Please check proofs: Let x be an integer. If 11x-5 is odd,

[shivers20]

Give a direct proof.
Let x be an integer. If 11x-5 is odd, then x is even.

Asuume 11x-5 is odd
11x-5= (2k+1)
so, x= 2k+1
11(2k+1)-5
22k+1-5 = 22k-4
Since 22k-4 is an integer 11x-5 is odd

Give an indirect proof.
Assume that x is odd. Then 2k+1 for some integer k
so, 11x-5= 11(2k+1)-5 = ? I am a bit confused with this one.

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Let x be an integer. Prove that 5x-11 is even if and only if x is odd.
Direct proof: Assume x is odd, then x=2k+1, for some integer k
so, 5(2k+1)-11 = 10k+5-11 = 10k-6= 2(5k-3)
Since (5k-3) is an integer, 5x-11 is even.

Assume x is even. Then x= 2l, for some integer l . Therefore, 5(2l)-11= 10l-11=
10l-12+1= 2(5l-6)+1
Since 5l-6 is an integer, 5x-11 is odd.

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Let x be an integer. Prove that x^3 is even if and only if x is even.

Proof: Assume that x is even. Then x=2k for some integer k.
Therefore, x^3=(2k)^3 = 8k^3= 2(4k^3)
Because 4k^3 is an integer, then integer k^3 is even.

For the Converse, assume x is odd. x=(2l+1), for some integer l.
x^3= (2l+1)^3= (2l+1)(2l+1)(2l+1)= 8l^3+12l^2+6l+1= 2(4l^3+6l^2+3l)+1
since (4l^3+6l^2+3l) is an integer, x^3 is even

 

[daon]

Give a direct proof.
Let x be an integer. If 11x-5 is odd, then x is even.

Asuume 11x-5 is odd
11x-5= (2k+1)
so, x= 2k+1 What justifies this? x is any integer, you cannot assume x is of the form 2k+1
11(2k+1)-5
22k+1-5 = 22k-4
Since 22k-4 is an integer 11x-5 is odd What did you just say here? This is very confusing since you say 11x-5 is odd and get that 11x-5 = 22k-4, which is even. Then you conclude that it is odd. Fix this up a bit, I'm sure your teacher will appreciate it..

Give an indirect proof.
Assume that x is odd. Then 2k+1 for some integer k
so, 11x-5= 11(2k+1)-5 = ? I am a bit confused with this one.

By indirect do you mean by contradiction? Then:

Let 11x-5 be an odd integer. Assume BWOC that x is odd. Then x = 2k+1 for some integer k. Thus 11x-5 = 11(2k+1)-5 = 22k + 6 = 2(11k+3). Since 11k+3 is an integer, 2(11k+3) is an even integer. Thus contradicting the fact that 11x-5 is odd, so x must be even.


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Let x be an integer. Prove that 5x-11 is even if and only if x is odd.
Direct proof: Assume x is odd, then x=2k+1, for some integer k
so, 5(2k+1)-11 = 10k+5-11 = 10k-6= 2(5k-3)
Since (5k-3) is an integer, 5x-11 is even.

You really should state that you are proving the Contrapositive of the statement "If 5x-11 is even then x is odd."
Assume x is even. Then x= 2l, for some integer l . Therefore, 5(2l)-11= 10l-11=
10l-12+1= 2(5l-6)+1
Since 5l-6 is an integer, 5x-11 is odd.
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Let x be an integer. Prove that x^3 is even if and only if x is even.

Proof: Assume that x is even. Then x=2k for some integer k.
Therefore, x^3=(2k)^3 = 8k^3= 2(4k^3)
Because 4k^3 is an integer, then integer k^3 is even.

I think you mean Contrapositive, not Converse.
For the Converse, assume x is odd. x=(2l+1), for some integer l.
x^3= (2l+1)^3= (2l+1)(2l+1)(2l+1)= 8l^3+12l^2+6l+1= 2(4l^3+6l^2+3l)+1
since (4l^3+6l^2+3l) is an integer, x^3 is evenThis is not very clear. You just showed that x^3 is odd, and yet you said that it is even. You should state that you have proved the contrapositive.

