Hello,
I am needing some help on the two problems below. Thank you in advance.
1) Prove by contradiction that there do not exist integers m and n such that 14m + 21n = 101.
and
2) Using the fact that an integer n, either n >= 14 or n <= 15, or otherwise prove 7 does not divide 100.
Thank you ThePerfectHacker. I made a typo when I first made my post. Sorry about that. So, it should be 14m + 21n = 100 not 101.
I am beening taught to turn this into a if/then statement and go from there.
If m and n are integers, then 14m + 21n = 100
Factor out a 7 givng us 7(2n +3m) = 100
From here I have trouble. Also, should the if/then statement be not equal to 100 instead of equal to? I haven't had that much experience with proof by contradiction. Thank You.
Let,
p= (there exists n,m such that 14m+21n=100).
Now if "p" is true we shown that 7 divides 100.
Thus,
q=(7 divides 100).
But this is false.
Meaning, (modens ponens)
[(p--> q) and (~q)]--> ~p
Thus,
~p=(there do not exists n,m such that 14m+21n=100)
This is mine 46th Post!!!
Hello, MathStudent1!
Here;s the first one . . .
1) Prove by contradiction that there do not exist integers m and n
. . .such that: .14m + 21n .= .101
Assume that there are integers m and n such that: .14m + 21n .= .101
. . . . . . . . . . . . . . . . . . .100
Divide by 7: . 2m + 3n .= .----
. . . . . . . . . . . . . . . . . . . .7
In the left side: m is an integer, so 2m is an integer. *
. . . . . . . . . . . . n is an integer, so 3n is an integer. *
Hence, 2m + 3n (the sum of two integers) is an integer. *
But the right side is an irreducible fraction (or decimal 14.2857...)
We have reached a contradicition, hence our assumption was incorrect.
. . There are no integers m and n that satisfy 14m + 21n .= .101
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
*
Reasons: The set of integers is closed under multiplication and addition.
Thank you Soroban. One question for you. When orginally I posted my thread I made a typo in that 14m + 21n = 100 NOT 101. Does that change the proof at all or does the logic remain the same. Following your proof, am I correct by stating this below:
Prove that there do not exist integers m and n such that: 14m + 21n = 100
Restating the problem - If m and n are integers, then 14m + 21n = 100
Factor out a 7 and we get 7(2m + 3n) = 100
Divide both sides by 7 and we get 2m + 3n = (100/7)
Since m is an integer then 2*m is an integer and since n is an integer then 3*n is and integer.
However, the RHS (100/7) is not an integer and therefore,
2m + 3n (IS NOT EQUAL TO) (100/7)
Hence, we have reached a contradiction of our assumption that m and n are integers.
Therefore, there do not exist integers m and n such that 14m + 21n = 100.
*****Thank you!*****