-0.000 000 000 742 147 676 627 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 627 6(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 676 627 6(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 147 676 627 6| = 0.000 000 000 742 147 676 627 6


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 147 676 627 6.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 742 147 676 627 6 × 2 = 0 + 0.000 000 001 484 295 353 255 2;
  • 2) 0.000 000 001 484 295 353 255 2 × 2 = 0 + 0.000 000 002 968 590 706 510 4;
  • 3) 0.000 000 002 968 590 706 510 4 × 2 = 0 + 0.000 000 005 937 181 413 020 8;
  • 4) 0.000 000 005 937 181 413 020 8 × 2 = 0 + 0.000 000 011 874 362 826 041 6;
  • 5) 0.000 000 011 874 362 826 041 6 × 2 = 0 + 0.000 000 023 748 725 652 083 2;
  • 6) 0.000 000 023 748 725 652 083 2 × 2 = 0 + 0.000 000 047 497 451 304 166 4;
  • 7) 0.000 000 047 497 451 304 166 4 × 2 = 0 + 0.000 000 094 994 902 608 332 8;
  • 8) 0.000 000 094 994 902 608 332 8 × 2 = 0 + 0.000 000 189 989 805 216 665 6;
  • 9) 0.000 000 189 989 805 216 665 6 × 2 = 0 + 0.000 000 379 979 610 433 331 2;
  • 10) 0.000 000 379 979 610 433 331 2 × 2 = 0 + 0.000 000 759 959 220 866 662 4;
  • 11) 0.000 000 759 959 220 866 662 4 × 2 = 0 + 0.000 001 519 918 441 733 324 8;
  • 12) 0.000 001 519 918 441 733 324 8 × 2 = 0 + 0.000 003 039 836 883 466 649 6;
  • 13) 0.000 003 039 836 883 466 649 6 × 2 = 0 + 0.000 006 079 673 766 933 299 2;
  • 14) 0.000 006 079 673 766 933 299 2 × 2 = 0 + 0.000 012 159 347 533 866 598 4;
  • 15) 0.000 012 159 347 533 866 598 4 × 2 = 0 + 0.000 024 318 695 067 733 196 8;
  • 16) 0.000 024 318 695 067 733 196 8 × 2 = 0 + 0.000 048 637 390 135 466 393 6;
  • 17) 0.000 048 637 390 135 466 393 6 × 2 = 0 + 0.000 097 274 780 270 932 787 2;
  • 18) 0.000 097 274 780 270 932 787 2 × 2 = 0 + 0.000 194 549 560 541 865 574 4;
  • 19) 0.000 194 549 560 541 865 574 4 × 2 = 0 + 0.000 389 099 121 083 731 148 8;
  • 20) 0.000 389 099 121 083 731 148 8 × 2 = 0 + 0.000 778 198 242 167 462 297 6;
  • 21) 0.000 778 198 242 167 462 297 6 × 2 = 0 + 0.001 556 396 484 334 924 595 2;
  • 22) 0.001 556 396 484 334 924 595 2 × 2 = 0 + 0.003 112 792 968 669 849 190 4;
  • 23) 0.003 112 792 968 669 849 190 4 × 2 = 0 + 0.006 225 585 937 339 698 380 8;
  • 24) 0.006 225 585 937 339 698 380 8 × 2 = 0 + 0.012 451 171 874 679 396 761 6;
  • 25) 0.012 451 171 874 679 396 761 6 × 2 = 0 + 0.024 902 343 749 358 793 523 2;
  • 26) 0.024 902 343 749 358 793 523 2 × 2 = 0 + 0.049 804 687 498 717 587 046 4;
  • 27) 0.049 804 687 498 717 587 046 4 × 2 = 0 + 0.099 609 374 997 435 174 092 8;
  • 28) 0.099 609 374 997 435 174 092 8 × 2 = 0 + 0.199 218 749 994 870 348 185 6;
  • 29) 0.199 218 749 994 870 348 185 6 × 2 = 0 + 0.398 437 499 989 740 696 371 2;
  • 30) 0.398 437 499 989 740 696 371 2 × 2 = 0 + 0.796 874 999 979 481 392 742 4;
  • 31) 0.796 874 999 979 481 392 742 4 × 2 = 1 + 0.593 749 999 958 962 785 484 8;
  • 32) 0.593 749 999 958 962 785 484 8 × 2 = 1 + 0.187 499 999 917 925 570 969 6;
  • 33) 0.187 499 999 917 925 570 969 6 × 2 = 0 + 0.374 999 999 835 851 141 939 2;
  • 34) 0.374 999 999 835 851 141 939 2 × 2 = 0 + 0.749 999 999 671 702 283 878 4;
  • 35) 0.749 999 999 671 702 283 878 4 × 2 = 1 + 0.499 999 999 343 404 567 756 8;
  • 36) 0.499 999 999 343 404 567 756 8 × 2 = 0 + 0.999 999 998 686 809 135 513 6;
  • 37) 0.999 999 998 686 809 135 513 6 × 2 = 1 + 0.999 999 997 373 618 271 027 2;
  • 38) 0.999 999 997 373 618 271 027 2 × 2 = 1 + 0.999 999 994 747 236 542 054 4;
  • 39) 0.999 999 994 747 236 542 054 4 × 2 = 1 + 0.999 999 989 494 473 084 108 8;
  • 40) 0.999 999 989 494 473 084 108 8 × 2 = 1 + 0.999 999 978 988 946 168 217 6;
  • 41) 0.999 999 978 988 946 168 217 6 × 2 = 1 + 0.999 999 957 977 892 336 435 2;
  • 42) 0.999 999 957 977 892 336 435 2 × 2 = 1 + 0.999 999 915 955 784 672 870 4;
  • 43) 0.999 999 915 955 784 672 870 4 × 2 = 1 + 0.999 999 831 911 569 345 740 8;
  • 44) 0.999 999 831 911 569 345 740 8 × 2 = 1 + 0.999 999 663 823 138 691 481 6;
  • 45) 0.999 999 663 823 138 691 481 6 × 2 = 1 + 0.999 999 327 646 277 382 963 2;
  • 46) 0.999 999 327 646 277 382 963 2 × 2 = 1 + 0.999 998 655 292 554 765 926 4;
  • 47) 0.999 998 655 292 554 765 926 4 × 2 = 1 + 0.999 997 310 585 109 531 852 8;
  • 48) 0.999 997 310 585 109 531 852 8 × 2 = 1 + 0.999 994 621 170 219 063 705 6;
  • 49) 0.999 994 621 170 219 063 705 6 × 2 = 1 + 0.999 989 242 340 438 127 411 2;
  • 50) 0.999 989 242 340 438 127 411 2 × 2 = 1 + 0.999 978 484 680 876 254 822 4;
  • 51) 0.999 978 484 680 876 254 822 4 × 2 = 1 + 0.999 956 969 361 752 509 644 8;
  • 52) 0.999 956 969 361 752 509 644 8 × 2 = 1 + 0.999 913 938 723 505 019 289 6;
  • 53) 0.999 913 938 723 505 019 289 6 × 2 = 1 + 0.999 827 877 447 010 038 579 2;
  • 54) 0.999 827 877 447 010 038 579 2 × 2 = 1 + 0.999 655 754 894 020 077 158 4;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 742 147 676 627 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 627 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 627 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 627 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111