I am not getting this line, any help please?
T Tayeeba New member Joined Feb 4, 2017 Messages 32 Mar 18, 2017 #1 I am not getting this line, any help please?
J JeffM Elite Member Joined Sep 14, 2012 Messages 7,872 Mar 18, 2017 #2 I am having trouble reading your attachment, but I think you are asking about logic something like this: 4k−1=3j, where j∈N+ ⟹ 4k=3j+1 ⟹ \displaystyle 4^k - 1 = 3j,\ where\ j \in \mathbb N^+ \implies 4^k = 3j + 1 \implies4k−1=3j, where j∈N+⟹4k=3j+1⟹ 4∗4k=4(3j+1) ⟹ 4(k+1)=12j+4=12j+3+1 ⟹ \displaystyle 4 * 4^k = 4(3j + 1) \implies 4^{(k+1)} = 12j + 4 = 12j + 3 + 1 \implies4∗4k=4(3j+1)⟹4(k+1)=12j+4=12j+3+1⟹ 4(k+1)−1=12j+3=3(4j+1), where 4j+1∈ N+∵j∈N+.\displaystyle 4^{(k+1)} - 1 = 12j + 3 = 3(4j + 1),\ where\ 4j + 1 \in\ \mathbb N^+ \because j \in \mathbb N^+.4(k+1)−1=12j+3=3(4j+1), where 4j+1∈ N+∵j∈N+. Last edited: Mar 18, 2017
I am having trouble reading your attachment, but I think you are asking about logic something like this: 4k−1=3j, where j∈N+ ⟹ 4k=3j+1 ⟹ \displaystyle 4^k - 1 = 3j,\ where\ j \in \mathbb N^+ \implies 4^k = 3j + 1 \implies4k−1=3j, where j∈N+⟹4k=3j+1⟹ 4∗4k=4(3j+1) ⟹ 4(k+1)=12j+4=12j+3+1 ⟹ \displaystyle 4 * 4^k = 4(3j + 1) \implies 4^{(k+1)} = 12j + 4 = 12j + 3 + 1 \implies4∗4k=4(3j+1)⟹4(k+1)=12j+4=12j+3+1⟹ 4(k+1)−1=12j+3=3(4j+1), where 4j+1∈ N+∵j∈N+.\displaystyle 4^{(k+1)} - 1 = 12j + 3 = 3(4j + 1),\ where\ 4j + 1 \in\ \mathbb N^+ \because j \in \mathbb N^+.4(k+1)−1=12j+3=3(4j+1), where 4j+1∈ N+∵j∈N+.
T Tayeeba New member Joined Feb 4, 2017 Messages 32 Mar 19, 2017 #3 JeffM said: I am having trouble reading your attachment, but I think you are asking about logic something like this: 4k−1=3j, where j∈N+ ⟹ 4k=3j+1 ⟹ \displaystyle 4^k - 1 = 3j,\ where\ j \in \mathbb N^+ \implies 4^k = 3j + 1 \implies4k−1=3j, where j∈N+⟹4k=3j+1⟹ 4∗4k=4(3j+1) ⟹ 4(k+1)=12j+4=12j+3+1 ⟹ \displaystyle 4 * 4^k = 4(3j + 1) \implies 4^{(k+1)} = 12j + 4 = 12j + 3 + 1 \implies4∗4k=4(3j+1)⟹4(k+1)=12j+4=12j+3+1⟹ 4(k+1)−1=12j+3=3(4j+1), where 4j+1∈ N+∵j∈N+.\displaystyle 4^{(k+1)} - 1 = 12j + 3 = 3(4j + 1),\ where\ 4j + 1 \in\ \mathbb N^+ \because j \in \mathbb N^+.4(k+1)−1=12j+3=3(4j+1), where 4j+1∈ N+∵j∈N+. Click to expand... Thanks a lot, I figured it out yesterday
JeffM said: I am having trouble reading your attachment, but I think you are asking about logic something like this: 4k−1=3j, where j∈N+ ⟹ 4k=3j+1 ⟹ \displaystyle 4^k - 1 = 3j,\ where\ j \in \mathbb N^+ \implies 4^k = 3j + 1 \implies4k−1=3j, where j∈N+⟹4k=3j+1⟹ 4∗4k=4(3j+1) ⟹ 4(k+1)=12j+4=12j+3+1 ⟹ \displaystyle 4 * 4^k = 4(3j + 1) \implies 4^{(k+1)} = 12j + 4 = 12j + 3 + 1 \implies4∗4k=4(3j+1)⟹4(k+1)=12j+4=12j+3+1⟹ 4(k+1)−1=12j+3=3(4j+1), where 4j+1∈ N+∵j∈N+.\displaystyle 4^{(k+1)} - 1 = 12j + 3 = 3(4j + 1),\ where\ 4j + 1 \in\ \mathbb N^+ \because j \in \mathbb N^+.4(k+1)−1=12j+3=3(4j+1), where 4j+1∈ N+∵j∈N+. Click to expand... Thanks a lot, I figured it out yesterday
