-0.000 000 000 744 49 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 744 49(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
-0.000 000 000 744 49(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:

|-0.000 000 000 744 49| = 0.000 000 000 744 49


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: 0.000 000 000 744 49.

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) 0.000 000 000 744 49 × 2 = 0 + 0.000 000 001 488 98;
  • 2) 0.000 000 001 488 98 × 2 = 0 + 0.000 000 002 977 96;
  • 3) 0.000 000 002 977 96 × 2 = 0 + 0.000 000 005 955 92;
  • 4) 0.000 000 005 955 92 × 2 = 0 + 0.000 000 011 911 84;
  • 5) 0.000 000 011 911 84 × 2 = 0 + 0.000 000 023 823 68;
  • 6) 0.000 000 023 823 68 × 2 = 0 + 0.000 000 047 647 36;
  • 7) 0.000 000 047 647 36 × 2 = 0 + 0.000 000 095 294 72;
  • 8) 0.000 000 095 294 72 × 2 = 0 + 0.000 000 190 589 44;
  • 9) 0.000 000 190 589 44 × 2 = 0 + 0.000 000 381 178 88;
  • 10) 0.000 000 381 178 88 × 2 = 0 + 0.000 000 762 357 76;
  • 11) 0.000 000 762 357 76 × 2 = 0 + 0.000 001 524 715 52;
  • 12) 0.000 001 524 715 52 × 2 = 0 + 0.000 003 049 431 04;
  • 13) 0.000 003 049 431 04 × 2 = 0 + 0.000 006 098 862 08;
  • 14) 0.000 006 098 862 08 × 2 = 0 + 0.000 012 197 724 16;
  • 15) 0.000 012 197 724 16 × 2 = 0 + 0.000 024 395 448 32;
  • 16) 0.000 024 395 448 32 × 2 = 0 + 0.000 048 790 896 64;
  • 17) 0.000 048 790 896 64 × 2 = 0 + 0.000 097 581 793 28;
  • 18) 0.000 097 581 793 28 × 2 = 0 + 0.000 195 163 586 56;
  • 19) 0.000 195 163 586 56 × 2 = 0 + 0.000 390 327 173 12;
  • 20) 0.000 390 327 173 12 × 2 = 0 + 0.000 780 654 346 24;
  • 21) 0.000 780 654 346 24 × 2 = 0 + 0.001 561 308 692 48;
  • 22) 0.001 561 308 692 48 × 2 = 0 + 0.003 122 617 384 96;
  • 23) 0.003 122 617 384 96 × 2 = 0 + 0.006 245 234 769 92;
  • 24) 0.006 245 234 769 92 × 2 = 0 + 0.012 490 469 539 84;
  • 25) 0.012 490 469 539 84 × 2 = 0 + 0.024 980 939 079 68;
  • 26) 0.024 980 939 079 68 × 2 = 0 + 0.049 961 878 159 36;
  • 27) 0.049 961 878 159 36 × 2 = 0 + 0.099 923 756 318 72;
  • 28) 0.099 923 756 318 72 × 2 = 0 + 0.199 847 512 637 44;
  • 29) 0.199 847 512 637 44 × 2 = 0 + 0.399 695 025 274 88;
  • 30) 0.399 695 025 274 88 × 2 = 0 + 0.799 390 050 549 76;
  • 31) 0.799 390 050 549 76 × 2 = 1 + 0.598 780 101 099 52;
  • 32) 0.598 780 101 099 52 × 2 = 1 + 0.197 560 202 199 04;
  • 33) 0.197 560 202 199 04 × 2 = 0 + 0.395 120 404 398 08;
  • 34) 0.395 120 404 398 08 × 2 = 0 + 0.790 240 808 796 16;
  • 35) 0.790 240 808 796 16 × 2 = 1 + 0.580 481 617 592 32;
  • 36) 0.580 481 617 592 32 × 2 = 1 + 0.160 963 235 184 64;
  • 37) 0.160 963 235 184 64 × 2 = 0 + 0.321 926 470 369 28;
  • 38) 0.321 926 470 369 28 × 2 = 0 + 0.643 852 940 738 56;
  • 39) 0.643 852 940 738 56 × 2 = 1 + 0.287 705 881 477 12;
  • 40) 0.287 705 881 477 12 × 2 = 0 + 0.575 411 762 954 24;
  • 41) 0.575 411 762 954 24 × 2 = 1 + 0.150 823 525 908 48;
  • 42) 0.150 823 525 908 48 × 2 = 0 + 0.301 647 051 816 96;
  • 43) 0.301 647 051 816 96 × 2 = 0 + 0.603 294 103 633 92;
  • 44) 0.603 294 103 633 92 × 2 = 1 + 0.206 588 207 267 84;
  • 45) 0.206 588 207 267 84 × 2 = 0 + 0.413 176 414 535 68;
  • 46) 0.413 176 414 535 68 × 2 = 0 + 0.826 352 829 071 36;
  • 47) 0.826 352 829 071 36 × 2 = 1 + 0.652 705 658 142 72;
  • 48) 0.652 705 658 142 72 × 2 = 1 + 0.305 411 316 285 44;
  • 49) 0.305 411 316 285 44 × 2 = 0 + 0.610 822 632 570 88;
  • 50) 0.610 822 632 570 88 × 2 = 1 + 0.221 645 265 141 76;
  • 51) 0.221 645 265 141 76 × 2 = 0 + 0.443 290 530 283 52;
  • 52) 0.443 290 530 283 52 × 2 = 0 + 0.886 581 060 567 04;
  • 53) 0.886 581 060 567 04 × 2 = 1 + 0.773 162 121 134 08;
  • 54) 0.773 162 121 134 08 × 2 = 1 + 0.546 324 242 268 16;

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:


0.000 000 000 744 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1001 0011 0100 11(2)

6. Positive number before normalization:

0.000 000 000 744 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1001 0011 0100 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:


0.000 000 000 744 49(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1001 0011 0100 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1001 0011 0100 11(2) × 20 =

1.1001 1001 0100 1001 1010 011(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 1001 0100 1001 1010 011


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 1010 0100 1101 0011 =

100 1100 1010 0100 1101 0011


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 1010 0100 1101 0011


Decimal number -0.000 000 000 744 49 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 1010 0100 1101 0011

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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