Proof by Contradiction

[Kailahernandez15]

Let m and n be non-negative integers. Prove that if mn>100, then m<10 and n<10 by contradiction

 

[Deleted member 4993]

Let m and n be non-negative integers. Prove that if mn>100, then m<10 and n<10 by contradiction

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[Dr.Peterson]

Let m and n be non-negative integers. Prove that if mn>100, then m<10 and n<10 by contradiction

I think you copied something wrong. It's easy to find m and n, both greater than 10, such that mn > 100. (I'm using contradiction to show that the claim is false!)

 

[JeffM]

Let m and n be non-negative integers. Prove that if mn>100, then m<10 and n<10 by contradiction

You cannot prove what is false

[MATH]m= 2 \text { and } n = 51 \implies mn = 102 > 100 \text { but } 51 \not < 10.[/MATH]
Of course you can prove that

[MATH]0 < m < 10 \text { and } 0 < n < 10 \implies mn < 100.[/MATH]
Is that what you are to prove?

 

[Kailahernandez15]

I think you copied something wrong. It's easy to find m and n, both greater than 10, such that mn > 100. (I'm using contradiction to show that the claim is false!)

It is the exact problem given by my professor. We should prove it using contradiction.
"Let m and n be non-negative integers. Prove that if mn>100, then m<10 and n<10"
I'm really having a hard time

 

[Dr.Peterson]

It is the exact problem given by my professor. We should prove it using contradiction.
"Let m and n be non-negative integers. Prove that if mn>100, then m<10 and n<10"
I'm really having a hard time

Are you positive that you (or your professor) didn't copy a symbol wrong? Something more like this could make sense:

Let m and n be non-negative integers. Prove by contradiction that if mn<100, then m<10 or n<10.​

 

[Steven G]

Let m and n be non-negative integers. Prove that if mn>100, then m<10 and n<10 by contradiction

We know that 10*10=100. If we decrease both numbers then the product will always be less than 100.
Just let m=9 and n=8. If the statement you made is correct then 9*8 >100 !

 

[Dr.Peterson]

We know that 10*10=100. If we decrease both numbers then the product will always be less than 100.
Just let m=9 and n=8. If the statement you made is correct then 9*8 >100 !

Not as the problem was stated; the conditional statement doesn't apply to m=9 and n=8, since then it is not true that mn > 100.

My counterexample would be something like m=10 and n=11. Then mn > 100, so the claim is that m<10 and n<10. That, of course, is not true.

 

[Steven G]

Not as the problem was stated; the conditional statement doesn't apply to m=9 and n=8, since then it is not true that mn > 100.

My counterexample would be something like m=10 and n=11. Then mn > 100, so the claim is that m<10 and n<10. That, of course, is not true.

Yes, you are correct. Is Otis still in the corner?