 

[shivers20]

Give a direct proof.
Let x be an integer. If 11x-5 is odd, then x is even.

Asuume 11x-5 is odd
11x-5= (2k+1)
so, x= 2k+1 What justifies this? x is any integer, you cannot assume x is of the form 2k+1 Odd number formula is 2k+1 and Even number is 2k.11(2k+1)-5
22k+1-5 = 22k-4
Since 22k-4 is an integer 11x-5 is odd What did you just say here? This is very confusing since you say 11x-5 is odd and get that 11x-5 = 22k-4, which is even. Then you conclude that it is odd. Fix this up a bit, I'm sure your teacher will appreciate it.. Is it suppose to be (22k-5)+1 is an element in a set of integers. Therefore, 11x-5 is odd.
Give an indirect proof.
Assume that x is odd. Then 2k+1 for some integer k
so, 11x-5= 11(2k+1)-5 = ? I am a bit confused with this one.

By indirect do you mean by contradiction? Then:

Let 11x-5 be an odd integer. Assume BWOC that x is odd. Then x = 2k+1 for some integer k. Thus 11x-5 = 11(2k+1)-5 = 22k + 6 = 2(11k+3). Since 11k+3 is an integer, 2(11k+3) is an even integer. Thus contradicting the fact that 11x-5 is odd, so x must be even. What does BWOC represent?


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Let x be an integer. Prove that 5x-11 is even if and only if x is odd.
Direct proof: Assume x is odd, then x=2k+1, for some integer k
so, 5(2k+1)-11 = 10k+5-11 = 10k-6= 2(5k-3)
Since (5k-3) is an integer, 5x-11 is even.

You really should state that you are proving the Contrapositive of the statement "If 5x-11 is even then x is odd." I agree.
Assume x is even. Then x= 2l, for some integer l . Therefore, 5(2l)-11= 10l-11=
10l-12+1= 2(5l-6)+1
Since 5l-6 is an integer, 5x-11 is odd.
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Let x be an integer. Prove that x^3 is even if and only if x is even.

Proof: Assume that x is even. Then x=2k for some integer k.
Therefore, x^3=(2k)^3 = 8k^3= 2(4k^3)
Because 4k^3 is an integer, then integer k^3 is even.

I think you mean Contrapositive, not Converse.
For the Converse, assume x is odd. x=(2l+1), for some integer l.
x^3= (2l+1)^3= (2l+1)(2l+1)(2l+1)= 8l^3+12l^2+6l+1= 2(4l^3+6l^2+3l)+1
since (4l^3+6l^2+3l) is an integer, x^3 is evenThis is not very clear. You just showed that x^3 is odd, and yet you said that it is even. You should state that you have proved the contrapositive.

I am sorry. x^3 is odd. How do I do that?

 

[Denis]

Re: Please check proofs: Let x be an integer. If 11x-5 is od

Let x be an integer. If 11x-5 is odd, then x is even.

Keep it a little simpler:
11x - 5 = 2k + 1
11x = 2k + 6
11x = 2(k + 3)

Since right side is even, then x has to be even.

 

[daon]

Odd number formula is 2k+1 and Even number is 2k

Yes, but you said "let 11x-5 be odd" not x. You need to use the assumption that 11x-5 is odd and derive that x is of the form 2k for some k. Denis has showed this for you.

What does BWOC represent?

By Way of Contradiction.

Is it suppose to be (22k-5)+1 is an element in a set of integers. Therefore, 11x-5 is odd.

(22k-5)+1 is not in the correct form to be called 'odd'.It must be 2*(something)+1. You have (something)+1

I am sorry. x^3 is odd. How do I do that?

You need to prove: if x^3 is even then x is even.

Pf by Contrapositive: If x is odd then x^3 is odd.
Let x be an odd integer. .....(your work is fine here)....
Hence, x^3 is odd. Therefore, this proves: if x^3 is even then x is even.