-0.000 000 000 742 068 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 068(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
-0.000 000 000 742 068(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:

|-0.000 000 000 742 068| = 0.000 000 000 742 068


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: 0.000 000 000 742 068.

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) 0.000 000 000 742 068 × 2 = 0 + 0.000 000 001 484 136;
  • 2) 0.000 000 001 484 136 × 2 = 0 + 0.000 000 002 968 272;
  • 3) 0.000 000 002 968 272 × 2 = 0 + 0.000 000 005 936 544;
  • 4) 0.000 000 005 936 544 × 2 = 0 + 0.000 000 011 873 088;
  • 5) 0.000 000 011 873 088 × 2 = 0 + 0.000 000 023 746 176;
  • 6) 0.000 000 023 746 176 × 2 = 0 + 0.000 000 047 492 352;
  • 7) 0.000 000 047 492 352 × 2 = 0 + 0.000 000 094 984 704;
  • 8) 0.000 000 094 984 704 × 2 = 0 + 0.000 000 189 969 408;
  • 9) 0.000 000 189 969 408 × 2 = 0 + 0.000 000 379 938 816;
  • 10) 0.000 000 379 938 816 × 2 = 0 + 0.000 000 759 877 632;
  • 11) 0.000 000 759 877 632 × 2 = 0 + 0.000 001 519 755 264;
  • 12) 0.000 001 519 755 264 × 2 = 0 + 0.000 003 039 510 528;
  • 13) 0.000 003 039 510 528 × 2 = 0 + 0.000 006 079 021 056;
  • 14) 0.000 006 079 021 056 × 2 = 0 + 0.000 012 158 042 112;
  • 15) 0.000 012 158 042 112 × 2 = 0 + 0.000 024 316 084 224;
  • 16) 0.000 024 316 084 224 × 2 = 0 + 0.000 048 632 168 448;
  • 17) 0.000 048 632 168 448 × 2 = 0 + 0.000 097 264 336 896;
  • 18) 0.000 097 264 336 896 × 2 = 0 + 0.000 194 528 673 792;
  • 19) 0.000 194 528 673 792 × 2 = 0 + 0.000 389 057 347 584;
  • 20) 0.000 389 057 347 584 × 2 = 0 + 0.000 778 114 695 168;
  • 21) 0.000 778 114 695 168 × 2 = 0 + 0.001 556 229 390 336;
  • 22) 0.001 556 229 390 336 × 2 = 0 + 0.003 112 458 780 672;
  • 23) 0.003 112 458 780 672 × 2 = 0 + 0.006 224 917 561 344;
  • 24) 0.006 224 917 561 344 × 2 = 0 + 0.012 449 835 122 688;
  • 25) 0.012 449 835 122 688 × 2 = 0 + 0.024 899 670 245 376;
  • 26) 0.024 899 670 245 376 × 2 = 0 + 0.049 799 340 490 752;
  • 27) 0.049 799 340 490 752 × 2 = 0 + 0.099 598 680 981 504;
  • 28) 0.099 598 680 981 504 × 2 = 0 + 0.199 197 361 963 008;
  • 29) 0.199 197 361 963 008 × 2 = 0 + 0.398 394 723 926 016;
  • 30) 0.398 394 723 926 016 × 2 = 0 + 0.796 789 447 852 032;
  • 31) 0.796 789 447 852 032 × 2 = 1 + 0.593 578 895 704 064;
  • 32) 0.593 578 895 704 064 × 2 = 1 + 0.187 157 791 408 128;
  • 33) 0.187 157 791 408 128 × 2 = 0 + 0.374 315 582 816 256;
  • 34) 0.374 315 582 816 256 × 2 = 0 + 0.748 631 165 632 512;
  • 35) 0.748 631 165 632 512 × 2 = 1 + 0.497 262 331 265 024;
  • 36) 0.497 262 331 265 024 × 2 = 0 + 0.994 524 662 530 048;
  • 37) 0.994 524 662 530 048 × 2 = 1 + 0.989 049 325 060 096;
  • 38) 0.989 049 325 060 096 × 2 = 1 + 0.978 098 650 120 192;
  • 39) 0.978 098 650 120 192 × 2 = 1 + 0.956 197 300 240 384;
  • 40) 0.956 197 300 240 384 × 2 = 1 + 0.912 394 600 480 768;
  • 41) 0.912 394 600 480 768 × 2 = 1 + 0.824 789 200 961 536;
  • 42) 0.824 789 200 961 536 × 2 = 1 + 0.649 578 401 923 072;
  • 43) 0.649 578 401 923 072 × 2 = 1 + 0.299 156 803 846 144;
  • 44) 0.299 156 803 846 144 × 2 = 0 + 0.598 313 607 692 288;
  • 45) 0.598 313 607 692 288 × 2 = 1 + 0.196 627 215 384 576;
  • 46) 0.196 627 215 384 576 × 2 = 0 + 0.393 254 430 769 152;
  • 47) 0.393 254 430 769 152 × 2 = 0 + 0.786 508 861 538 304;
  • 48) 0.786 508 861 538 304 × 2 = 1 + 0.573 017 723 076 608;
  • 49) 0.573 017 723 076 608 × 2 = 1 + 0.146 035 446 153 216;
  • 50) 0.146 035 446 153 216 × 2 = 0 + 0.292 070 892 306 432;
  • 51) 0.292 070 892 306 432 × 2 = 0 + 0.584 141 784 612 864;
  • 52) 0.584 141 784 612 864 × 2 = 1 + 0.168 283 569 225 728;
  • 53) 0.168 283 569 225 728 × 2 = 0 + 0.336 567 138 451 456;
  • 54) 0.336 567 138 451 456 × 2 = 0 + 0.673 134 276 902 912;

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:


0.000 000 000 742 068(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1001 1001 00(2)

6. Positive number before normalization:

0.000 000 000 742 068(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1001 1001 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:


0.000 000 000 742 068(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1001 1001 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1001 1001 00(2) × 20 =

1.1001 0111 1111 0100 1100 100(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 0100 1100 100


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 0110 0100 =

100 1011 1111 1010 0110 0100


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 0110 0100


Decimal number -0.000 000 000 742 068 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1010 0110 0100

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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