-0.000 000 000 742 147 676 639 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 639 7(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
-0.000 000 000 742 147 676 639 7(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:

|-0.000 000 000 742 147 676 639 7| = 0.000 000 000 742 147 676 639 7


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: 0.000 000 000 742 147 676 639 7.

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) 0.000 000 000 742 147 676 639 7 × 2 = 0 + 0.000 000 001 484 295 353 279 4;
  • 2) 0.000 000 001 484 295 353 279 4 × 2 = 0 + 0.000 000 002 968 590 706 558 8;
  • 3) 0.000 000 002 968 590 706 558 8 × 2 = 0 + 0.000 000 005 937 181 413 117 6;
  • 4) 0.000 000 005 937 181 413 117 6 × 2 = 0 + 0.000 000 011 874 362 826 235 2;
  • 5) 0.000 000 011 874 362 826 235 2 × 2 = 0 + 0.000 000 023 748 725 652 470 4;
  • 6) 0.000 000 023 748 725 652 470 4 × 2 = 0 + 0.000 000 047 497 451 304 940 8;
  • 7) 0.000 000 047 497 451 304 940 8 × 2 = 0 + 0.000 000 094 994 902 609 881 6;
  • 8) 0.000 000 094 994 902 609 881 6 × 2 = 0 + 0.000 000 189 989 805 219 763 2;
  • 9) 0.000 000 189 989 805 219 763 2 × 2 = 0 + 0.000 000 379 979 610 439 526 4;
  • 10) 0.000 000 379 979 610 439 526 4 × 2 = 0 + 0.000 000 759 959 220 879 052 8;
  • 11) 0.000 000 759 959 220 879 052 8 × 2 = 0 + 0.000 001 519 918 441 758 105 6;
  • 12) 0.000 001 519 918 441 758 105 6 × 2 = 0 + 0.000 003 039 836 883 516 211 2;
  • 13) 0.000 003 039 836 883 516 211 2 × 2 = 0 + 0.000 006 079 673 767 032 422 4;
  • 14) 0.000 006 079 673 767 032 422 4 × 2 = 0 + 0.000 012 159 347 534 064 844 8;
  • 15) 0.000 012 159 347 534 064 844 8 × 2 = 0 + 0.000 024 318 695 068 129 689 6;
  • 16) 0.000 024 318 695 068 129 689 6 × 2 = 0 + 0.000 048 637 390 136 259 379 2;
  • 17) 0.000 048 637 390 136 259 379 2 × 2 = 0 + 0.000 097 274 780 272 518 758 4;
  • 18) 0.000 097 274 780 272 518 758 4 × 2 = 0 + 0.000 194 549 560 545 037 516 8;
  • 19) 0.000 194 549 560 545 037 516 8 × 2 = 0 + 0.000 389 099 121 090 075 033 6;
  • 20) 0.000 389 099 121 090 075 033 6 × 2 = 0 + 0.000 778 198 242 180 150 067 2;
  • 21) 0.000 778 198 242 180 150 067 2 × 2 = 0 + 0.001 556 396 484 360 300 134 4;
  • 22) 0.001 556 396 484 360 300 134 4 × 2 = 0 + 0.003 112 792 968 720 600 268 8;
  • 23) 0.003 112 792 968 720 600 268 8 × 2 = 0 + 0.006 225 585 937 441 200 537 6;
  • 24) 0.006 225 585 937 441 200 537 6 × 2 = 0 + 0.012 451 171 874 882 401 075 2;
  • 25) 0.012 451 171 874 882 401 075 2 × 2 = 0 + 0.024 902 343 749 764 802 150 4;
  • 26) 0.024 902 343 749 764 802 150 4 × 2 = 0 + 0.049 804 687 499 529 604 300 8;
  • 27) 0.049 804 687 499 529 604 300 8 × 2 = 0 + 0.099 609 374 999 059 208 601 6;
  • 28) 0.099 609 374 999 059 208 601 6 × 2 = 0 + 0.199 218 749 998 118 417 203 2;
  • 29) 0.199 218 749 998 118 417 203 2 × 2 = 0 + 0.398 437 499 996 236 834 406 4;
  • 30) 0.398 437 499 996 236 834 406 4 × 2 = 0 + 0.796 874 999 992 473 668 812 8;
  • 31) 0.796 874 999 992 473 668 812 8 × 2 = 1 + 0.593 749 999 984 947 337 625 6;
  • 32) 0.593 749 999 984 947 337 625 6 × 2 = 1 + 0.187 499 999 969 894 675 251 2;
  • 33) 0.187 499 999 969 894 675 251 2 × 2 = 0 + 0.374 999 999 939 789 350 502 4;
  • 34) 0.374 999 999 939 789 350 502 4 × 2 = 0 + 0.749 999 999 879 578 701 004 8;
  • 35) 0.749 999 999 879 578 701 004 8 × 2 = 1 + 0.499 999 999 759 157 402 009 6;
  • 36) 0.499 999 999 759 157 402 009 6 × 2 = 0 + 0.999 999 999 518 314 804 019 2;
  • 37) 0.999 999 999 518 314 804 019 2 × 2 = 1 + 0.999 999 999 036 629 608 038 4;
  • 38) 0.999 999 999 036 629 608 038 4 × 2 = 1 + 0.999 999 998 073 259 216 076 8;
  • 39) 0.999 999 998 073 259 216 076 8 × 2 = 1 + 0.999 999 996 146 518 432 153 6;
  • 40) 0.999 999 996 146 518 432 153 6 × 2 = 1 + 0.999 999 992 293 036 864 307 2;
  • 41) 0.999 999 992 293 036 864 307 2 × 2 = 1 + 0.999 999 984 586 073 728 614 4;
  • 42) 0.999 999 984 586 073 728 614 4 × 2 = 1 + 0.999 999 969 172 147 457 228 8;
  • 43) 0.999 999 969 172 147 457 228 8 × 2 = 1 + 0.999 999 938 344 294 914 457 6;
  • 44) 0.999 999 938 344 294 914 457 6 × 2 = 1 + 0.999 999 876 688 589 828 915 2;
  • 45) 0.999 999 876 688 589 828 915 2 × 2 = 1 + 0.999 999 753 377 179 657 830 4;
  • 46) 0.999 999 753 377 179 657 830 4 × 2 = 1 + 0.999 999 506 754 359 315 660 8;
  • 47) 0.999 999 506 754 359 315 660 8 × 2 = 1 + 0.999 999 013 508 718 631 321 6;
  • 48) 0.999 999 013 508 718 631 321 6 × 2 = 1 + 0.999 998 027 017 437 262 643 2;
  • 49) 0.999 998 027 017 437 262 643 2 × 2 = 1 + 0.999 996 054 034 874 525 286 4;
  • 50) 0.999 996 054 034 874 525 286 4 × 2 = 1 + 0.999 992 108 069 749 050 572 8;
  • 51) 0.999 992 108 069 749 050 572 8 × 2 = 1 + 0.999 984 216 139 498 101 145 6;
  • 52) 0.999 984 216 139 498 101 145 6 × 2 = 1 + 0.999 968 432 278 996 202 291 2;
  • 53) 0.999 968 432 278 996 202 291 2 × 2 = 1 + 0.999 936 864 557 992 404 582 4;
  • 54) 0.999 936 864 557 992 404 582 4 × 2 = 1 + 0.999 873 729 115 984 809 164 8;

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:


0.000 000 000 742 147 676 639 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 639 7(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:


0.000 000 000 742 147 676 639 7(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 -0.000 000 000 742 147 676 639 7 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