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Use deductive reasoning and order the sentences to show that the conjecture is true.

= Then (2m+1) and (2n+1) are odd integers.
= An even integer plus one is an odd integer. equiv 2(2mn+m+n) is divisible by 2 and is therefore an even integer.
= Let m and n be integers.
= (2m+1)(2n+1)=4mn+2m+2n+1=2(2mn+m+n)+1

User Atmorell
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Final answer:

By using deductive reasoning and ordering the given statements, we show that when m and n are integers, the expressions (2m+1) and (2n+1) are indeed odd integers.

Step-by-step explanation:

To demonstrate that the conjecture 'if m and n are integers, then (2m+1) and (2n+1) are odd integers' is true, we must use deductive reasoning. Let's order the sentences step-by-step to validate the conjecture:

  1. Let m and n be integers.
  2. An even integer plus one is an odd integer.
  3. (2m+1) is an even integer (2m) plus one, so (2m+1) is an odd integer.
  4. (2n+1) is an even integer (2n) plus one, so (2n+1) is an odd integer.
  5. From the distributive property, (2m+1)(2n+1) = 4mn + 2m + 2n + 1.
  6. Grouping like terms gives us: 2(2mn + m + n) + 1.
  7. The term 2(2mn + m + n) is divisible by 2, which makes it an even integer.
  8. Therefore, (2m+1)(2n+1) is an even integer plus one, which is classified as an odd integer.

Through this ordered series of statements, we confirm the original conjecture that the product of two numbers each of the form (2 times an integer plus one) yields another odd integer.

User Lockhrt
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