-0.000 000 000 742 147 676 646 693 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 693 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
-0.000 000 000 742 147 676 646 693 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:

|-0.000 000 000 742 147 676 646 693 6| = 0.000 000 000 742 147 676 646 693 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: 0.000 000 000 742 147 676 646 693 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) 0.000 000 000 742 147 676 646 693 6 × 2 = 0 + 0.000 000 001 484 295 353 293 387 2;
  • 2) 0.000 000 001 484 295 353 293 387 2 × 2 = 0 + 0.000 000 002 968 590 706 586 774 4;
  • 3) 0.000 000 002 968 590 706 586 774 4 × 2 = 0 + 0.000 000 005 937 181 413 173 548 8;
  • 4) 0.000 000 005 937 181 413 173 548 8 × 2 = 0 + 0.000 000 011 874 362 826 347 097 6;
  • 5) 0.000 000 011 874 362 826 347 097 6 × 2 = 0 + 0.000 000 023 748 725 652 694 195 2;
  • 6) 0.000 000 023 748 725 652 694 195 2 × 2 = 0 + 0.000 000 047 497 451 305 388 390 4;
  • 7) 0.000 000 047 497 451 305 388 390 4 × 2 = 0 + 0.000 000 094 994 902 610 776 780 8;
  • 8) 0.000 000 094 994 902 610 776 780 8 × 2 = 0 + 0.000 000 189 989 805 221 553 561 6;
  • 9) 0.000 000 189 989 805 221 553 561 6 × 2 = 0 + 0.000 000 379 979 610 443 107 123 2;
  • 10) 0.000 000 379 979 610 443 107 123 2 × 2 = 0 + 0.000 000 759 959 220 886 214 246 4;
  • 11) 0.000 000 759 959 220 886 214 246 4 × 2 = 0 + 0.000 001 519 918 441 772 428 492 8;
  • 12) 0.000 001 519 918 441 772 428 492 8 × 2 = 0 + 0.000 003 039 836 883 544 856 985 6;
  • 13) 0.000 003 039 836 883 544 856 985 6 × 2 = 0 + 0.000 006 079 673 767 089 713 971 2;
  • 14) 0.000 006 079 673 767 089 713 971 2 × 2 = 0 + 0.000 012 159 347 534 179 427 942 4;
  • 15) 0.000 012 159 347 534 179 427 942 4 × 2 = 0 + 0.000 024 318 695 068 358 855 884 8;
  • 16) 0.000 024 318 695 068 358 855 884 8 × 2 = 0 + 0.000 048 637 390 136 717 711 769 6;
  • 17) 0.000 048 637 390 136 717 711 769 6 × 2 = 0 + 0.000 097 274 780 273 435 423 539 2;
  • 18) 0.000 097 274 780 273 435 423 539 2 × 2 = 0 + 0.000 194 549 560 546 870 847 078 4;
  • 19) 0.000 194 549 560 546 870 847 078 4 × 2 = 0 + 0.000 389 099 121 093 741 694 156 8;
  • 20) 0.000 389 099 121 093 741 694 156 8 × 2 = 0 + 0.000 778 198 242 187 483 388 313 6;
  • 21) 0.000 778 198 242 187 483 388 313 6 × 2 = 0 + 0.001 556 396 484 374 966 776 627 2;
  • 22) 0.001 556 396 484 374 966 776 627 2 × 2 = 0 + 0.003 112 792 968 749 933 553 254 4;
  • 23) 0.003 112 792 968 749 933 553 254 4 × 2 = 0 + 0.006 225 585 937 499 867 106 508 8;
  • 24) 0.006 225 585 937 499 867 106 508 8 × 2 = 0 + 0.012 451 171 874 999 734 213 017 6;
  • 25) 0.012 451 171 874 999 734 213 017 6 × 2 = 0 + 0.024 902 343 749 999 468 426 035 2;
  • 26) 0.024 902 343 749 999 468 426 035 2 × 2 = 0 + 0.049 804 687 499 998 936 852 070 4;
  • 27) 0.049 804 687 499 998 936 852 070 4 × 2 = 0 + 0.099 609 374 999 997 873 704 140 8;
