-0.000 000 000 742 147 676 645 34 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 645 34(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
-0.000 000 000 742 147 676 645 34(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:

|-0.000 000 000 742 147 676 645 34| = 0.000 000 000 742 147 676 645 34


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: 0.000 000 000 742 147 676 645 34.

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) 0.000 000 000 742 147 676 645 34 × 2 = 0 + 0.000 000 001 484 295 353 290 68;
  • 2) 0.000 000 001 484 295 353 290 68 × 2 = 0 + 0.000 000 002 968 590 706 581 36;
  • 3) 0.000 000 002 968 590 706 581 36 × 2 = 0 + 0.000 000 005 937 181 413 162 72;
  • 4) 0.000 000 005 937 181 413 162 72 × 2 = 0 + 0.000 000 011 874 362 826 325 44;
  • 5) 0.000 000 011 874 362 826 325 44 × 2 = 0 + 0.000 000 023 748 725 652 650 88;
  • 6) 0.000 000 023 748 725 652 650 88 × 2 = 0 + 0.000 000 047 497 451 305 301 76;
  • 7) 0.000 000 047 497 451 305 301 76 × 2 = 0 + 0.000 000 094 994 902 610 603 52;
  • 8) 0.000 000 094 994 902 610 603 52 × 2 = 0 + 0.000 000 189 989 805 221 207 04;
  • 9) 0.000 000 189 989 805 221 207 04 × 2 = 0 + 0.000 000 379 979 610 442 414 08;
  • 10) 0.000 000 379 979 610 442 414 08 × 2 = 0 + 0.000 000 759 959 220 884 828 16;
  • 11) 0.000 000 759 959 220 884 828 16 × 2 = 0 + 0.000 001 519 918 441 769 656 32;
  • 12) 0.000 001 519 918 441 769 656 32 × 2 = 0 + 0.000 003 039 836 883 539 312 64;
  • 13) 0.000 003 039 836 883 539 312 64 × 2 = 0 + 0.000 006 079 673 767 078 625 28;
  • 14) 0.000 006 079 673 767 078 625 28 × 2 = 0 + 0.000 012 159 347 534 157 250 56;
  • 15) 0.000 012 159 347 534 157 250 56 × 2 = 0 + 0.000 024 318 695 068 314 501 12;
  • 16) 0.000 024 318 695 068 314 501 12 × 2 = 0 + 0.000 048 637 390 136 629 002 24;
  • 17) 0.000 048 637 390 136 629 002 24 × 2 = 0 + 0.000 097 274 780 273 258 004 48;
  • 18) 0.000 097 274 780 273 258 004 48 × 2 = 0 + 0.000 194 549 560 546 516 008 96;
  • 19) 0.000 194 549 560 546 516 008 96 × 2 = 0 + 0.000 389 099 121 093 032 017 92;
  • 20) 0.000 389 099 121 093 032 017 92 × 2 = 0 + 0.000 778 198 242 186 064 035 84;
  • 21) 0.000 778 198 242 186 064 035 84 × 2 = 0 + 0.001 556 396 484 372 128 071 68;
  • 22) 0.001 556 396 484 372 128 071 68 × 2 = 0 + 0.003 112 792 968 744 256 143 36;
  • 23) 0.003 112 792 968 744 256 143 36 × 2 = 0 + 0.006 225 585 937 488 512 286 72;
  • 24) 0.006 225 585 937 488 512 286 72 × 2 = 0 + 0.012 451 171 874 977 024 573 44;
  • 25) 0.012 451 171 874 977 024 573 44 × 2 = 0 + 0.024 902 343 749 954 049 146 88;
  • 26) 0.024 902 343 749 954 049 146 88 × 2 = 0 + 0.049 804 687 499 908 098 293 76;
  • 27) 0.049 804 687 499 908 098 293 76 × 2 = 0 + 0.099 609 374 999 816 196 587 52;
  • 28) 0.099 609 374 999 816 196 587 52 × 2 = 0 + 0.199 218 749 999 632 393 175 04;
  • 29) 0.199 218 749 999 632 393 175 04 × 2 = 0 + 0.398 437 499 999 264 786 350 08;
  • 30) 0.398 437 499 999 264 786 350 08 × 2 = 0 + 0.796 874 999 998 529 572 700 16;
  • 31) 0.796 874 999 998 529 572 700 16 × 2 = 1 + 0.593 749 999 997 059 145 400 32;
  • 32) 0.593 749 999 997 059 145 400 32 × 2 = 1 + 0.187 499 999 994 118 290 800 64;
  • 33) 0.187 499 999 994 118 290 800 64 × 2 = 0 + 0.374 999 999 988 236 581 601 28;
  • 34) 0.374 999 999 988 236 581 601 28 × 2 = 0 + 0.749 999 999 976 473 163 202 56;
  • 35) 0.749 999 999 976 473 163 202 56 × 2 = 1 + 0.499 999 999 952 946 326 405 12;
  • 36) 0.499 999 999 952 946 326 405 12 × 2 = 0 + 0.999 999 999 905 892 652 810 24;
  • 37) 0.999 999 999 905 892 652 810 24 × 2 = 1 + 0.999 999 999 811 785 305 620 48;
  • 38) 0.999 999 999 811 785 305 620 48 × 2 = 1 + 0.999 999 999 623 570 611 240 96;
  • 39) 0.999 999 999 623 570 611 240 96 × 2 = 1 + 0.999 999 999 247 141 222 481 92;
  • 40) 0.999 999 999 247 141 222 481 92 × 2 = 1 + 0.999 999 998 494 282 444 963 84;
  • 41) 0.999 999 998 494 282 444 963 84 × 2 = 1 + 0.999 999 996 988 564 889 927 68;
  • 42) 0.999 999 996 988 564 889 927 68 × 2 = 1 + 0.999 999 993 977 129 779 855 36;
  • 43) 0.999 999 993 977 129 779 855 36 × 2 = 1 + 0.999 999 987 954 259 559 710 72;
  • 44) 0.999 999 987 954 259 559 710 72 × 2 = 1 + 0.999 999 975 908 519 119 421 44;
  • 45) 0.999 999 975 908 519 119 421 44 × 2 = 1 + 0.999 999 951 817 038 238 842 88;
  • 46) 0.999 999 951 817 038 238 842 88 × 2 = 1 + 0.999 999 903 634 076 477 685 76;
  • 47) 0.999 999 903 634 076 477 685 76 × 2 = 1 + 0.999 999 807 268 152 955 371 52;
  • 48) 0.999 999 807 268 152 955 371 52 × 2 = 1 + 0.999 999 614 536 305 910 743 04;
  • 49) 0.999 999 614 536 305 910 743 04 × 2 = 1 + 0.999 999 229 072 611 821 486 08;
  • 50) 0.999 999 229 072 611 821 486 08 × 2 = 1 + 0.999 998 458 145 223 642 972 16;
  • 51) 0.999 998 458 145 223 642 972 16 × 2 = 1 + 0.999 996 916 290 447 285 944 32;
  • 52) 0.999 996 916 290 447 285 944 32 × 2 = 1 + 0.999 993 832 580 894 571 888 64;
  • 53) 0.999 993 832 580 894 571 888 64 × 2 = 1 + 0.999 987 665 161 789 143 777 28;
  • 54) 0.999 987 665 161 789 143 777 28 × 2 = 1 + 0.999 975 330 323 578 287 554 56;

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:


0.000 000 000 742 147 676 645 34(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 645 34(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:


0.000 000 000 742 147 676 645 34(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 -0.000 000 000 742 147 676 645 34 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