Prove if $s<x$ and $t<z$, then $st<xz$. [closed]
Assuming $s,t,x,z$ are natural numbers, if $s < x$ and $t < z$, then $st < xz$. Prove this.
Do I need induction? Please help.
I am very confused.
Thank you.
inequality induction
edited Dec 2 '18 at 13:40
AryanSonwatikar
31312
asked Nov 29 '18 at 3:35
qq2657
12
closed as unclear what you're asking by Shailesh, max_zorn, Paul Plummer, Lord Shark the Unknown, choco_addicted Nov 29 '18 at 6:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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Assuming $s,t,x,z$ are natural numbers, if $s < x$ and $t < z$, then $st < xz$. Prove this.
Do I need induction? Please help.
I am very confused.
Thank you.
inequality induction
edited Dec 2 '18 at 13:40
AryanSonwatikar
31312
asked Nov 29 '18 at 3:35
qq2657
12
closed as unclear what you're asking by Shailesh, max_zorn, Paul Plummer, Lord Shark the Unknown, choco_addicted Nov 29 '18 at 6:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
5
Perhaps you can elaborate a bit. Without any more context, the statement is false: consider $s,t=-2$ and $x,z=1$
– Y. Forman
Nov 29 '18 at 3:37
but wouldn't that still make st < xz? -2<1??
– qq2657
Nov 29 '18 at 3:38
2
This is not true. $-1<0$ and $0<1$, but you can’t conclude that $0<0$. You probably want to consider only positive numbers.
– MPW
Nov 29 '18 at 3:39
@qq2657 $st=4$ and $xz=1$
– Y. Forman
Nov 29 '18 at 3:40
1
@fleablood That may or may not be an axiom. There are lots of different axiom systems.
– Robert Israel
Nov 29 '18 at 4:06
|
show 4 more comments
Assuming $s,t,x,z$ are natural numbers, if $s < x$ and $t < z$, then $st < xz$. Prove this.
Do I need induction? Please help.
I am very confused.
Thank you.
inequality induction
edited Dec 2 '18 at 13:40
AryanSonwatikar
31312
asked Nov 29 '18 at 3:35
qq2657
12
Assuming $s,t,x,z$ are natural numbers, if $s < x$ and $t < z$, then $st < xz$. Prove this.
Do I need induction? Please help.
I am very confused.
Thank you.
inequality induction
inequality induction
edited Dec 2 '18 at 13:40
AryanSonwatikar
31312
asked Nov 29 '18 at 3:35
qq2657
12
edited Dec 2 '18 at 13:40
AryanSonwatikar
31312
asked Nov 29 '18 at 3:35
qq2657
12
edited Dec 2 '18 at 13:40
AryanSonwatikar
31312
edited Dec 2 '18 at 13:40
AryanSonwatikar
31312
edited Dec 2 '18 at 13:40
AryanSonwatikar
31312
31312
asked Nov 29 '18 at 3:35
qq2657
12
asked Nov 29 '18 at 3:35
qq2657
12
asked Nov 29 '18 at 3:35
qq2657
12
12
closed as unclear what you're asking by Shailesh, max_zorn, Paul Plummer, Lord Shark the Unknown, choco_addicted Nov 29 '18 at 6:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Shailesh, max_zorn, Paul Plummer, Lord Shark the Unknown, choco_addicted Nov 29 '18 at 6:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
5
Perhaps you can elaborate a bit. Without any more context, the statement is false: consider $s,t=-2$ and $x,z=1$
– Y. Forman
Nov 29 '18 at 3:37
but wouldn't that still make st < xz? -2<1??
– qq2657
Nov 29 '18 at 3:38
2
This is not true. $-1<0$ and $0<1$, but you can’t conclude that $0<0$. You probably want to consider only positive numbers.
– MPW
Nov 29 '18 at 3:39
@qq2657 $st=4$ and $xz=1$
– Y. Forman
Nov 29 '18 at 3:40
1
@fleablood That may or may not be an axiom. There are lots of different axiom systems.
– Robert Israel
Nov 29 '18 at 4:06
|
show 4 more comments
5
Perhaps you can elaborate a bit. Without any more context, the statement is false: consider $s,t=-2$ and $x,z=1$
– Y. Forman
Nov 29 '18 at 3:37
but wouldn't that still make st < xz? -2<1??
– qq2657
Nov 29 '18 at 3:38
2
This is not true. $-1<0$ and $0<1$, but you can’t conclude that $0<0$. You probably want to consider only positive numbers.
– MPW
Nov 29 '18 at 3:39
@qq2657 $st=4$ and $xz=1$
– Y. Forman
Nov 29 '18 at 3:40
1
@fleablood That may or may not be an axiom. There are lots of different axiom systems.
