-0.000 000 000 742 147 692 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 692 4(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
-0.000 000 000 742 147 692 4(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:

|-0.000 000 000 742 147 692 4| = 0.000 000 000 742 147 692 4


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: 0.000 000 000 742 147 692 4.

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) 0.000 000 000 742 147 692 4 × 2 = 0 + 0.000 000 001 484 295 384 8;
  • 2) 0.000 000 001 484 295 384 8 × 2 = 0 + 0.000 000 002 968 590 769 6;
  • 3) 0.000 000 002 968 590 769 6 × 2 = 0 + 0.000 000 005 937 181 539 2;
  • 4) 0.000 000 005 937 181 539 2 × 2 = 0 + 0.000 000 011 874 363 078 4;
  • 5) 0.000 000 011 874 363 078 4 × 2 = 0 + 0.000 000 023 748 726 156 8;
  • 6) 0.000 000 023 748 726 156 8 × 2 = 0 + 0.000 000 047 497 452 313 6;
  • 7) 0.000 000 047 497 452 313 6 × 2 = 0 + 0.000 000 094 994 904 627 2;
  • 8) 0.000 000 094 994 904 627 2 × 2 = 0 + 0.000 000 189 989 809 254 4;
  • 9) 0.000 000 189 989 809 254 4 × 2 = 0 + 0.000 000 379 979 618 508 8;
  • 10) 0.000 000 379 979 618 508 8 × 2 = 0 + 0.000 000 759 959 237 017 6;
  • 11) 0.000 000 759 959 237 017 6 × 2 = 0 + 0.000 001 519 918 474 035 2;
  • 12) 0.000 001 519 918 474 035 2 × 2 = 0 + 0.000 003 039 836 948 070 4;
  • 13) 0.000 003 039 836 948 070 4 × 2 = 0 + 0.000 006 079 673 896 140 8;
  • 14) 0.000 006 079 673 896 140 8 × 2 = 0 + 0.000 012 159 347 792 281 6;
  • 15) 0.000 012 159 347 792 281 6 × 2 = 0 + 0.000 024 318 695 584 563 2;
  • 16) 0.000 024 318 695 584 563 2 × 2 = 0 + 0.000 048 637 391 169 126 4;
  • 17) 0.000 048 637 391 169 126 4 × 2 = 0 + 0.000 097 274 782 338 252 8;
  • 18) 0.000 097 274 782 338 252 8 × 2 = 0 + 0.000 194 549 564 676 505 6;
  • 19) 0.000 194 549 564 676 505 6 × 2 = 0 + 0.000 389 099 129 353 011 2;
  • 20) 0.000 389 099 129 353 011 2 × 2 = 0 + 0.000 778 198 258 706 022 4;
  • 21) 0.000 778 198 258 706 022 4 × 2 = 0 + 0.001 556 396 517 412 044 8;
  • 22) 0.001 556 396 517 412 044 8 × 2 = 0 + 0.003 112 793 034 824 089 6;
  • 23) 0.003 112 793 034 824 089 6 × 2 = 0 + 0.006 225 586 069 648 179 2;
  • 24) 0.006 225 586 069 648 179 2 × 2 = 0 + 0.012 451 172 139 296 358 4;
  • 25) 0.012 451 172 139 296 358 4 × 2 = 0 + 0.024 902 344 278 592 716 8;
  • 26) 0.024 902 344 278 592 716 8 × 2 = 0 + 0.049 804 688 557 185 433 6;
  • 27) 0.049 804 688 557 185 433 6 × 2 = 0 + 0.099 609 377 114 370 867 2;
  • 28) 0.099 609 377 114 370 867 2 × 2 = 0 + 0.199 218 754 228 741 734 4;
  • 29) 0.199 218 754 228 741 734 4 × 2 = 0 + 0.398 437 508 457 483 468 8;
  • 30) 0.398 437 508 457 483 468 8 × 2 = 0 + 0.796 875 016 914 966 937 6;
  • 31) 0.796 875 016 914 966 937 6 × 2 = 1 + 0.593 750 033 829 933 875 2;
  • 32) 0.593 750 033 829 933 875 2 × 2 = 1 + 0.187 500 067 659 867 750 4;
  • 33) 0.187 500 067 659 867 750 4 × 2 = 0 + 0.375 000 135 319 735 500 8;
  • 34) 0.375 000 135 319 735 500 8 × 2 = 0 + 0.750 000 270 639 471 001 6;
  • 35) 0.750 000 270 639 471 001 6 × 2 = 1 + 0.500 000 541 278 942 003 2;
  • 36) 0.500 000 541 278 942 003 2 × 2 = 1 + 0.000 001 082 557 884 006 4;
  • 37) 0.000 001 082 557 884 006 4 × 2 = 0 + 0.000 002 165 115 768 012 8;
  • 38) 0.000 002 165 115 768 012 8 × 2 = 0 + 0.000 004 330 231 536 025 6;
  • 39) 0.000 004 330 231 536 025 6 × 2 = 0 + 0.000 008 660 463 072 051 2;
  • 40) 0.000 008 660 463 072 051 2 × 2 = 0 + 0.000 017 320 926 144 102 4;
  • 41) 0.000 017 320 926 144 102 4 × 2 = 0 + 0.000 034 641 852 288 204 8;
  • 42) 0.000 034 641 852 288 204 8 × 2 = 0 + 0.000 069 283 704 576 409 6;
  • 43) 0.000 069 283 704 576 409 6 × 2 = 0 + 0.000 138 567 409 152 819 2;
  • 44) 0.000 138 567 409 152 819 2 × 2 = 0 + 0.000 277 134 818 305 638 4;
  • 45) 0.000 277 134 818 305 638 4 × 2 = 0 + 0.000 554 269 636 611 276 8;
  • 46) 0.000 554 269 636 611 276 8 × 2 = 0 + 0.001 108 539 273 222 553 6;
  • 47) 0.001 108 539 273 222 553 6 × 2 = 0 + 0.002 217 078 546 445 107 2;
  • 48) 0.002 217 078 546 445 107 2 × 2 = 0 + 0.004 434 157 092 890 214 4;
  • 49) 0.004 434 157 092 890 214 4 × 2 = 0 + 0.008 868 314 185 780 428 8;
  • 50) 0.008 868 314 185 780 428 8 × 2 = 0 + 0.017 736 628 371 560 857 6;
  • 51) 0.017 736 628 371 560 857 6 × 2 = 0 + 0.035 473 256 743 121 715 2;
  • 52) 0.035 473 256 743 121 715 2 × 2 = 0 + 0.070 946 513 486 243 430 4;
  • 53) 0.070 946 513 486 243 430 4 × 2 = 0 + 0.141 893 026 972 486 860 8;
  • 54) 0.141 893 026 972 486 860 8 × 2 = 0 + 0.283 786 053 944 973 721 6;

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:


0.000 000 000 742 147 692 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 692 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 692 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 692 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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