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In the following question: " Let a, b ∈ Z. Show that if a ≡ 5 (mod 6) and b ≡ 3 (mod 4), then 4a + 6b ≡ 6 (mod 8)." why do we ignore the -6 when substituting the variables of a and b, if we expand "4a + 6b ≡ 6 (mod 8)" we get (4a+6b)-6=8y, when we substitute we have to end up with the 8y right? Im just confused why the -6 is ignored?

In the following question: " Let a, b ∈ Z. Show that if a ≡ 5 (mod 6) and b ≡ 3 (mod 4), then 4a + 6b ≡ 6 (mod 8)." why do we ignore the -6 when substituting the variables of a and b, if we expand "4a + 6b ≡ 6 (mod 8)" we get (4a+6b)-6=8y, when we substitute we have to end up with the 8y right? Im just confused why the -6 is ignored?

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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In the following question:

" Let a, b ∈ Z. Show that if a ≡ 5 (mod 6) and b ≡ 3 (mod 4), then 4a + 6b ≡ 6 (mod 8)."

why do we ignore the -6 when substituting the variables of a and b, if we expand "4a + 6b ≡ 6 (mod 8)" we get (4a+6b)-6=8y, when we substitute we have to end up with the 8y right? Im just confused why the -6 is ignored?

Proof. Assume that a = 5 (mod 6) and b = 3 (mod 4). Then 6|(a - 5) and 4|(b − 3). Therefore a 5= 6x and b- 3 = 4y, where x, y E Z. So a = 6x +5 and b = 4y + 3. Observe that 4a + 6b = 4(6x + 5) + 6(4y + 3) = 24x +20 +24y + 18 = 24x + 24y + 38 = 8(3x + 3y + 4) + 6. Since 3x + 3y + 4 is an integer, 8|(4a + 6b − 6) and so 4a + 6b = 6 (mod 8)
Transcribed Image Text:Proof. Assume that a = 5 (mod 6) and b = 3 (mod 4). Then 6|(a - 5) and 4|(b − 3). Therefore a 5= 6x and b- 3 = 4y, where x, y E Z. So a = 6x +5 and b = 4y + 3. Observe that 4a + 6b = 4(6x + 5) + 6(4y + 3) = 24x +20 +24y + 18 = 24x + 24y + 38 = 8(3x + 3y + 4) + 6. Since 3x + 3y + 4 is an integer, 8|(4a + 6b − 6) and so 4a + 6b = 6 (mod 8)
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