  • 28) 0.099 609 374 999 997 873 704 140 8 × 2 = 0 + 0.199 218 749 999 995 747 408 281 6;
  • 29) 0.199 218 749 999 995 747 408 281 6 × 2 = 0 + 0.398 437 499 999 991 494 816 563 2;
  • 30) 0.398 437 499 999 991 494 816 563 2 × 2 = 0 + 0.796 874 999 999 982 989 633 126 4;
  • 31) 0.796 874 999 999 982 989 633 126 4 × 2 = 1 + 0.593 749 999 999 965 979 266 252 8;
  • 32) 0.593 749 999 999 965 979 266 252 8 × 2 = 1 + 0.187 499 999 999 931 958 532 505 6;
  • 33) 0.187 499 999 999 931 958 532 505 6 × 2 = 0 + 0.374 999 999 999 863 917 065 011 2;
  • 34) 0.374 999 999 999 863 917 065 011 2 × 2 = 0 + 0.749 999 999 999 727 834 130 022 4;
  • 35) 0.749 999 999 999 727 834 130 022 4 × 2 = 1 + 0.499 999 999 999 455 668 260 044 8;
  • 36) 0.499 999 999 999 455 668 260 044 8 × 2 = 0 + 0.999 999 999 998 911 336 520 089 6;
  • 37) 0.999 999 999 998 911 336 520 089 6 × 2 = 1 + 0.999 999 999 997 822 673 040 179 2;
  • 38) 0.999 999 999 997 822 673 040 179 2 × 2 = 1 + 0.999 999 999 995 645 346 080 358 4;
  • 39) 0.999 999 999 995 645 346 080 358 4 × 2 = 1 + 0.999 999 999 991 290 692 160 716 8;
  • 40) 0.999 999 999 991 290 692 160 716 8 × 2 = 1 + 0.999 999 999 982 581 384 321 433 6;
  • 41) 0.999 999 999 982 581 384 321 433 6 × 2 = 1 + 0.999 999 999 965 162 768 642 867 2;
  • 42) 0.999 999 999 965 162 768 642 867 2 × 2 = 1 + 0.999 999 999 930 325 537 285 734 4;
  • 43) 0.999 999 999 930 325 537 285 734 4 × 2 = 1 + 0.999 999 999 860 651 074 571 468 8;
  • 44) 0.999 999 999 860 651 074 571 468 8 × 2 = 1 + 0.999 999 999 721 302 149 142 937 6;
  • 45) 0.999 999 999 721 302 149 142 937 6 × 2 = 1 + 0.999 999 999 442 604 298 285 875 2;
  • 46) 0.999 999 999 442 604 298 285 875 2 × 2 = 1 + 0.999 999 998 885 208 596 571 750 4;
  • 47) 0.999 999 998 885 208 596 571 750 4 × 2 = 1 + 0.999 999 997 770 417 193 143 500 8;
  • 48) 0.999 999 997 770 417 193 143 500 8 × 2 = 1 + 0.999 999 995 540 834 386 287 001 6;
  • 49) 0.999 999 995 540 834 386 287 001 6 × 2 = 1 + 0.999 999 991 081 668 772 574 003 2;
  • 50) 0.999 999 991 081 668 772 574 003 2 × 2 = 1 + 0.999 999 982 163 337 545 148 006 4;
  • 51) 0.999 999 982 163 337 545 148 006 4 × 2 = 1 + 0.999 999 964 326 675 090 296 012 8;
  • 52) 0.999 999 964 326 675 090 296 012 8 × 2 = 1 + 0.999 999 928 653 350 180 592 025 6;
  • 53) 0.999 999 928 653 350 180 592 025 6 × 2 = 1 + 0.999 999 857 306 700 361 184 051 2;
  • 54) 0.999 999 857 306 700 361 184 051 2 × 2 = 1 + 0.999 999 714 613 400 722 368 102 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:


0.000 000 000 742 147 676 646 693 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 693 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:


0.000 000 000 742 147 676 646 693 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 -0.000 000 000 742 147 676 646 693 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