-0.000 000 000 742 147 676 621 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 621 8(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
-0.000 000 000 742 147 676 621 8(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:

|-0.000 000 000 742 147 676 621 8| = 0.000 000 000 742 147 676 621 8


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: 0.000 000 000 742 147 676 621 8.

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) 0.000 000 000 742 147 676 621 8 × 2 = 0 + 0.000 000 001 484 295 353 243 6;
  • 2) 0.000 000 001 484 295 353 243 6 × 2 = 0 + 0.000 000 002 968 590 706 487 2;
  • 3) 0.000 000 002 968 590 706 487 2 × 2 = 0 + 0.000 000 005 937 181 412 974 4;
  • 4) 0.000 000 005 937 181 412 974 4 × 2 = 0 + 0.000 000 011 874 362 825 948 8;
  • 5) 0.000 000 011 874 362 825 948 8 × 2 = 0 + 0.000 000 023 748 725 651 897 6;
  • 6) 0.000 000 023 748 725 651 897 6 × 2 = 0 + 0.000 000 047 497 451 303 795 2;
  • 7) 0.000 000 047 497 451 303 795 2 × 2 = 0 + 0.000 000 094 994 902 607 590 4;
  • 8) 0.000 000 094 994 902 607 590 4 × 2 = 0 + 0.000 000 189 989 805 215 180 8;
  • 9) 0.000 000 189 989 805 215 180 8 × 2 = 0 + 0.000 000 379 979 610 430 361 6;
  • 10) 0.000 000 379 979 610 430 361 6 × 2 = 0 + 0.000 000 759 959 220 860 723 2;
  • 11) 0.000 000 759 959 220 860 723 2 × 2 = 0 + 0.000 001 519 918 441 721 446 4;
  • 12) 0.000 001 519 918 441 721 446 4 × 2 = 0 + 0.000 003 039 836 883 442 892 8;
  • 13) 0.000 003 039 836 883 442 892 8 × 2 = 0 + 0.000 006 079 673 766 885 785 6;
  • 14) 0.000 006 079 673 766 885 785 6 × 2 = 0 + 0.000 012 159 347 533 771 571 2;
  • 15) 0.000 012 159 347 533 771 571 2 × 2 = 0 + 0.000 024 318 695 067 543 142 4;
  • 16) 0.000 024 318 695 067 543 142 4 × 2 = 0 + 0.000 048 637 390 135 086 284 8;
  • 17) 0.000 048 637 390 135 086 284 8 × 2 = 0 + 0.000 097 274 780 270 172 569 6;
  • 18) 0.000 097 274 780 270 172 569 6 × 2 = 0 + 0.000 194 549 560 540 345 139 2;
  • 19) 0.000 194 549 560 540 345 139 2 × 2 = 0 + 0.000 389 099 121 080 690 278 4;
  • 20) 0.000 389 099 121 080 690 278 4 × 2 = 0 + 0.000 778 198 242 161 380 556 8;
  • 21) 0.000 778 198 242 161 380 556 8 × 2 = 0 + 0.001 556 396 484 322 761 113 6;
  • 22) 0.001 556 396 484 322 761 113 6 × 2 = 0 + 0.003 112 792 968 645 522 227 2;
  • 23) 0.003 112 792 968 645 522 227 2 × 2 = 0 + 0.006 225 585 937 291 044 454 4;
  • 24) 0.006 225 585 937 291 044 454 4 × 2 = 0 + 0.012 451 171 874 582 088 908 8;
  • 25) 0.012 451 171 874 582 088 908 8 × 2 = 0 + 0.024 902 343 749 164 177 817 6;
  • 26) 0.024 902 343 749 164 177 817 6 × 2 = 0 + 0.049 804 687 498 328 355 635 2;
  • 27) 0.049 804 687 498 328 355 635 2 × 2 = 0 + 0.099 609 374 996 656 711 270 4;
  • 28) 0.099 609 374 996 656 711 270 4 × 2 = 0 + 0.199 218 749 993 313 422 540 8;
  • 29) 0.199 218 749 993 313 422 540 8 × 2 = 0 + 0.398 437 499 986 626 845 081 6;
  • 30) 0.398 437 499 986 626 845 081 6 × 2 = 0 + 0.796 874 999 973 253 690 163 2;
  • 31) 0.796 874 999 973 253 690 163 2 × 2 = 1 + 0.593 749 999 946 507 380 326 4;
  • 32) 0.593 749 999 946 507 380 326 4 × 2 = 1 + 0.187 499 999 893 014 760 652 8;
  • 33) 0.187 499 999 893 014 760 652 8 × 2 = 0 + 0.374 999 999 786 029 521 305 6;
  • 34) 0.374 999 999 786 029 521 305 6 × 2 = 0 + 0.749 999 999 572 059 042 611 2;
  • 35) 0.749 999 999 572 059 042 611 2 × 2 = 1 + 0.499 999 999 144 118 085 222 4;
  • 36) 0.499 999 999 144 118 085 222 4 × 2 = 0 + 0.999 999 998 288 236 170 444 8;
  • 37) 0.999 999 998 288 236 170 444 8 × 2 = 1 + 0.999 999 996 576 472 340 889 6;
  • 38) 0.999 999 996 576 472 340 889 6 × 2 = 1 + 0.999 999 993 152 944 681 779 2;
  • 39) 0.999 999 993 152 944 681 779 2 × 2 = 1 + 0.999 999 986 305 889 363 558 4;
  • 40) 0.999 999 986 305 889 363 558 4 × 2 = 1 + 0.999 999 972 611 778 727 116 8;
  • 41) 0.999 999 972 611 778 727 116 8 × 2 = 1 + 0.999 999 945 223 557 454 233 6;
  • 42) 0.999 999 945 223 557 454 233 6 × 2 = 1 + 0.999 999 890 447 114 908 467 2;
  • 43) 0.999 999 890 447 114 908 467 2 × 2 = 1 + 0.999 999 780 894 229 816 934 4;
  • 44) 0.999 999 780 894 229 816 934 4 × 2 = 1 + 0.999 999 561 788 459 633 868 8;
  • 45) 0.999 999 561 788 459 633 868 8 × 2 = 1 + 0.999 999 123 576 919 267 737 6;
  • 46) 0.999 999 123 576 919 267 737 6 × 2 = 1 + 0.999 998 247 153 838 535 475 2;
  • 47) 0.999 998 247 153 838 535 475 2 × 2 = 1 + 0.999 996 494 307 677 070 950 4;
  • 48) 0.999 996 494 307 677 070 950 4 × 2 = 1 + 0.999 992 988 615 354 141 900 8;
  • 49) 0.999 992 988 615 354 141 900 8 × 2 = 1 + 0.999 985 977 230 708 283 801 6;
  • 50) 0.999 985 977 230 708 283 801 6 × 2 = 1 + 0.999 971 954 461 416 567 603 2;
  • 51) 0.999 971 954 461 416 567 603 2 × 2 = 1 + 0.999 943 908 922 833 135 206 4;
  • 52) 0.999 943 908 922 833 135 206 4 × 2 = 1 + 0.999 887 817 845 666 270 412 8;
  • 53) 0.999 887 817 845 666 270 412 8 × 2 = 1 + 0.999 775 635 691 332 540 825 6;
  • 54) 0.999 775 635 691 332 540 825 6 × 2 = 1 + 0.999 551 271 382 665 081 651 2;

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:


0.000 000 000 742 147 676 621 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 621 8(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:


0.000 000 000 742 147 676 621 8(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 -0.000 000 000 742 147 676 621 8 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