-0.000 000 000 742 147 676 61 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 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
-0.000 000 000 742 147 676 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:

|-0.000 000 000 742 147 676 61| = 0.000 000 000 742 147 676 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: 0.000 000 000 742 147 676 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) 0.000 000 000 742 147 676 61 × 2 = 0 + 0.000 000 001 484 295 353 22;
  • 2) 0.000 000 001 484 295 353 22 × 2 = 0 + 0.000 000 002 968 590 706 44;
  • 3) 0.000 000 002 968 590 706 44 × 2 = 0 + 0.000 000 005 937 181 412 88;
  • 4) 0.000 000 005 937 181 412 88 × 2 = 0 + 0.000 000 011 874 362 825 76;
  • 5) 0.000 000 011 874 362 825 76 × 2 = 0 + 0.000 000 023 748 725 651 52;
  • 6) 0.000 000 023 748 725 651 52 × 2 = 0 + 0.000 000 047 497 451 303 04;
  • 7) 0.000 000 047 497 451 303 04 × 2 = 0 + 0.000 000 094 994 902 606 08;
  • 8) 0.000 000 094 994 902 606 08 × 2 = 0 + 0.000 000 189 989 805 212 16;
  • 9) 0.000 000 189 989 805 212 16 × 2 = 0 + 0.000 000 379 979 610 424 32;
  • 10) 0.000 000 379 979 610 424 32 × 2 = 0 + 0.000 000 759 959 220 848 64;
  • 11) 0.000 000 759 959 220 848 64 × 2 = 0 + 0.000 001 519 918 441 697 28;
  • 12) 0.000 001 519 918 441 697 28 × 2 = 0 + 0.000 003 039 836 883 394 56;
  • 13) 0.000 003 039 836 883 394 56 × 2 = 0 + 0.000 006 079 673 766 789 12;
  • 14) 0.000 006 079 673 766 789 12 × 2 = 0 + 0.000 012 159 347 533 578 24;
  • 15) 0.000 012 159 347 533 578 24 × 2 = 0 + 0.000 024 318 695 067 156 48;
  • 16) 0.000 024 318 695 067 156 48 × 2 = 0 + 0.000 048 637 390 134 312 96;
  • 17) 0.000 048 637 390 134 312 96 × 2 = 0 + 0.000 097 274 780 268 625 92;
  • 18) 0.000 097 274 780 268 625 92 × 2 = 0 + 0.000 194 549 560 537 251 84;
  • 19) 0.000 194 549 560 537 251 84 × 2 = 0 + 0.000 389 099 121 074 503 68;
  • 20) 0.000 389 099 121 074 503 68 × 2 = 0 + 0.000 778 198 242 149 007 36;
  • 21) 0.000 778 198 242 149 007 36 × 2 = 0 + 0.001 556 396 484 298 014 72;
  • 22) 0.001 556 396 484 298 014 72 × 2 = 0 + 0.003 112 792 968 596 029 44;
  • 23) 0.003 112 792 968 596 029 44 × 2 = 0 + 0.006 225 585 937 192 058 88;
  • 24) 0.006 225 585 937 192 058 88 × 2 = 0 + 0.012 451 171 874 384 117 76;
  • 25) 0.012 451 171 874 384 117 76 × 2 = 0 + 0.024 902 343 748 768 235 52;
  • 26) 0.024 902 343 748 768 235 52 × 2 = 0 + 0.049 804 687 497 536 471 04;
  • 27) 0.049 804 687 497 536 471 04 × 2 = 0 + 0.099 609 374 995 072 942 08;
  • 28) 0.099 609 374 995 072 942 08 × 2 = 0 + 0.199 218 749 990 145 884 16;
  • 29) 0.199 218 749 990 145 884 16 × 2 = 0 + 0.398 437 499 980 291 768 32;
  • 30) 0.398 437 499 980 291 768 32 × 2 = 0 + 0.796 874 999 960 583 536 64;
  • 31) 0.796 874 999 960 583 536 64 × 2 = 1 + 0.593 749 999 921 167 073 28;
  • 32) 0.593 749 999 921 167 073 28 × 2 = 1 + 0.187 499 999 842 334 146 56;
  • 33) 0.187 499 999 842 334 146 56 × 2 = 0 + 0.374 999 999 684 668 293 12;
  • 34) 0.374 999 999 684 668 293 12 × 2 = 0 + 0.749 999 999 369 336 586 24;
  • 35) 0.749 999 999 369 336 586 24 × 2 = 1 + 0.499 999 998 738 673 172 48;
  • 36) 0.499 999 998 738 673 172 48 × 2 = 0 + 0.999 999 997 477 346 344 96;
  • 37) 0.999 999 997 477 346 344 96 × 2 = 1 + 0.999 999 994 954 692 689 92;
  • 38) 0.999 999 994 954 692 689 92 × 2 = 1 + 0.999 999 989 909 385 379 84;
  • 39) 0.999 999 989 909 385 379 84 × 2 = 1 + 0.999 999 979 818 770 759 68;
  • 40) 0.999 999 979 818 770 759 68 × 2 = 1 + 0.999 999 959 637 541 519 36;
  • 41) 0.999 999 959 637 541 519 36 × 2 = 1 + 0.999 999 919 275 083 038 72;
  • 42) 0.999 999 919 275 083 038 72 × 2 = 1 + 0.999 999 838 550 166 077 44;
  • 43) 0.999 999 838 550 166 077 44 × 2 = 1 + 0.999 999 677 100 332 154 88;
  • 44) 0.999 999 677 100 332 154 88 × 2 = 1 + 0.999 999 354 200 664 309 76;
  • 45) 0.999 999 354 200 664 309 76 × 2 = 1 + 0.999 998 708 401 328 619 52;
  • 46) 0.999 998 708 401 328 619 52 × 2 = 1 + 0.999 997 416 802 657 239 04;
  • 47) 0.999 997 416 802 657 239 04 × 2 = 1 + 0.999 994 833 605 314 478 08;
  • 48) 0.999 994 833 605 314 478 08 × 2 = 1 + 0.999 989 667 210 628 956 16;
  • 49) 0.999 989 667 210 628 956 16 × 2 = 1 + 0.999 979 334 421 257 912 32;
  • 50) 0.999 979 334 421 257 912 32 × 2 = 1 + 0.999 958 668 842 515 824 64;
  • 51) 0.999 958 668 842 515 824 64 × 2 = 1 + 0.999 917 337 685 031 649 28;
  • 52) 0.999 917 337 685 031 649 28 × 2 = 1 + 0.999 834 675 370 063 298 56;
  • 53) 0.999 834 675 370 063 298 56 × 2 = 1 + 0.999 669 350 740 126 597 12;
  • 54) 0.999 669 350 740 126 597 12 × 2 = 1 + 0.999 338 701 480 253 194 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:


0.000 000 000 742 147 676 61(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 61(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:


0.000 000 000 742 147 676 61(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 -0.000 000 000 742 147 676 61 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