-0.000 000 000 742 61 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 61(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 61(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 61| = 0.000 000 000 742 61


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 61.

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 61 × 2 = 0 + 0.000 000 001 485 22;
  • 2) 0.000 000 001 485 22 × 2 = 0 + 0.000 000 002 970 44;
  • 3) 0.000 000 002 970 44 × 2 = 0 + 0.000 000 005 940 88;
  • 4) 0.000 000 005 940 88 × 2 = 0 + 0.000 000 011 881 76;
  • 5) 0.000 000 011 881 76 × 2 = 0 + 0.000 000 023 763 52;
  • 6) 0.000 000 023 763 52 × 2 = 0 + 0.000 000 047 527 04;
  • 7) 0.000 000 047 527 04 × 2 = 0 + 0.000 000 095 054 08;
  • 8) 0.000 000 095 054 08 × 2 = 0 + 0.000 000 190 108 16;
  • 9) 0.000 000 190 108 16 × 2 = 0 + 0.000 000 380 216 32;
  • 10) 0.000 000 380 216 32 × 2 = 0 + 0.000 000 760 432 64;
  • 11) 0.000 000 760 432 64 × 2 = 0 + 0.000 001 520 865 28;
  • 12) 0.000 001 520 865 28 × 2 = 0 + 0.000 003 041 730 56;
  • 13) 0.000 003 041 730 56 × 2 = 0 + 0.000 006 083 461 12;
  • 14) 0.000 006 083 461 12 × 2 = 0 + 0.000 012 166 922 24;
  • 15) 0.000 012 166 922 24 × 2 = 0 + 0.000 024 333 844 48;
  • 16) 0.000 024 333 844 48 × 2 = 0 + 0.000 048 667 688 96;
  • 17) 0.000 048 667 688 96 × 2 = 0 + 0.000 097 335 377 92;
  • 18) 0.000 097 335 377 92 × 2 = 0 + 0.000 194 670 755 84;
  • 19) 0.000 194 670 755 84 × 2 = 0 + 0.000 389 341 511 68;
  • 20) 0.000 389 341 511 68 × 2 = 0 + 0.000 778 683 023 36;
  • 21) 0.000 778 683 023 36 × 2 = 0 + 0.001 557 366 046 72;
  • 22) 0.001 557 366 046 72 × 2 = 0 + 0.003 114 732 093 44;
  • 23) 0.003 114 732 093 44 × 2 = 0 + 0.006 229 464 186 88;
  • 24) 0.006 229 464 186 88 × 2 = 0 + 0.012 458 928 373 76;
  • 25) 0.012 458 928 373 76 × 2 = 0 + 0.024 917 856 747 52;
  • 26) 0.024 917 856 747 52 × 2 = 0 + 0.049 835 713 495 04;
  • 27) 0.049 835 713 495 04 × 2 = 0 + 0.099 671 426 990 08;
  • 28) 0.099 671 426 990 08 × 2 = 0 + 0.199 342 853 980 16;
  • 29) 0.199 342 853 980 16 × 2 = 0 + 0.398 685 707 960 32;
  • 30) 0.398 685 707 960 32 × 2 = 0 + 0.797 371 415 920 64;
  • 31) 0.797 371 415 920 64 × 2 = 1 + 0.594 742 831 841 28;
  • 32) 0.594 742 831 841 28 × 2 = 1 + 0.189 485 663 682 56;
  • 33) 0.189 485 663 682 56 × 2 = 0 + 0.378 971 327 365 12;
  • 34) 0.378 971 327 365 12 × 2 = 0 + 0.757 942 654 730 24;
  • 35) 0.757 942 654 730 24 × 2 = 1 + 0.515 885 309 460 48;
  • 36) 0.515 885 309 460 48 × 2 = 1 + 0.031 770 618 920 96;
  • 37) 0.031 770 618 920 96 × 2 = 0 + 0.063 541 237 841 92;
  • 38) 0.063 541 237 841 92 × 2 = 0 + 0.127 082 475 683 84;
  • 39) 0.127 082 475 683 84 × 2 = 0 + 0.254 164 951 367 68;
  • 40) 0.254 164 951 367 68 × 2 = 0 + 0.508 329 902 735 36;
  • 41) 0.508 329 902 735 36 × 2 = 1 + 0.016 659 805 470 72;
  • 42) 0.016 659 805 470 72 × 2 = 0 + 0.033 319 610 941 44;
  • 43) 0.033 319 610 941 44 × 2 = 0 + 0.066 639 221 882 88;
  • 44) 0.066 639 221 882 88 × 2 = 0 + 0.133 278 443 765 76;
  • 45) 0.133 278 443 765 76 × 2 = 0 + 0.266 556 887 531 52;
  • 46) 0.266 556 887 531 52 × 2 = 0 + 0.533 113 775 063 04;
  • 47) 0.533 113 775 063 04 × 2 = 1 + 0.066 227 550 126 08;
  • 48) 0.066 227 550 126 08 × 2 = 0 + 0.132 455 100 252 16;
  • 49) 0.132 455 100 252 16 × 2 = 0 + 0.264 910 200 504 32;
  • 50) 0.264 910 200 504 32 × 2 = 0 + 0.529 820 401 008 64;
  • 51) 0.529 820 401 008 64 × 2 = 1 + 0.059 640 802 017 28;
  • 52) 0.059 640 802 017 28 × 2 = 0 + 0.119 281 604 034 56;
  • 53) 0.119 281 604 034 56 × 2 = 0 + 0.238 563 208 069 12;
  • 54) 0.238 563 208 069 12 × 2 = 0 + 0.477 126 416 138 24;

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 61(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1000 0010 0010 00(2)

6. Positive number before normalization:

0.000 000 000 742 61(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1000 0010 0010 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 61(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1000 0010 0010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1000 0010 0010 00(2) × 20 =

1.1001 1000 0100 0001 0001 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 0100 0001 0001 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 0010 0000 1000 1000 =

100 1100 0010 0000 1000 1000


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 0010 0000 1000 1000


Decimal number -0.000 000 000 742 61 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0010 0000 1000 1000

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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