-0.000 000 000 742 147 676 646 688 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 688 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
-0.000 000 000 742 147 676 646 688 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:

|-0.000 000 000 742 147 676 646 688 6| = 0.000 000 000 742 147 676 646 688 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: 0.000 000 000 742 147 676 646 688 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) 0.000 000 000 742 147 676 646 688 6 × 2 = 0 + 0.000 000 001 484 295 353 293 377 2;
  • 2) 0.000 000 001 484 295 353 293 377 2 × 2 = 0 + 0.000 000 002 968 590 706 586 754 4;
  • 3) 0.000 000 002 968 590 706 586 754 4 × 2 = 0 + 0.000 000 005 937 181 413 173 508 8;
  • 4) 0.000 000 005 937 181 413 173 508 8 × 2 = 0 + 0.000 000 011 874 362 826 347 017 6;
  • 5) 0.000 000 011 874 362 826 347 017 6 × 2 = 0 + 0.000 000 023 748 725 652 694 035 2;
  • 6) 0.000 000 023 748 725 652 694 035 2 × 2 = 0 + 0.000 000 047 497 451 305 388 070 4;
  • 7) 0.000 000 047 497 451 305 388 070 4 × 2 = 0 + 0.000 000 094 994 902 610 776 140 8;
  • 8) 0.000 000 094 994 902 610 776 140 8 × 2 = 0 + 0.000 000 189 989 805 221 552 281 6;
  • 9) 0.000 000 189 989 805 221 552 281 6 × 2 = 0 + 0.000 000 379 979 610 443 104 563 2;
  • 10) 0.000 000 379 979 610 443 104 563 2 × 2 = 0 + 0.000 000 759 959 220 886 209 126 4;
  • 11) 0.000 000 759 959 220 886 209 126 4 × 2 = 0 + 0.000 001 519 918 441 772 418 252 8;
  • 12) 0.000 001 519 918 441 772 418 252 8 × 2 = 0 + 0.000 003 039 836 883 544 836 505 6;
  • 13) 0.000 003 039 836 883 544 836 505 6 × 2 = 0 + 0.000 006 079 673 767 089 673 011 2;
  • 14) 0.000 006 079 673 767 089 673 011 2 × 2 = 0 + 0.000 012 159 347 534 179 346 022 4;
  • 15) 0.000 012 159 347 534 179 346 022 4 × 2 = 0 + 0.000 024 318 695 068 358 692 044 8;
  • 16) 0.000 024 318 695 068 358 692 044 8 × 2 = 0 + 0.000 048 637 390 136 717 384 089 6;
  • 17) 0.000 048 637 390 136 717 384 089 6 × 2 = 0 + 0.000 097 274 780 273 434 768 179 2;
  • 18) 0.000 097 274 780 273 434 768 179 2 × 2 = 0 + 0.000 194 549 560 546 869 536 358 4;
  • 19) 0.000 194 549 560 546 869 536 358 4 × 2 = 0 + 0.000 389 099 121 093 739 072 716 8;
  • 20) 0.000 389 099 121 093 739 072 716 8 × 2 = 0 + 0.000 778 198 242 187 478 145 433 6;
  • 21) 0.000 778 198 242 187 478 145 433 6 × 2 = 0 + 0.001 556 396 484 374 956 290 867 2;
  • 22) 0.001 556 396 484 374 956 290 867 2 × 2 = 0 + 0.003 112 792 968 749 912 581 734 4;
  • 23) 0.003 112 792 968 749 912 581 734 4 × 2 = 0 + 0.006 225 585 937 499 825 163 468 8;
  • 24) 0.006 225 585 937 499 825 163 468 8 × 2 = 0 + 0.012 451 171 874 999 650 326 937 6;
  • 25) 0.012 451 171 874 999 650 326 937 6 × 2 = 0 + 0.024 902 343 749 999 300 653 875 2;
  • 26) 0.024 902 343 749 999 300 653 875 2 × 2 = 0 + 0.049 804 687 499 998 601 307 750 4;
  • 27) 0.049 804 687 499 998 601 307 750 4 × 2 = 0 + 0.099 609 374 999 997 202 615 500 8;