– Robert Israel
Nov 29 '18 at 4:06
5
5
Perhaps you can elaborate a bit. Without any more context, the statement is false: consider $s,t=-2$ and $x,z=1$
– Y. Forman
Nov 29 '18 at 3:37
Perhaps you can elaborate a bit. Without any more context, the statement is false: consider $s,t=-2$ and $x,z=1$
– Y. Forman
Nov 29 '18 at 3:37
but wouldn't that still make st < xz? -2<1??
– qq2657
Nov 29 '18 at 3:38
but wouldn't that still make st < xz? -2<1??
– qq2657
Nov 29 '18 at 3:38
2
2
This is not true. $-1<0$ and $0<1$, but you can’t conclude that $0<0$. You probably want to consider only positive numbers.
– MPW
Nov 29 '18 at 3:39
This is not true. $-1<0$ and $0<1$, but you can’t conclude that $0<0$. You probably want to consider only positive numbers.
– MPW
Nov 29 '18 at 3:39
@qq2657 $st=4$ and $xz=1$
– Y. Forman
Nov 29 '18 at 3:40
@qq2657 $st=4$ and $xz=1$
– Y. Forman
Nov 29 '18 at 3:40
1
1
@fleablood That may or may not be an axiom. There are lots of different axiom systems.
– Robert Israel
Nov 29 '18 at 4:06
@fleablood That may or may not be an axiom. There are lots of different axiom systems.
– Robert Israel
Nov 29 '18 at 4:06
|
show 4 more comments
1 Answer
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oldest
votes
This is false; a counterexample would be
$$1 < 2 ;;; text{and} ;;; -2 < -1 ;;; text{but} ;;; 1(-2) = -2 not < -2 = 2(-1)$$
In general, though, unless you are limited to have $s,t,x,z in mathbb{Z}$, you wouldn't even try to use induction on this. Induction doesn't quite work with real numbers.
answered Nov 29 '18 at 3:39
Eevee Trainer
4,9971734
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is false; a counterexample would be
$$1 < 2 ;;; text{and} ;;; -2 < -1 ;;; text{but} ;;; 1(-2) = -2 not < -2 = 2(-1)$$
In general, though, unless you are limited to have $s,t,x,z in mathbb{Z}$, you wouldn't even try to use induction on this. Induction doesn't quite work with real numbers.
answered Nov 29 '18 at 3:39
Eevee Trainer
4,9971734
add a comment |
This is false; a counterexample would be
$$1 < 2 ;;; text{and} ;;; -2 < -1 ;;; text{but} ;;; 1(-2) = -2 not < -2 = 2(-1)$$
In general, though, unless you are limited to have $s,t,x,z in mathbb{Z}$, you wouldn't even try to use induction on this. Induction doesn't quite work with real numbers.
answered Nov 29 '18 at 3:39
Eevee Trainer
4,9971734
add a comment |
This is false; a counterexample would be
$$1 < 2 ;;; text{and} ;;; -2 < -1 ;;; text{but} ;;; 1(-2) = -2 not < -2 = 2(-1)$$
In general, though, unless you are limited to have $s,t,x,z in mathbb{Z}$, you wouldn't even try to use induction on this. Induction doesn't quite work with real numbers.
answered Nov 29 '18 at 3:39
Eevee Trainer
4,9971734
This is false; a counterexample would be
$$1 < 2 ;;; text{and} ;;; -2 < -1 ;;; text{but} ;;; 1(-2) = -2 not < -2 = 2(-1)$$
In general, though, unless you are limited to have $s,t,x,z in mathbb{Z}$, you wouldn't even try to use induction on this. Induction doesn't quite work with real numbers.
answered Nov 29 '18 at 3:39
Eevee Trainer
4,9971734
answered Nov 29 '18 at 3:39
Eevee Trainer
4,9971734
answered Nov 29 '18 at 3:39
Eevee Trainer
4,9971734
answered Nov 29 '18 at 3:39
Eevee Trainer
4,9971734
4,9971734
add a comment |
add a comment |
5
Perhaps you can elaborate a bit. Without any more context, the statement is false: consider $s,t=-2$ and $x,z=1$
– Y. Forman
Nov 29 '18 at 3:37
but wouldn't that still make st < xz? -2<1??
– qq2657
Nov 29 '18 at 3:38
2
This is not true. $-1<0$ and $0<1$, but you can’t conclude that $0<0$. You probably want to consider only positive numbers.
– MPW
Nov 29 '18 at 3:39
@qq2657 $st=4$ and $xz=1$
– Y. Forman
Nov 29 '18 at 3:40
1
@fleablood That may or may not be an axiom. There are lots of different axiom systems.
– Robert Israel
Nov 29 '18 at 4:06