Theorem: permit a, b, and also c be integers with a \\ne 0 and also b \\ne 0. If a|b and also b|c, climate a|c.

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

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I have a different lesson mentioning the meaning of a|b.

To review, the mathematics notation a|b is check out as “a divides b “. The assumption is the both a and also b space integers but a doesn’t same zero, a \\ne 0. In addition, the upright bar in a|b is dubbed pipe.


As that stands, the notation a|b is not helpful to us because in its existing form, there’s no method that we can algebraically manipulate it. We must convert it in one equation form.

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


For example, in 2|10, we understand that 2 evenly divides 10. That means there is an integer when multiplied to 2 provides a product of 10.

What could that number be? it is \\colorred5 because 2 \\times 5 = 10.

Thus, us say 2|10 indicates 10 = 2\\left( 5 \\right)



Note: The purpose of brainstorming in composing proof is for us to recognize what the organize is trying to convey; and gather sufficient information to affix the dots, which will certainly be provided to bridge the hypothesis and the conclusion.

Since we are using the method of direct proof, we want to display that we can manipulate the hypothesis to arrive at the conclusion.

Hypothesis: a divides b and b divides c

Conclusion: a divides c


Now, let’s express every notation right into an equation. We hope that by doing therefore will reveal an opportunity so we have the right to proceed through our line of reasoning.

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

What have to we carry out next? Well, we have the right to substitute the expression because that b of Equation #1 right into the b of Equation #2.


After substitution, we acquire the one below.

c = \\left( am \\right)n

Apply the Associative building of Multiplication. An alert that the grouping symbol (parenthesis) moves from to be to mn.

The Associative home of Multiplication guarantees that as soon as multiplying numbers, the product is constantly the very same no matter how we team the numbers. Thus, \\left( am \\right)n = a\\left( mn \\right).

This property permits us to rewrite the equation there is no breaking any math laws since the 2 equations might look different however they are basically 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 \\right)n → c = a\\left( mn \\right)

After us substitute the expression of \\largeb from Equation #1 into the \\largeb that Equation #2, and apply the Associative residential property of Multiplication, us are all set to relocate to the following step.

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

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

The residential or commercial property is dubbed the Closure home of Multiplication. It says that if m and also n are integers climate the product the m and also n is also an integer. Therefore, m \\times n \\in \\mathbbZ.

From whereby we left off, we have

c = a\\left( mn \\right).

Since mn is just another integer using the Closure property of Multiplication, that means we can let mn = k whereby k is an integer.

We can rewrite c = a\\left( mn \\right) as c = a\\left( k \\right).

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The equation c = a\\left( k \\right) can be to express in notation form as a|c which way that a divides c.

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


THEOREM: let a, b, and also c it is in integers with a \\ne 0 and b \\ne 0. If a|b and also b|c, climate a|c.

PROOF: expect a, b, and c space integers whereby both a and also b execute not same to zero. Due to the fact that a divides b, a|b, climate there exist an creature m such that b = to be (Equation #1). Similarly, because b divides c, b|c, over there exists an essence n such that c=bn (Equation #2). Now, instead of the expression the b indigenous Equation #1 into the b in Equation #2. By law so, the equation c=bm is reinvented to c=(am)n. Next, use the Associative building of Multiplication top top the equation c=(am)n to acquire c=a(mn). Due to the fact that m and also n space integers, your product must likewise be an essence by the Closure residential property of Multiplication; the is, m \\times n \\in \\mathbbZ. Allow k = m \\times n. In the equation c=a(mn), instead of mn through k to acquire c=ak.The equation c=ak means that a divides c or once written in shorthand we have actually a|c. Therefore, we have proved that a divides c. ◾️