  • 28) 0.099 609 374 999 997 202 615 500 8 × 2 = 0 + 0.199 218 749 999 994 405 231 001 6;
  • 29) 0.199 218 749 999 994 405 231 001 6 × 2 = 0 + 0.398 437 499 999 988 810 462 003 2;
  • 30) 0.398 437 499 999 988 810 462 003 2 × 2 = 0 + 0.796 874 999 999 977 620 924 006 4;
  • 31) 0.796 874 999 999 977 620 924 006 4 × 2 = 1 + 0.593 749 999 999 955 241 848 012 8;
  • 32) 0.593 749 999 999 955 241 848 012 8 × 2 = 1 + 0.187 499 999 999 910 483 696 025 6;
  • 33) 0.187 499 999 999 910 483 696 025 6 × 2 = 0 + 0.374 999 999 999 820 967 392 051 2;
  • 34) 0.374 999 999 999 820 967 392 051 2 × 2 = 0 + 0.749 999 999 999 641 934 784 102 4;
  • 35) 0.749 999 999 999 641 934 784 102 4 × 2 = 1 + 0.499 999 999 999 283 869 568 204 8;
  • 36) 0.499 999 999 999 283 869 568 204 8 × 2 = 0 + 0.999 999 999 998 567 739 136 409 6;
  • 37) 0.999 999 999 998 567 739 136 409 6 × 2 = 1 + 0.999 999 999 997 135 478 272 819 2;
  • 38) 0.999 999 999 997 135 478 272 819 2 × 2 = 1 + 0.999 999 999 994 270 956 545 638 4;
  • 39) 0.999 999 999 994 270 956 545 638 4 × 2 = 1 + 0.999 999 999 988 541 913 091 276 8;
  • 40) 0.999 999 999 988 541 913 091 276 8 × 2 = 1 + 0.999 999 999 977 083 826 182 553 6;
  • 41) 0.999 999 999 977 083 826 182 553 6 × 2 = 1 + 0.999 999 999 954 167 652 365 107 2;
  • 42) 0.999 999 999 954 167 652 365 107 2 × 2 = 1 + 0.999 999 999 908 335 304 730 214 4;
  • 43) 0.999 999 999 908 335 304 730 214 4 × 2 = 1 + 0.999 999 999 816 670 609 460 428 8;
  • 44) 0.999 999 999 816 670 609 460 428 8 × 2 = 1 + 0.999 999 999 633 341 218 920 857 6;
  • 45) 0.999 999 999 633 341 218 920 857 6 × 2 = 1 + 0.999 999 999 266 682 437 841 715 2;
  • 46) 0.999 999 999 266 682 437 841 715 2 × 2 = 1 + 0.999 999 998 533 364 875 683 430 4;
  • 47) 0.999 999 998 533 364 875 683 430 4 × 2 = 1 + 0.999 999 997 066 729 751 366 860 8;
  • 48) 0.999 999 997 066 729 751 366 860 8 × 2 = 1 + 0.999 999 994 133 459 502 733 721 6;
  • 49) 0.999 999 994 133 459 502 733 721 6 × 2 = 1 + 0.999 999 988 266 919 005 467 443 2;
  • 50) 0.999 999 988 266 919 005 467 443 2 × 2 = 1 + 0.999 999 976 533 838 010 934 886 4;
  • 51) 0.999 999 976 533 838 010 934 886 4 × 2 = 1 + 0.999 999 953 067 676 021 869 772 8;
  • 52) 0.999 999 953 067 676 021 869 772 8 × 2 = 1 + 0.999 999 906 135 352 043 739 545 6;
  • 53) 0.999 999 906 135 352 043 739 545 6 × 2 = 1 + 0.999 999 812 270 704 087 479 091 2;
  • 54) 0.999 999 812 270 704 087 479 091 2 × 2 = 1 + 0.999 999 624 541 408 174 958 182 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:


0.000 000 000 742 147 676 646 688 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 688 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:


0.000 000 000 742 147 676 646 688 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 -0.000 000 000 742 147 676 646 688 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