-0.000 000 000 742 073 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 073(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
-0.000 000 000 742 073(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:

|-0.000 000 000 742 073| = 0.000 000 000 742 073


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: 0.000 000 000 742 073.

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) 0.000 000 000 742 073 × 2 = 0 + 0.000 000 001 484 146;
  • 2) 0.000 000 001 484 146 × 2 = 0 + 0.000 000 002 968 292;
  • 3) 0.000 000 002 968 292 × 2 = 0 + 0.000 000 005 936 584;
  • 4) 0.000 000 005 936 584 × 2 = 0 + 0.000 000 011 873 168;
  • 5) 0.000 000 011 873 168 × 2 = 0 + 0.000 000 023 746 336;
  • 6) 0.000 000 023 746 336 × 2 = 0 + 0.000 000 047 492 672;
  • 7) 0.000 000 047 492 672 × 2 = 0 + 0.000 000 094 985 344;
  • 8) 0.000 000 094 985 344 × 2 = 0 + 0.000 000 189 970 688;
  • 9) 0.000 000 189 970 688 × 2 = 0 + 0.000 000 379 941 376;
  • 10) 0.000 000 379 941 376 × 2 = 0 + 0.000 000 759 882 752;
  • 11) 0.000 000 759 882 752 × 2 = 0 + 0.000 001 519 765 504;
  • 12) 0.000 001 519 765 504 × 2 = 0 + 0.000 003 039 531 008;
  • 13) 0.000 003 039 531 008 × 2 = 0 + 0.000 006 079 062 016;
  • 14) 0.000 006 079 062 016 × 2 = 0 + 0.000 012 158 124 032;
  • 15) 0.000 012 158 124 032 × 2 = 0 + 0.000 024 316 248 064;
  • 16) 0.000 024 316 248 064 × 2 = 0 + 0.000 048 632 496 128;
  • 17) 0.000 048 632 496 128 × 2 = 0 + 0.000 097 264 992 256;
  • 18) 0.000 097 264 992 256 × 2 = 0 + 0.000 194 529 984 512;
  • 19) 0.000 194 529 984 512 × 2 = 0 + 0.000 389 059 969 024;
  • 20) 0.000 389 059 969 024 × 2 = 0 + 0.000 778 119 938 048;
  • 21) 0.000 778 119 938 048 × 2 = 0 + 0.001 556 239 876 096;
  • 22) 0.001 556 239 876 096 × 2 = 0 + 0.003 112 479 752 192;
  • 23) 0.003 112 479 752 192 × 2 = 0 + 0.006 224 959 504 384;
  • 24) 0.006 224 959 504 384 × 2 = 0 + 0.012 449 919 008 768;
  • 25) 0.012 449 919 008 768 × 2 = 0 + 0.024 899 838 017 536;
  • 26) 0.024 899 838 017 536 × 2 = 0 + 0.049 799 676 035 072;
  • 27) 0.049 799 676 035 072 × 2 = 0 + 0.099 599 352 070 144;
  • 28) 0.099 599 352 070 144 × 2 = 0 + 0.199 198 704 140 288;
  • 29) 0.199 198 704 140 288 × 2 = 0 + 0.398 397 408 280 576;
  • 30) 0.398 397 408 280 576 × 2 = 0 + 0.796 794 816 561 152;
  • 31) 0.796 794 816 561 152 × 2 = 1 + 0.593 589 633 122 304;
  • 32) 0.593 589 633 122 304 × 2 = 1 + 0.187 179 266 244 608;
  • 33) 0.187 179 266 244 608 × 2 = 0 + 0.374 358 532 489 216;
  • 34) 0.374 358 532 489 216 × 2 = 0 + 0.748 717 064 978 432;
  • 35) 0.748 717 064 978 432 × 2 = 1 + 0.497 434 129 956 864;
  • 36) 0.497 434 129 956 864 × 2 = 0 + 0.994 868 259 913 728;
  • 37) 0.994 868 259 913 728 × 2 = 1 + 0.989 736 519 827 456;
  • 38) 0.989 736 519 827 456 × 2 = 1 + 0.979 473 039 654 912;
  • 39) 0.979 473 039 654 912 × 2 = 1 + 0.958 946 079 309 824;
  • 40) 0.958 946 079 309 824 × 2 = 1 + 0.917 892 158 619 648;
  • 41) 0.917 892 158 619 648 × 2 = 1 + 0.835 784 317 239 296;
  • 42) 0.835 784 317 239 296 × 2 = 1 + 0.671 568 634 478 592;
  • 43) 0.671 568 634 478 592 × 2 = 1 + 0.343 137 268 957 184;
  • 44) 0.343 137 268 957 184 × 2 = 0 + 0.686 274 537 914 368;
  • 45) 0.686 274 537 914 368 × 2 = 1 + 0.372 549 075 828 736;
  • 46) 0.372 549 075 828 736 × 2 = 0 + 0.745 098 151 657 472;
  • 47) 0.745 098 151 657 472 × 2 = 1 + 0.490 196 303 314 944;
  • 48) 0.490 196 303 314 944 × 2 = 0 + 0.980 392 606 629 888;
  • 49) 0.980 392 606 629 888 × 2 = 1 + 0.960 785 213 259 776;
  • 50) 0.960 785 213 259 776 × 2 = 1 + 0.921 570 426 519 552;
  • 51) 0.921 570 426 519 552 × 2 = 1 + 0.843 140 853 039 104;
  • 52) 0.843 140 853 039 104 × 2 = 1 + 0.686 281 706 078 208;
  • 53) 0.686 281 706 078 208 × 2 = 1 + 0.372 563 412 156 416;
  • 54) 0.372 563 412 156 416 × 2 = 0 + 0.745 126 824 312 832;

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:


0.000 000 000 742 073(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 1111 10(2)

6. Positive number before normalization:

0.000 000 000 742 073(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 1111 10(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:


0.000 000 000 742 073(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 1111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 1111 10(2) × 20 =

1.1001 0111 1111 0101 0111 110(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 0101 0111 110


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 1010 1011 1110 =

100 1011 1111 1010 1011 1110


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 1010 1011 1110


Decimal number -0.000 000 000 742 073 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1010 1011 1110

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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