-0.000 000 000 963 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 963(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
-0.000 000 000 963(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:

|-0.000 000 000 963| = 0.000 000 000 963


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: 0.000 000 000 963.

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) 0.000 000 000 963 × 2 = 0 + 0.000 000 001 926;
  • 2) 0.000 000 001 926 × 2 = 0 + 0.000 000 003 852;
  • 3) 0.000 000 003 852 × 2 = 0 + 0.000 000 007 704;
  • 4) 0.000 000 007 704 × 2 = 0 + 0.000 000 015 408;
  • 5) 0.000 000 015 408 × 2 = 0 + 0.000 000 030 816;
  • 6) 0.000 000 030 816 × 2 = 0 + 0.000 000 061 632;
  • 7) 0.000 000 061 632 × 2 = 0 + 0.000 000 123 264;
  • 8) 0.000 000 123 264 × 2 = 0 + 0.000 000 246 528;
  • 9) 0.000 000 246 528 × 2 = 0 + 0.000 000 493 056;
  • 10) 0.000 000 493 056 × 2 = 0 + 0.000 000 986 112;
  • 11) 0.000 000 986 112 × 2 = 0 + 0.000 001 972 224;
  • 12) 0.000 001 972 224 × 2 = 0 + 0.000 003 944 448;
  • 13) 0.000 003 944 448 × 2 = 0 + 0.000 007 888 896;
  • 14) 0.000 007 888 896 × 2 = 0 + 0.000 015 777 792;
  • 15) 0.000 015 777 792 × 2 = 0 + 0.000 031 555 584;
  • 16) 0.000 031 555 584 × 2 = 0 + 0.000 063 111 168;
  • 17) 0.000 063 111 168 × 2 = 0 + 0.000 126 222 336;
  • 18) 0.000 126 222 336 × 2 = 0 + 0.000 252 444 672;
  • 19) 0.000 252 444 672 × 2 = 0 + 0.000 504 889 344;
  • 20) 0.000 504 889 344 × 2 = 0 + 0.001 009 778 688;
  • 21) 0.001 009 778 688 × 2 = 0 + 0.002 019 557 376;
  • 22) 0.002 019 557 376 × 2 = 0 + 0.004 039 114 752;
  • 23) 0.004 039 114 752 × 2 = 0 + 0.008 078 229 504;
  • 24) 0.008 078 229 504 × 2 = 0 + 0.016 156 459 008;
  • 25) 0.016 156 459 008 × 2 = 0 + 0.032 312 918 016;
  • 26) 0.032 312 918 016 × 2 = 0 + 0.064 625 836 032;
  • 27) 0.064 625 836 032 × 2 = 0 + 0.129 251 672 064;
  • 28) 0.129 251 672 064 × 2 = 0 + 0.258 503 344 128;
  • 29) 0.258 503 344 128 × 2 = 0 + 0.517 006 688 256;
  • 30) 0.517 006 688 256 × 2 = 1 + 0.034 013 376 512;
  • 31) 0.034 013 376 512 × 2 = 0 + 0.068 026 753 024;
  • 32) 0.068 026 753 024 × 2 = 0 + 0.136 053 506 048;
  • 33) 0.136 053 506 048 × 2 = 0 + 0.272 107 012 096;
  • 34) 0.272 107 012 096 × 2 = 0 + 0.544 214 024 192;
  • 35) 0.544 214 024 192 × 2 = 1 + 0.088 428 048 384;
  • 36) 0.088 428 048 384 × 2 = 0 + 0.176 856 096 768;
  • 37) 0.176 856 096 768 × 2 = 0 + 0.353 712 193 536;
  • 38) 0.353 712 193 536 × 2 = 0 + 0.707 424 387 072;
  • 39) 0.707 424 387 072 × 2 = 1 + 0.414 848 774 144;
  • 40) 0.414 848 774 144 × 2 = 0 + 0.829 697 548 288;
  • 41) 0.829 697 548 288 × 2 = 1 + 0.659 395 096 576;
  • 42) 0.659 395 096 576 × 2 = 1 + 0.318 790 193 152;
  • 43) 0.318 790 193 152 × 2 = 0 + 0.637 580 386 304;
  • 44) 0.637 580 386 304 × 2 = 1 + 0.275 160 772 608;
  • 45) 0.275 160 772 608 × 2 = 0 + 0.550 321 545 216;
  • 46) 0.550 321 545 216 × 2 = 1 + 0.100 643 090 432;
  • 47) 0.100 643 090 432 × 2 = 0 + 0.201 286 180 864;
  • 48) 0.201 286 180 864 × 2 = 0 + 0.402 572 361 728;
  • 49) 0.402 572 361 728 × 2 = 0 + 0.805 144 723 456;
  • 50) 0.805 144 723 456 × 2 = 1 + 0.610 289 446 912;
  • 51) 0.610 289 446 912 × 2 = 1 + 0.220 578 893 824;
  • 52) 0.220 578 893 824 × 2 = 0 + 0.441 157 787 648;
  • 53) 0.441 157 787 648 × 2 = 0 + 0.882 315 575 296;

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:


0.000 000 000 963(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0010 0010 1101 0100 0110 0(2)

6. Positive number before normalization:

0.000 000 000 963(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0010 0010 1101 0100 0110 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 30 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 963(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0010 0010 1101 0100 0110 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0010 0010 1101 0100 0110 0(2) × 20 =

1.0000 1000 1011 0101 0001 100(2) × 2-30


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): -30

Mantissa (not normalized):
1.0000 1000 1011 0101 0001 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-30 + 2(8-1) - 1 =

(-30 + 127)(10) =

97(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;
  • 97 ÷ 2 = 48 + 1;
  • 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) =

97(10) =

0110 0001(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. 000 0100 0101 1010 1000 1100 =

000 0100 0101 1010 1000 1100


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 0001

Mantissa (23 bits) =
000 0100 0101 1010 1000 1100


Decimal number -0.000 000 000 963 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0001 - 000 0100 0101 1010 1000 1100

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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