-0.000 000 000 742 147 675 21 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 21(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
-0.000 000 000 742 147 675 21(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:

|-0.000 000 000 742 147 675 21| = 0.000 000 000 742 147 675 21


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: 0.000 000 000 742 147 675 21.

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) 0.000 000 000 742 147 675 21 × 2 = 0 + 0.000 000 001 484 295 350 42;
  • 2) 0.000 000 001 484 295 350 42 × 2 = 0 + 0.000 000 002 968 590 700 84;
  • 3) 0.000 000 002 968 590 700 84 × 2 = 0 + 0.000 000 005 937 181 401 68;
  • 4) 0.000 000 005 937 181 401 68 × 2 = 0 + 0.000 000 011 874 362 803 36;
  • 5) 0.000 000 011 874 362 803 36 × 2 = 0 + 0.000 000 023 748 725 606 72;
  • 6) 0.000 000 023 748 725 606 72 × 2 = 0 + 0.000 000 047 497 451 213 44;
  • 7) 0.000 000 047 497 451 213 44 × 2 = 0 + 0.000 000 094 994 902 426 88;
  • 8) 0.000 000 094 994 902 426 88 × 2 = 0 + 0.000 000 189 989 804 853 76;
  • 9) 0.000 000 189 989 804 853 76 × 2 = 0 + 0.000 000 379 979 609 707 52;
  • 10) 0.000 000 379 979 609 707 52 × 2 = 0 + 0.000 000 759 959 219 415 04;
  • 11) 0.000 000 759 959 219 415 04 × 2 = 0 + 0.000 001 519 918 438 830 08;
  • 12) 0.000 001 519 918 438 830 08 × 2 = 0 + 0.000 003 039 836 877 660 16;
  • 13) 0.000 003 039 836 877 660 16 × 2 = 0 + 0.000 006 079 673 755 320 32;
  • 14) 0.000 006 079 673 755 320 32 × 2 = 0 + 0.000 012 159 347 510 640 64;
  • 15) 0.000 012 159 347 510 640 64 × 2 = 0 + 0.000 024 318 695 021 281 28;
  • 16) 0.000 024 318 695 021 281 28 × 2 = 0 + 0.000 048 637 390 042 562 56;
  • 17) 0.000 048 637 390 042 562 56 × 2 = 0 + 0.000 097 274 780 085 125 12;
  • 18) 0.000 097 274 780 085 125 12 × 2 = 0 + 0.000 194 549 560 170 250 24;
  • 19) 0.000 194 549 560 170 250 24 × 2 = 0 + 0.000 389 099 120 340 500 48;
  • 20) 0.000 389 099 120 340 500 48 × 2 = 0 + 0.000 778 198 240 681 000 96;
  • 21) 0.000 778 198 240 681 000 96 × 2 = 0 + 0.001 556 396 481 362 001 92;
  • 22) 0.001 556 396 481 362 001 92 × 2 = 0 + 0.003 112 792 962 724 003 84;
  • 23) 0.003 112 792 962 724 003 84 × 2 = 0 + 0.006 225 585 925 448 007 68;
  • 24) 0.006 225 585 925 448 007 68 × 2 = 0 + 0.012 451 171 850 896 015 36;
  • 25) 0.012 451 171 850 896 015 36 × 2 = 0 + 0.024 902 343 701 792 030 72;
  • 26) 0.024 902 343 701 792 030 72 × 2 = 0 + 0.049 804 687 403 584 061 44;
  • 27) 0.049 804 687 403 584 061 44 × 2 = 0 + 0.099 609 374 807 168 122 88;
  • 28) 0.099 609 374 807 168 122 88 × 2 = 0 + 0.199 218 749 614 336 245 76;
  • 29) 0.199 218 749 614 336 245 76 × 2 = 0 + 0.398 437 499 228 672 491 52;
  • 30) 0.398 437 499 228 672 491 52 × 2 = 0 + 0.796 874 998 457 344 983 04;
  • 31) 0.796 874 998 457 344 983 04 × 2 = 1 + 0.593 749 996 914 689 966 08;
  • 32) 0.593 749 996 914 689 966 08 × 2 = 1 + 0.187 499 993 829 379 932 16;
  • 33) 0.187 499 993 829 379 932 16 × 2 = 0 + 0.374 999 987 658 759 864 32;
  • 34) 0.374 999 987 658 759 864 32 × 2 = 0 + 0.749 999 975 317 519 728 64;
  • 35) 0.749 999 975 317 519 728 64 × 2 = 1 + 0.499 999 950 635 039 457 28;
  • 36) 0.499 999 950 635 039 457 28 × 2 = 0 + 0.999 999 901 270 078 914 56;
  • 37) 0.999 999 901 270 078 914 56 × 2 = 1 + 0.999 999 802 540 157 829 12;
  • 38) 0.999 999 802 540 157 829 12 × 2 = 1 + 0.999 999 605 080 315 658 24;
  • 39) 0.999 999 605 080 315 658 24 × 2 = 1 + 0.999 999 210 160 631 316 48;
  • 40) 0.999 999 210 160 631 316 48 × 2 = 1 + 0.999 998 420 321 262 632 96;
  • 41) 0.999 998 420 321 262 632 96 × 2 = 1 + 0.999 996 840 642 525 265 92;
  • 42) 0.999 996 840 642 525 265 92 × 2 = 1 + 0.999 993 681 285 050 531 84;
  • 43) 0.999 993 681 285 050 531 84 × 2 = 1 + 0.999 987 362 570 101 063 68;
  • 44) 0.999 987 362 570 101 063 68 × 2 = 1 + 0.999 974 725 140 202 127 36;
  • 45) 0.999 974 725 140 202 127 36 × 2 = 1 + 0.999 949 450 280 404 254 72;
  • 46) 0.999 949 450 280 404 254 72 × 2 = 1 + 0.999 898 900 560 808 509 44;
  • 47) 0.999 898 900 560 808 509 44 × 2 = 1 + 0.999 797 801 121 617 018 88;
  • 48) 0.999 797 801 121 617 018 88 × 2 = 1 + 0.999 595 602 243 234 037 76;
  • 49) 0.999 595 602 243 234 037 76 × 2 = 1 + 0.999 191 204 486 468 075 52;
  • 50) 0.999 191 204 486 468 075 52 × 2 = 1 + 0.998 382 408 972 936 151 04;
  • 51) 0.998 382 408 972 936 151 04 × 2 = 1 + 0.996 764 817 945 872 302 08;
  • 52) 0.996 764 817 945 872 302 08 × 2 = 1 + 0.993 529 635 891 744 604 16;
  • 53) 0.993 529 635 891 744 604 16 × 2 = 1 + 0.987 059 271 783 489 208 32;
  • 54) 0.987 059 271 783 489 208 32 × 2 = 1 + 0.974 118 543 566 978 416 64;

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:


0.000 000 000 742 147 675 21(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 675 21(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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:


0.000 000 000 742 147 675 21(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 1111 1111 111


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 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1111 1111 1111


Decimal number -0.000 000 000 742 147 675 21 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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