Asked by Shayla
Consider the following pattern involving fractions: 15/20=3/4, 15-3/20-4=12/16=3/4
18/15=6/5, 18-6/15-5=12/10=6/5
1.) Make a conjecture about the pattern. Show that the conjecture is true for another example.
2.) Explain how inductive reasoning can be used to support the conjecture.
3.) Find a counter example to the conjecture. What does the counter example tell about the conjecture?
OR
If a counter example cannot be found, describe what it would need to look like. What does not finding a counter example tell about the conjecture?
I need help with these questions. Thanks in advance.
18/15=6/5, 18-6/15-5=12/10=6/5
1.) Make a conjecture about the pattern. Show that the conjecture is true for another example.
2.) Explain how inductive reasoning can be used to support the conjecture.
3.) Find a counter example to the conjecture. What does the counter example tell about the conjecture?
OR
If a counter example cannot be found, describe what it would need to look like. What does not finding a counter example tell about the conjecture?
I need help with these questions. Thanks in advance.
Answers
Answered by
Reiny
You have to have brackets for your conjecture to be true
15/20 = 3/4 , then (15-3)/(20-4) = 12/16 = 3/4
18/15 = 6/5, then (18-6)/(15-5) = 12/10 = 6/5
suppose a fraction a/b is not in lowest terms and can be reduced to c/d
then a/b = c/d = k
is (a-c)/(b-d) = k ??
from c/d = k, and a/b = k
then c = dk and a = bk
(a-c)/(b-d)
= (bk - dk)/(b - d)
= k(b-d)/(b-d)
= k
so I have proven you conjecture to be true for all cases, except b = d, or else we would have divided by zero
15/20 = 3/4 , then (15-3)/(20-4) = 12/16 = 3/4
18/15 = 6/5, then (18-6)/(15-5) = 12/10 = 6/5
suppose a fraction a/b is not in lowest terms and can be reduced to c/d
then a/b = c/d = k
is (a-c)/(b-d) = k ??
from c/d = k, and a/b = k
then c = dk and a = bk
(a-c)/(b-d)
= (bk - dk)/(b - d)
= k(b-d)/(b-d)
= k
so I have proven you conjecture to be true for all cases, except b = d, or else we would have divided by zero
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