the largeb of Equation #2, and apply the Associative residential or commercial property of Multiplication, us are ready to move to the next step.

Notice the inside the parenthesis are two arbitrarily integers that are being multiplied.

If girlfriend remember, over there is a straightforward yet an extremely useful property of the set of Integers ( the symbol because that the collection of integers is mathbbZ ).

The home is dubbed the Closure residential or commercial property of Multiplication. It says that if m and also n space integers then the product that m and n is likewise an integer. Therefore, m imes n in mathbbZ.

From wherein we left off, we have

c = aleft( mn ight).

Since mn is just one more integer using the Closure building of Multiplication, that way we can let mn = k whereby k is one integer.

We have the right to rewrite c = aleft( mn ight) as c = aleft( k ight).

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The equation c = aleft( k ight) have the right to be express in notation form as a|c which means that a divides c.

This is exactly where we desire to show! now it’s time to create the actual proof.


THEOREM: allow a, b, and also c it is in integers through a e 0 and b e 0. If a|b and also b|c, climate a|c.

PROOF: mean a, b, and c space integers whereby both a and also b perform not equal to zero. Due to the fact that a divides b, a|b, then there exists an integer m such that b = to be (Equation #1). Similarly, since b divides c, b|c, over there exists an integer n such that c=bn (Equation #2). Now, substitute the expression that b native Equation #1 into the b in Equation #2. By law so, the equation c=bm is changed to c=(am)n. Next, apply the Associative home of Multiplication on the equation c=(am)n to acquire c=a(mn). Because m and also n space integers, their product must also be an creature by the Closure residential property of Multiplication; that is, m imes n in mathbbZ. Allow k = m imes n. In the equation c=a(mn), instead of mn through k to obtain c=ak.The equation c=ak implies that a divides c or when written in shorthand we have actually a|c. Therefore, we have actually proved the a divides c. ◾️