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Does an odd number plus an odd number equal an odd number? - Answers
Well, let's see . . . . .3 + 5 = 85 + 15 = 207 + 19 = 2631 + 27 = 5849 + 11 = 60It's looking bad for your proposed rule.This can actually be proven.Any odd number can be expressed as 2p + 1 where pis any integer.Let a = 2p + 1 and b = 2q + 1 where p and q are any two integers. Then, by definition, a and b are both odd, so lets add them up.We know that any two integers added together equals another integer, the question is whether or not that integer is even or odd.Let m be the integer such that a + b = m.Substituting, we get (2p + 1) + (2q + 1) = m or 2p + 2q + 2 = m.Dividing by 2 we get p + q = m/2 - 1.Since p and q are both integers, m/2 - 1 must also be an integer. Therefore, m must be an even number.Substituting the expression for p into the expression for a from above we get:a = 2(m/2 - 1 - q) + 1 = m - 2 - 2q + 1 = m - (2q + 1) = m - b.Therefore, a + b = m where a and b were both defined to be odd and m was previously shown to be even.Thus, an odd + an odd = an even.Q.E.D.
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Does an odd number plus an odd number equal an odd number? - Answers
Well, let's see . . . . .3 + 5 = 85 + 15 = 207 + 19 = 2631 + 27 = 5849 + 11 = 60It's looking bad for your proposed rule.This can actually be proven.Any odd number can be expressed as 2p + 1 where pis any integer.Let a = 2p + 1 and b = 2q + 1 where p and q are any two integers. Then, by definition, a and b are both odd, so lets add them up.We know that any two integers added together equals another integer, the question is whether or not that integer is even or odd.Let m be the integer such that a + b = m.Substituting, we get (2p + 1) + (2q + 1) = m or 2p + 2q + 2 = m.Dividing by 2 we get p + q = m/2 - 1.Since p and q are both integers, m/2 - 1 must also be an integer. Therefore, m must be an even number.Substituting the expression for p into the expression for a from above we get:a = 2(m/2 - 1 - q) + 1 = m - 2 - 2q + 1 = m - (2q + 1) = m - b.Therefore, a + b = m where a and b were both defined to be odd and m was previously shown to be even.Thus, an odd + an odd = an even.Q.E.D.
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Does an odd number plus an odd number equal an odd number? - Answers
Well, let's see . . . . .3 + 5 = 85 + 15 = 207 + 19 = 2631 + 27 = 5849 + 11 = 60It's looking bad for your proposed rule.This can actually be proven.Any odd number can be expressed as 2p + 1 where pis any integer.Let a = 2p + 1 and b = 2q + 1 where p and q are any two integers. Then, by definition, a and b are both odd, so lets add them up.We know that any two integers added together equals another integer, the question is whether or not that integer is even or odd.Let m be the integer such that a + b = m.Substituting, we get (2p + 1) + (2q + 1) = m or 2p + 2q + 2 = m.Dividing by 2 we get p + q = m/2 - 1.Since p and q are both integers, m/2 - 1 must also be an integer. Therefore, m must be an even number.Substituting the expression for p into the expression for a from above we get:a = 2(m/2 - 1 - q) + 1 = m - 2 - 2q + 1 = m - (2q + 1) = m - b.Therefore, a + b = m where a and b were both defined to be odd and m was previously shown to be even.Thus, an odd + an odd = an even.Q.E.D.
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- og:descriptionWell, let's see . . . . .3 + 5 = 85 + 15 = 207 + 19 = 2631 + 27 = 5849 + 11 = 60It's looking bad for your proposed rule.This can actually be proven.Any odd number can be expressed as 2p + 1 where pis any integer.Let a = 2p + 1 and b = 2q + 1 where p and q are any two integers. Then, by definition, a and b are both odd, so lets add them up.We know that any two integers added together equals another integer, the question is whether or not that integer is even or odd.Let m be the integer such that a + b = m.Substituting, we get (2p + 1) + (2q + 1) = m or 2p + 2q + 2 = m.Dividing by 2 we get p + q = m/2 - 1.Since p and q are both integers, m/2 - 1 must also be an integer. Therefore, m must be an even number.Substituting the expression for p into the expression for a from above we get:a = 2(m/2 - 1 - q) + 1 = m - 2 - 2q + 1 = m - (2q + 1) = m - b.Therefore, a + b = m where a and b were both defined to be odd and m was previously shown to be even.Thus, an odd + an odd = an even.Q.E.D.
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