-0.000 000 000 742 148 16 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 16(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
-0.000 000 000 742 148 16(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:

|-0.000 000 000 742 148 16| = 0.000 000 000 742 148 16


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: 0.000 000 000 742 148 16.

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) 0.000 000 000 742 148 16 × 2 = 0 + 0.000 000 001 484 296 32;
  • 2) 0.000 000 001 484 296 32 × 2 = 0 + 0.000 000 002 968 592 64;
  • 3) 0.000 000 002 968 592 64 × 2 = 0 + 0.000 000 005 937 185 28;
  • 4) 0.000 000 005 937 185 28 × 2 = 0 + 0.000 000 011 874 370 56;
  • 5) 0.000 000 011 874 370 56 × 2 = 0 + 0.000 000 023 748 741 12;
  • 6) 0.000 000 023 748 741 12 × 2 = 0 + 0.000 000 047 497 482 24;
  • 7) 0.000 000 047 497 482 24 × 2 = 0 + 0.000 000 094 994 964 48;
  • 8) 0.000 000 094 994 964 48 × 2 = 0 + 0.000 000 189 989 928 96;
  • 9) 0.000 000 189 989 928 96 × 2 = 0 + 0.000 000 379 979 857 92;
  • 10) 0.000 000 379 979 857 92 × 2 = 0 + 0.000 000 759 959 715 84;
  • 11) 0.000 000 759 959 715 84 × 2 = 0 + 0.000 001 519 919 431 68;
  • 12) 0.000 001 519 919 431 68 × 2 = 0 + 0.000 003 039 838 863 36;
  • 13) 0.000 003 039 838 863 36 × 2 = 0 + 0.000 006 079 677 726 72;
  • 14) 0.000 006 079 677 726 72 × 2 = 0 + 0.000 012 159 355 453 44;
  • 15) 0.000 012 159 355 453 44 × 2 = 0 + 0.000 024 318 710 906 88;
  • 16) 0.000 024 318 710 906 88 × 2 = 0 + 0.000 048 637 421 813 76;
  • 17) 0.000 048 637 421 813 76 × 2 = 0 + 0.000 097 274 843 627 52;
  • 18) 0.000 097 274 843 627 52 × 2 = 0 + 0.000 194 549 687 255 04;
  • 19) 0.000 194 549 687 255 04 × 2 = 0 + 0.000 389 099 374 510 08;
  • 20) 0.000 389 099 374 510 08 × 2 = 0 + 0.000 778 198 749 020 16;
  • 21) 0.000 778 198 749 020 16 × 2 = 0 + 0.001 556 397 498 040 32;
  • 22) 0.001 556 397 498 040 32 × 2 = 0 + 0.003 112 794 996 080 64;
  • 23) 0.003 112 794 996 080 64 × 2 = 0 + 0.006 225 589 992 161 28;
  • 24) 0.006 225 589 992 161 28 × 2 = 0 + 0.012 451 179 984 322 56;
  • 25) 0.012 451 179 984 322 56 × 2 = 0 + 0.024 902 359 968 645 12;
  • 26) 0.024 902 359 968 645 12 × 2 = 0 + 0.049 804 719 937 290 24;
  • 27) 0.049 804 719 937 290 24 × 2 = 0 + 0.099 609 439 874 580 48;
  • 28) 0.099 609 439 874 580 48 × 2 = 0 + 0.199 218 879 749 160 96;
  • 29) 0.199 218 879 749 160 96 × 2 = 0 + 0.398 437 759 498 321 92;
  • 30) 0.398 437 759 498 321 92 × 2 = 0 + 0.796 875 518 996 643 84;
  • 31) 0.796 875 518 996 643 84 × 2 = 1 + 0.593 751 037 993 287 68;
  • 32) 0.593 751 037 993 287 68 × 2 = 1 + 0.187 502 075 986 575 36;
  • 33) 0.187 502 075 986 575 36 × 2 = 0 + 0.375 004 151 973 150 72;
  • 34) 0.375 004 151 973 150 72 × 2 = 0 + 0.750 008 303 946 301 44;
  • 35) 0.750 008 303 946 301 44 × 2 = 1 + 0.500 016 607 892 602 88;
  • 36) 0.500 016 607 892 602 88 × 2 = 1 + 0.000 033 215 785 205 76;
  • 37) 0.000 033 215 785 205 76 × 2 = 0 + 0.000 066 431 570 411 52;
  • 38) 0.000 066 431 570 411 52 × 2 = 0 + 0.000 132 863 140 823 04;
  • 39) 0.000 132 863 140 823 04 × 2 = 0 + 0.000 265 726 281 646 08;
  • 40) 0.000 265 726 281 646 08 × 2 = 0 + 0.000 531 452 563 292 16;
  • 41) 0.000 531 452 563 292 16 × 2 = 0 + 0.001 062 905 126 584 32;
  • 42) 0.001 062 905 126 584 32 × 2 = 0 + 0.002 125 810 253 168 64;
  • 43) 0.002 125 810 253 168 64 × 2 = 0 + 0.004 251 620 506 337 28;
  • 44) 0.004 251 620 506 337 28 × 2 = 0 + 0.008 503 241 012 674 56;
  • 45) 0.008 503 241 012 674 56 × 2 = 0 + 0.017 006 482 025 349 12;
  • 46) 0.017 006 482 025 349 12 × 2 = 0 + 0.034 012 964 050 698 24;
  • 47) 0.034 012 964 050 698 24 × 2 = 0 + 0.068 025 928 101 396 48;
  • 48) 0.068 025 928 101 396 48 × 2 = 0 + 0.136 051 856 202 792 96;
  • 49) 0.136 051 856 202 792 96 × 2 = 0 + 0.272 103 712 405 585 92;
  • 50) 0.272 103 712 405 585 92 × 2 = 0 + 0.544 207 424 811 171 84;
  • 51) 0.544 207 424 811 171 84 × 2 = 1 + 0.088 414 849 622 343 68;
  • 52) 0.088 414 849 622 343 68 × 2 = 0 + 0.176 829 699 244 687 36;
  • 53) 0.176 829 699 244 687 36 × 2 = 0 + 0.353 659 398 489 374 72;
  • 54) 0.353 659 398 489 374 72 × 2 = 0 + 0.707 318 796 978 749 44;

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:


0.000 000 000 742 148 16(10) =

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

6. Positive number before normalization:

0.000 000 000 742 148 16(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 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:


0.000 000 000 742 148 16(10) =

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

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

1.1001 1000 0000 0000 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 0000 0000 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 0000 0000 0000 1000 =

100 1100 0000 0000 0000 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 0000 0000 0000 1000


Decimal number -0.000 000 000 742 148 16 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 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